3.209 \(\int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=239 \[ \frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {472 i a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {92 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {1576 a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}} \]
[Out]
(4+4*I)*a^(5/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d-1576/315*a^2*(a+I*a*tan(d*x
+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/9*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(9/2)-38/63*I*a^2*(a+I*a*tan(d*x+c
))^(1/2)/d/tan(d*x+c)^(7/2)+92/105*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)+472/315*I*a^2*(a+I*a*tan(d*
x+c))^(1/2)/d/tan(d*x+c)^(3/2)
________________________________________________________________________________________
Rubi [A] time = 0.74, antiderivative size = 239, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used =
{3553, 3598, 12, 3544, 205} \[ \frac {472 i a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {92 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {1576 a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}+\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(11/2),x]
[Out]
((4 + 4*I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*a^2*Sqrt[a
+ I*a*Tan[c + d*x]])/(9*d*Tan[c + d*x]^(9/2)) - (((38*I)/63)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^
(7/2)) + (92*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(5/2)) + (((472*I)/315)*a^2*Sqrt[a + I*a*Tan[
c + d*x]])/(d*Tan[c + d*x]^(3/2)) - (1576*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*Sqrt[Tan[c + d*x]])
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 3544
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Rule 3553
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Rule 3598
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {11}{2}}(c+d x)} \, dx &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2}{9} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {19 i a^2}{2}+\frac {17}{2} a^2 \tan (c+d x)\right )}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {4 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {69 a^3}{2}+\frac {57}{2} i a^3 \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx}{63 a}\\ &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {92 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {177 i a^4}{2}-69 a^4 \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{315 a^2}\\ &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {92 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {472 i a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {16 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {591 a^5}{4}-\frac {177}{2} i a^5 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{945 a^3}\\ &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {92 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {472 i a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {1576 a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {32 \int -\frac {945 i a^6 \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{945 a^4}\\ &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {92 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {472 i a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {1576 a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}+\left (4 i a^2\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {92 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {472 i a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {1576 a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}+\frac {\left (8 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {38 i a^2 \sqrt {a+i a \tan (c+d x)}}{63 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {92 a^2 \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {472 i a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {1576 a^2 \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 4.64, size = 207, normalized size = 0.87 \[ \frac {a^2 \sqrt {a+i a \tan (c+d x)} \left (\frac {10080 e^{-i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{\sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-\sqrt {\tan (c+d x)} \csc ^5(c+d x) (-282 i \sin (c+d x)+49 i \sin (3 (c+d x))+331 i \sin (5 (c+d x))+1650 \cos (c+d x)-2051 \cos (3 (c+d x))+961 \cos (5 (c+d x)))\right )}{2520 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(11/2),x]
[Out]
(a^2*((10080*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/(E^(I*(c
+ d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]) - Csc[c + d*x]^5*(1650*Cos[c + d*x]
- 2051*Cos[3*(c + d*x)] + 961*Cos[5*(c + d*x)] - (282*I)*Sin[c + d*x] + (49*I)*Sin[3*(c + d*x)] + (331*I)*Sin
[5*(c + d*x)])*Sqrt[Tan[c + d*x]])*Sqrt[a + I*a*Tan[c + d*x]])/(2520*d)
________________________________________________________________________________________
fricas [B] time = 0.59, size = 565, normalized size = 2.36 \[ \frac {\sqrt {2} {\left (-5168 i \, a^{2} e^{\left (11 i \, d x + 11 i \, c\right )} + 8008 i \, a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} - 5472 i \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} - 7728 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 8400 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 2520 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - 315 \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + i \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) + 315 \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - i \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{630 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x, algorithm="fricas")
[Out]
1/630*(sqrt(2)*(-5168*I*a^2*e^(11*I*d*x + 11*I*c) + 8008*I*a^2*e^(9*I*d*x + 9*I*c) - 5472*I*a^2*e^(7*I*d*x + 7
*I*c) - 7728*I*a^2*e^(5*I*d*x + 5*I*c) + 8400*I*a^2*e^(3*I*d*x + 3*I*c) - 2520*I*a^2*e^(I*d*x + I*c))*sqrt(a/(
e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) - 315*sqrt(32*I*a^5/d^2
)*(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5
*d*e^(2*I*d*x + 2*I*c) - d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1
))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + I*sqrt(32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-
I*d*x - I*c)/a^2) + 315*sqrt(32*I*a^5/d^2)*(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*
x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) - d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*
c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) - I*s
qrt(32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2))/(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c)
+ 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) - d)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^(5/2)/tan(d*x + c)^(11/2), x)
________________________________________________________________________________________
maple [B] time = 0.22, size = 501, normalized size = 2.10 \[ -\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (315 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{5}\left (d x +c \right )\right ) a +315 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{5}\left (d x +c \right )\right ) a +1260 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, \left (\tan ^{5}\left (d x +c \right )\right ) a -472 i \left (\tan ^{3}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+1576 \left (\tan ^{4}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+190 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )-276 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )+70 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{315 d \tan \left (d x +c \right )^{\frac {9}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x)
[Out]
-1/315/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(315*I*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^5*a+315*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(
1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^5*a+126
0*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*ta
n(d*x+c)^5*a-472*I*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+1576*tan(d*x+c)
^4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+190*I*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+
c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-276*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x
+c)^2+70*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2))/tan(d*x+c)^(9/2)/(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)
________________________________________________________________________________________
maxima [B] time = 1.98, size = 3464, normalized size = 14.49 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(11/2),x, algorithm="maxima")
[Out]
1/1587600*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(6350400*I - 6350400)*a^2
*cos(7*d*x + 7*c) + (21168000*I - 21168000)*a^2*cos(5*d*x + 5*c) - (25824960*I - 25824960)*a^2*cos(3*d*x + 3*c
) + (11429460*I - 11429460)*a^2*cos(d*x + c) + (6350400*I + 6350400)*a^2*sin(7*d*x + 7*c) - (21168000*I + 2116
8000)*a^2*sin(5*d*x + 5*c) + (25824960*I + 25824960)*a^2*sin(3*d*x + 3*c) - (11429460*I + 11429460)*a^2*sin(d*
x + c))*cos(7/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (((1219680*I - 1219680)*a^2*cos(d*x + c) -
(1219680*I + 1219680)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (1219680*I - 1219680)*a^2*cos(d*x + c) + ((12196
80*I - 1219680)*a^2*cos(d*x + c) - (1219680*I + 1219680)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^2 - (1219680*I + 1
219680)*a^2*sin(d*x + c) + (-(6350400*I - 6350400)*a^2*cos(2*d*x + 2*c)^2 - (6350400*I - 6350400)*a^2*sin(2*d*
x + 2*c)^2 + (12700800*I - 12700800)*a^2*cos(2*d*x + 2*c) - (6350400*I - 6350400)*a^2)*cos(3*d*x + 3*c) + (-(2
439360*I - 2439360)*a^2*cos(d*x + c) + (2439360*I + 2439360)*a^2*sin(d*x + c))*cos(2*d*x + 2*c) + ((6350400*I
+ 6350400)*a^2*cos(2*d*x + 2*c)^2 + (6350400*I + 6350400)*a^2*sin(2*d*x + 2*c)^2 - (12700800*I + 12700800)*a^2
*cos(2*d*x + 2*c) + (6350400*I + 6350400)*a^2)*sin(3*d*x + 3*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x
+ 2*c) + 1)) + ((6350400*I + 6350400)*a^2*cos(7*d*x + 7*c) - (21168000*I + 21168000)*a^2*cos(5*d*x + 5*c) + (2
5824960*I + 25824960)*a^2*cos(3*d*x + 3*c) - (11429460*I + 11429460)*a^2*cos(d*x + c) + (6350400*I - 6350400)*
a^2*sin(7*d*x + 7*c) - (21168000*I - 21168000)*a^2*sin(5*d*x + 5*c) + (25824960*I - 25824960)*a^2*sin(3*d*x +
3*c) - (11429460*I - 11429460)*a^2*sin(d*x + c))*sin(7/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (
(-(1219680*I + 1219680)*a^2*cos(d*x + c) - (1219680*I - 1219680)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (12196
80*I + 1219680)*a^2*cos(d*x + c) + (-(1219680*I + 1219680)*a^2*cos(d*x + c) - (1219680*I - 1219680)*a^2*sin(d*
x + c))*sin(2*d*x + 2*c)^2 - (1219680*I - 1219680)*a^2*sin(d*x + c) + ((6350400*I + 6350400)*a^2*cos(2*d*x + 2
*c)^2 + (6350400*I + 6350400)*a^2*sin(2*d*x + 2*c)^2 - (12700800*I + 12700800)*a^2*cos(2*d*x + 2*c) + (6350400
*I + 6350400)*a^2)*cos(3*d*x + 3*c) + ((2439360*I + 2439360)*a^2*cos(d*x + c) + (2439360*I - 2439360)*a^2*sin(
d*x + c))*cos(2*d*x + 2*c) + ((6350400*I - 6350400)*a^2*cos(2*d*x + 2*c)^2 + (6350400*I - 6350400)*a^2*sin(2*d
*x + 2*c)^2 - (12700800*I - 12700800)*a^2*cos(2*d*x + 2*c) + (6350400*I - 6350400)*a^2)*sin(3*d*x + 3*c))*sin(
3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a) + (((6350400*I - 6350400)*a^2*cos(2*d*x + 2*c)^
4 + (6350400*I - 6350400)*a^2*sin(2*d*x + 2*c)^4 - (25401600*I - 25401600)*a^2*cos(2*d*x + 2*c)^3 + (38102400*
I - 38102400)*a^2*cos(2*d*x + 2*c)^2 - (25401600*I - 25401600)*a^2*cos(2*d*x + 2*c) + ((12700800*I - 12700800)
*a^2*cos(2*d*x + 2*c)^2 - (25401600*I - 25401600)*a^2*cos(2*d*x + 2*c) + (12700800*I - 12700800)*a^2)*sin(2*d*
x + 2*c)^2 + (6350400*I - 6350400)*a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c)
+ 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - cos(d*x + c), (cos(2*d*x + 2*c)^2 + sin
(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - si
n(d*x + c)) + (-(3175200*I + 3175200)*a^2*cos(2*d*x + 2*c)^4 - (3175200*I + 3175200)*a^2*sin(2*d*x + 2*c)^4 +
(12700800*I + 12700800)*a^2*cos(2*d*x + 2*c)^3 - (19051200*I + 19051200)*a^2*cos(2*d*x + 2*c)^2 + (12700800*I
+ 12700800)*a^2*cos(2*d*x + 2*c) + (-(6350400*I + 6350400)*a^2*cos(2*d*x + 2*c)^2 + (12700800*I + 12700800)*a^
2*cos(2*d*x + 2*c) - (6350400*I + 6350400)*a^2)*sin(2*d*x + 2*c)^2 - (3175200*I + 3175200)*a^2)*log(cos(d*x +
c)^2 + sin(d*x + c)^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan
2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2) -
2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c),
-cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))
)))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + (((6350400*I - 6350400)
*a^2*cos(9*d*x + 9*c) - (7408800*I - 7408800)*a^2*cos(7*d*x + 7*c) + (8414280*I - 8414280)*a^2*cos(5*d*x + 5*c
) - (631260*I - 631260)*a^2*cos(3*d*x + 3*c) - (1079820*I - 1079820)*a^2*cos(d*x + c) - (6350400*I + 6350400)*
a^2*sin(9*d*x + 9*c) + (7408800*I + 7408800)*a^2*sin(7*d*x + 7*c) - (8414280*I + 8414280)*a^2*sin(5*d*x + 5*c)
+ (631260*I + 631260)*a^2*sin(3*d*x + 3*c) + (1079820*I + 1079820)*a^2*sin(d*x + c))*cos(9/2*arctan2(sin(2*d*
x + 2*c), -cos(2*d*x + 2*c) + 1)) + (((7159320*I - 7159320)*a^2*cos(d*x + c) - (7159320*I + 7159320)*a^2*sin(d
*x + c))*cos(2*d*x + 2*c)^2 + (7159320*I - 7159320)*a^2*cos(d*x + c) + ((7159320*I - 7159320)*a^2*cos(d*x + c)
- (7159320*I + 7159320)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^2 - (7159320*I + 7159320)*a^2*sin(d*x + c) + ((635
0400*I - 6350400)*a^2*cos(2*d*x + 2*c)^2 + (6350400*I - 6350400)*a^2*sin(2*d*x + 2*c)^2 - (12700800*I - 127008
00)*a^2*cos(2*d*x + 2*c) + (6350400*I - 6350400)*a^2)*cos(5*d*x + 5*c) + (-(14817600*I - 14817600)*a^2*cos(2*d
*x + 2*c)^2 - (14817600*I - 14817600)*a^2*sin(2*d*x + 2*c)^2 + (29635200*I - 29635200)*a^2*cos(2*d*x + 2*c) -
(14817600*I - 14817600)*a^2)*cos(3*d*x + 3*c) + (-(14318640*I - 14318640)*a^2*cos(d*x + c) + (14318640*I + 143
18640)*a^2*sin(d*x + c))*cos(2*d*x + 2*c) + (-(6350400*I + 6350400)*a^2*cos(2*d*x + 2*c)^2 - (6350400*I + 6350
400)*a^2*sin(2*d*x + 2*c)^2 + (12700800*I + 12700800)*a^2*cos(2*d*x + 2*c) - (6350400*I + 6350400)*a^2)*sin(5*
d*x + 5*c) + ((14817600*I + 14817600)*a^2*cos(2*d*x + 2*c)^2 + (14817600*I + 14817600)*a^2*sin(2*d*x + 2*c)^2
- (29635200*I + 29635200)*a^2*cos(2*d*x + 2*c) + (14817600*I + 14817600)*a^2)*sin(3*d*x + 3*c))*cos(5/2*arctan
2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + ((-(12378240*I - 12378240)*a^2*cos(d*x + c) + (12378240*I + 1237
8240)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^4 + (-(12378240*I - 12378240)*a^2*cos(d*x + c) + (12378240*I + 123782
40)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^4 + ((49512960*I - 49512960)*a^2*cos(d*x + c) - (49512960*I + 49512960)
*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^3 + (-(74269440*I - 74269440)*a^2*cos(d*x + c) + (74269440*I + 74269440)*a
^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (12378240*I - 12378240)*a^2*cos(d*x + c) + ((-(24756480*I - 24756480)*a^
2*cos(d*x + c) + (24756480*I + 24756480)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (24756480*I - 24756480)*a^2*co
s(d*x + c) + (24756480*I + 24756480)*a^2*sin(d*x + c) + ((49512960*I - 49512960)*a^2*cos(d*x + c) - (49512960*
I + 49512960)*a^2*sin(d*x + c))*cos(2*d*x + 2*c))*sin(2*d*x + 2*c)^2 + (12378240*I + 12378240)*a^2*sin(d*x + c
) + ((49512960*I - 49512960)*a^2*cos(d*x + c) - (49512960*I + 49512960)*a^2*sin(d*x + c))*cos(2*d*x + 2*c))*co
s(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(6350400*I + 6350400)*a^2*cos(9*d*x + 9*c) + (7408
800*I + 7408800)*a^2*cos(7*d*x + 7*c) - (8414280*I + 8414280)*a^2*cos(5*d*x + 5*c) + (631260*I + 631260)*a^2*c
os(3*d*x + 3*c) + (1079820*I + 1079820)*a^2*cos(d*x + c) - (6350400*I - 6350400)*a^2*sin(9*d*x + 9*c) + (74088
00*I - 7408800)*a^2*sin(7*d*x + 7*c) - (8414280*I - 8414280)*a^2*sin(5*d*x + 5*c) + (631260*I - 631260)*a^2*si
n(3*d*x + 3*c) + (1079820*I - 1079820)*a^2*sin(d*x + c))*sin(9/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) +
1)) + ((-(7159320*I + 7159320)*a^2*cos(d*x + c) - (7159320*I - 7159320)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2
- (7159320*I + 7159320)*a^2*cos(d*x + c) + (-(7159320*I + 7159320)*a^2*cos(d*x + c) - (7159320*I - 7159320)*a^
2*sin(d*x + c))*sin(2*d*x + 2*c)^2 - (7159320*I - 7159320)*a^2*sin(d*x + c) + (-(6350400*I + 6350400)*a^2*cos(
2*d*x + 2*c)^2 - (6350400*I + 6350400)*a^2*sin(2*d*x + 2*c)^2 + (12700800*I + 12700800)*a^2*cos(2*d*x + 2*c) -
(6350400*I + 6350400)*a^2)*cos(5*d*x + 5*c) + ((14817600*I + 14817600)*a^2*cos(2*d*x + 2*c)^2 + (14817600*I +
14817600)*a^2*sin(2*d*x + 2*c)^2 - (29635200*I + 29635200)*a^2*cos(2*d*x + 2*c) + (14817600*I + 14817600)*a^2
)*cos(3*d*x + 3*c) + ((14318640*I + 14318640)*a^2*cos(d*x + c) + (14318640*I - 14318640)*a^2*sin(d*x + c))*cos
(2*d*x + 2*c) + (-(6350400*I - 6350400)*a^2*cos(2*d*x + 2*c)^2 - (6350400*I - 6350400)*a^2*sin(2*d*x + 2*c)^2
+ (12700800*I - 12700800)*a^2*cos(2*d*x + 2*c) - (6350400*I - 6350400)*a^2)*sin(5*d*x + 5*c) + ((14817600*I -
14817600)*a^2*cos(2*d*x + 2*c)^2 + (14817600*I - 14817600)*a^2*sin(2*d*x + 2*c)^2 - (29635200*I - 29635200)*a^
2*cos(2*d*x + 2*c) + (14817600*I - 14817600)*a^2)*sin(3*d*x + 3*c))*sin(5/2*arctan2(sin(2*d*x + 2*c), -cos(2*d
*x + 2*c) + 1)) + (((12378240*I + 12378240)*a^2*cos(d*x + c) + (12378240*I - 12378240)*a^2*sin(d*x + c))*cos(2
*d*x + 2*c)^4 + ((12378240*I + 12378240)*a^2*cos(d*x + c) + (12378240*I - 12378240)*a^2*sin(d*x + c))*sin(2*d*
x + 2*c)^4 + (-(49512960*I + 49512960)*a^2*cos(d*x + c) - (49512960*I - 49512960)*a^2*sin(d*x + c))*cos(2*d*x
+ 2*c)^3 + ((74269440*I + 74269440)*a^2*cos(d*x + c) + (74269440*I - 74269440)*a^2*sin(d*x + c))*cos(2*d*x + 2
*c)^2 + (12378240*I + 12378240)*a^2*cos(d*x + c) + (((24756480*I + 24756480)*a^2*cos(d*x + c) + (24756480*I -
24756480)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (24756480*I + 24756480)*a^2*cos(d*x + c) + (24756480*I - 2475
6480)*a^2*sin(d*x + c) + (-(49512960*I + 49512960)*a^2*cos(d*x + c) - (49512960*I - 49512960)*a^2*sin(d*x + c)
)*cos(2*d*x + 2*c))*sin(2*d*x + 2*c)^2 + (12378240*I - 12378240)*a^2*sin(d*x + c) + (-(49512960*I + 49512960)*
a^2*cos(d*x + c) - (49512960*I - 49512960)*a^2*sin(d*x + c))*cos(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c
), -cos(2*d*x + 2*c) + 1)))*sqrt(a))/((cos(2*d*x + 2*c)^4 + sin(2*d*x + 2*c)^4 - 4*cos(2*d*x + 2*c)^3 + 2*(cos
(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 6*cos(2*d*x + 2*c)^2 - 4*cos(2*d*x + 2*c) + 1)*
(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*d)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{11/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(11/2),x)
[Out]
int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(11/2), x)
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))**(5/2)/tan(d*x+c)**(11/2),x)
[Out]
Timed out
________________________________________________________________________________________