3.208 \(\int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=202 \[ \frac {(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {32 a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {104 i a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \sqrt {\tan (c+d x)}} \]
[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+104/21*I*a^2*(a+I*a*tan(d*x
+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/7*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(7/2)-6/7*I*a^2*(a+I*a*tan(d*x+c))
^(1/2)/d/tan(d*x+c)^(5/2)+32/21*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)
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Rubi [A] time = 0.57, antiderivative size = 202, normalized size of antiderivative = 1.00,
number of steps used = 7, 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} \[ \frac {32 a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {104 i a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \sqrt {\tan (c+d x)}}+\frac {(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(9/2),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]])/(7*d*Tan[c + d*x]^(7/2)) - (((6*I)/7)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(5
/2)) + (32*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(21*d*Tan[c + d*x]^(3/2)) + (((104*I)/21)*a^2*Sqrt[a + I*a*Tan[c +
d*x]])/(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 {9}{2}}(c+d x)} \, dx &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2}{7} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {15 i a^2}{2}+\frac {13}{2} a^2 \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (20 a^3+15 i a^3 \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{35 a}\\ &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {32 a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {65 i a^4}{2}-20 a^4 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{105 a^2}\\ &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {32 a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {104 i a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \sqrt {\tan (c+d x)}}-\frac {16 \int -\frac {105 a^5 \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {\tan (c+d x)}} \, dx}{105 a^3}\\ &=-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {32 a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {104 i a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \sqrt {\tan (c+d x)}}+\left (4 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)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {32 a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {104 i a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \sqrt {\tan (c+d x)}}-\frac {\left (8 i 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)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {32 a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {104 i a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 2.88, size = 188, normalized size = 0.93 \[ \frac {4 i \sqrt {2} a^2 e^{-i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (e^{i (c+d x)} \left (70 e^{2 i (c+d x)}-77 e^{4 i (c+d x)}+40 e^{6 i (c+d x)}-21\right )-21 \left (-1+e^{2 i (c+d x)}\right )^{7/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{21 d \left (-1+e^{2 i (c+d x)}\right )^3 \sqrt {\tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(9/2),x]
[Out]
(((4*I)/21)*Sqrt[2]*a^2*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(E^(I*(c + d*x))*(-21 + 70*E^(
(2*I)*(c + d*x)) - 77*E^((4*I)*(c + d*x)) + 40*E^((6*I)*(c + d*x))) - 21*(-1 + E^((2*I)*(c + d*x)))^(7/2)*ArcT
anh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^3*Sqrt[Tan
[c + d*x]])
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fricas [B] time = 0.73, size = 509, normalized size = 2.52 \[ -\frac {8 \, \sqrt {2} {\left (40 \, a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} - 37 \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} - 7 \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 49 \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 21 \, 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}} - 21 \, \sqrt {-\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, 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}} + \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 ) + 21 \, \sqrt {-\frac {32 i \, a^{5}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, 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}} - \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 )}{42 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, 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)^(9/2),x, algorithm="fricas")
[Out]
-1/42*(8*sqrt(2)*(40*a^2*e^(9*I*d*x + 9*I*c) - 37*a^2*e^(7*I*d*x + 7*I*c) - 7*a^2*e^(5*I*d*x + 5*I*c) + 49*a^2
*e^(3*I*d*x + 3*I*c) - 21*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)) - 21*sqrt(-32*I*a^5/d^2)*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*
d*e^(4*I*d*x + 4*I*c) - 4*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)) + sqrt(-32*I*a^5/d^2)
*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) + 21*sqrt(-32*I*a^5/d^2)*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6
*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*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)) - sqrt(-32*
I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^
(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]
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)^(9/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^(5/2)/tan(d*x + c)^(9/2), x)
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maple [B] time = 0.22, size = 459, normalized size = 2.27 \[ \frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (21 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{4}\left (d x +c \right )\right ) a +84 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, \left (\tan ^{4}\left (d x +c \right )\right ) a -21 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{4}\left (d x +c \right )\right ) a +32 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )+104 i \left (\tan ^{3}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-18 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )-6 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{21 d \tan \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}} \]
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)^(9/2),x)
[Out]
1/21/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(7/2)*(21*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)^4*a+84*I*ln(1/2*(2*I*a*ta
n(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)*tan(d*x+c)^4*a-21*(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)^4*a+32*(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+104*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)-18*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)-6*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)
)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)
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maxima [B] time = 1.69, size = 3140, normalized size = 15.54 \[ \text {result too large to display} \]
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)^(9/2),x, algorithm="maxima")
[Out]
1/11025*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(44100*I + 44100)*a^2*cos(7
*d*x + 7*c) + (44100*I + 44100)*a^2*cos(5*d*x + 5*c) - (26460*I + 26460)*a^2*cos(3*d*x + 3*c) + (1260*I + 1260
)*a^2*cos(d*x + c) - (44100*I - 44100)*a^2*sin(7*d*x + 7*c) + (44100*I - 44100)*a^2*sin(5*d*x + 5*c) - (26460*
I - 26460)*a^2*sin(3*d*x + 3*c) + (1260*I - 1260)*a^2*sin(d*x + c))*cos(7/2*arctan2(sin(2*d*x + 2*c), -cos(2*d
*x + 2*c) + 1)) + (((27300*I + 27300)*a^2*cos(d*x + c) + (27300*I - 27300)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^
2 + (27300*I + 27300)*a^2*cos(d*x + c) + ((27300*I + 27300)*a^2*cos(d*x + c) + (27300*I - 27300)*a^2*sin(d*x +
c))*sin(2*d*x + 2*c)^2 + (27300*I - 27300)*a^2*sin(d*x + c) + (-(44100*I + 44100)*a^2*cos(2*d*x + 2*c)^2 - (4
4100*I + 44100)*a^2*sin(2*d*x + 2*c)^2 + (88200*I + 88200)*a^2*cos(2*d*x + 2*c) - (44100*I + 44100)*a^2)*cos(3
*d*x + 3*c) + (-(54600*I + 54600)*a^2*cos(d*x + c) - (54600*I - 54600)*a^2*sin(d*x + c))*cos(2*d*x + 2*c) + (-
(44100*I - 44100)*a^2*cos(2*d*x + 2*c)^2 - (44100*I - 44100)*a^2*sin(2*d*x + 2*c)^2 + (88200*I - 88200)*a^2*co
s(2*d*x + 2*c) - (44100*I - 44100)*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)) + (-(44100*I - 44100)*a^2*cos(7*d*x + 7*c) + (44100*I - 44100)*a^2*cos(5*d*x + 5*c) - (26460*I - 26460)*
a^2*cos(3*d*x + 3*c) + (1260*I - 1260)*a^2*cos(d*x + c) + (44100*I + 44100)*a^2*sin(7*d*x + 7*c) - (44100*I +
44100)*a^2*sin(5*d*x + 5*c) + (26460*I + 26460)*a^2*sin(3*d*x + 3*c) - (1260*I + 1260)*a^2*sin(d*x + c))*sin(7
/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (((27300*I - 27300)*a^2*cos(d*x + c) - (27300*I + 27300
)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (27300*I - 27300)*a^2*cos(d*x + c) + ((27300*I - 27300)*a^2*cos(d*x +
c) - (27300*I + 27300)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^2 - (27300*I + 27300)*a^2*sin(d*x + c) + (-(44100*I
- 44100)*a^2*cos(2*d*x + 2*c)^2 - (44100*I - 44100)*a^2*sin(2*d*x + 2*c)^2 + (88200*I - 88200)*a^2*cos(2*d*x
+ 2*c) - (44100*I - 44100)*a^2)*cos(3*d*x + 3*c) + (-(54600*I - 54600)*a^2*cos(d*x + c) + (54600*I + 54600)*a^
2*sin(d*x + c))*cos(2*d*x + 2*c) + ((44100*I + 44100)*a^2*cos(2*d*x + 2*c)^2 + (44100*I + 44100)*a^2*sin(2*d*x
+ 2*c)^2 - (88200*I + 88200)*a^2*cos(2*d*x + 2*c) + (44100*I + 44100)*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) + (((44100*I + 44100)*a^2*cos(2*d*x + 2*c)^4 + (44100*I + 4
4100)*a^2*sin(2*d*x + 2*c)^4 - (176400*I + 176400)*a^2*cos(2*d*x + 2*c)^3 + (264600*I + 264600)*a^2*cos(2*d*x
+ 2*c)^2 - (176400*I + 176400)*a^2*cos(2*d*x + 2*c) + ((88200*I + 88200)*a^2*cos(2*d*x + 2*c)^2 - (176400*I +
176400)*a^2*cos(2*d*x + 2*c) + (88200*I + 88200)*a^2)*sin(2*d*x + 2*c)^2 + (44100*I + 44100)*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)) - sin(d*x + c)) + ((22050*I - 22050)*a^2*cos(2*d*x + 2*c
)^4 + (22050*I - 22050)*a^2*sin(2*d*x + 2*c)^4 - (88200*I - 88200)*a^2*cos(2*d*x + 2*c)^3 + (132300*I - 132300
)*a^2*cos(2*d*x + 2*c)^2 - (88200*I - 88200)*a^2*cos(2*d*x + 2*c) + ((44100*I - 44100)*a^2*cos(2*d*x + 2*c)^2
- (88200*I - 88200)*a^2*cos(2*d*x + 2*c) + (44100*I - 44100)*a^2)*sin(2*d*x + 2*c)^2 + (22050*I - 22050)*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*arctan2(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) + ((((63840
*I + 63840)*a^2*cos(d*x + c) + (63840*I - 63840)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (63840*I + 63840)*a^2*
cos(d*x + c) + ((63840*I + 63840)*a^2*cos(d*x + c) + (63840*I - 63840)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^2 +
(63840*I - 63840)*a^2*sin(d*x + c) + ((44100*I + 44100)*a^2*cos(2*d*x + 2*c)^2 + (44100*I + 44100)*a^2*sin(2*d
*x + 2*c)^2 - (88200*I + 88200)*a^2*cos(2*d*x + 2*c) + (44100*I + 44100)*a^2)*cos(5*d*x + 5*c) + (-(102900*I +
102900)*a^2*cos(2*d*x + 2*c)^2 - (102900*I + 102900)*a^2*sin(2*d*x + 2*c)^2 + (205800*I + 205800)*a^2*cos(2*d
*x + 2*c) - (102900*I + 102900)*a^2)*cos(3*d*x + 3*c) + (-(127680*I + 127680)*a^2*cos(d*x + c) - (127680*I - 1
27680)*a^2*sin(d*x + c))*cos(2*d*x + 2*c) + ((44100*I - 44100)*a^2*cos(2*d*x + 2*c)^2 + (44100*I - 44100)*a^2*
sin(2*d*x + 2*c)^2 - (88200*I - 88200)*a^2*cos(2*d*x + 2*c) + (44100*I - 44100)*a^2)*sin(5*d*x + 5*c) + (-(102
900*I - 102900)*a^2*cos(2*d*x + 2*c)^2 - (102900*I - 102900)*a^2*sin(2*d*x + 2*c)^2 + (205800*I - 205800)*a^2*
cos(2*d*x + 2*c) - (102900*I - 102900)*a^2)*sin(3*d*x + 3*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2
*c) + 1)) + ((-(48300*I + 48300)*a^2*cos(d*x + c) - (48300*I - 48300)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^4 + (
-(48300*I + 48300)*a^2*cos(d*x + c) - (48300*I - 48300)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^4 + ((193200*I + 19
3200)*a^2*cos(d*x + c) + (193200*I - 193200)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^3 + (-(289800*I + 289800)*a^2*
cos(d*x + c) - (289800*I - 289800)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (48300*I + 48300)*a^2*cos(d*x + c) +
((-(96600*I + 96600)*a^2*cos(d*x + c) - (96600*I - 96600)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (96600*I + 9
6600)*a^2*cos(d*x + c) - (96600*I - 96600)*a^2*sin(d*x + c) + ((193200*I + 193200)*a^2*cos(d*x + c) + (193200*
I - 193200)*a^2*sin(d*x + c))*cos(2*d*x + 2*c))*sin(2*d*x + 2*c)^2 - (48300*I - 48300)*a^2*sin(d*x + c) + ((19
3200*I + 193200)*a^2*cos(d*x + c) + (193200*I - 193200)*a^2*sin(d*x + c))*cos(2*d*x + 2*c))*cos(1/2*arctan2(si
n(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (((63840*I - 63840)*a^2*cos(d*x + c) - (63840*I + 63840)*a^2*sin(d*x
+ c))*cos(2*d*x + 2*c)^2 + (63840*I - 63840)*a^2*cos(d*x + c) + ((63840*I - 63840)*a^2*cos(d*x + c) - (63840*
I + 63840)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^2 - (63840*I + 63840)*a^2*sin(d*x + c) + ((44100*I - 44100)*a^2*
cos(2*d*x + 2*c)^2 + (44100*I - 44100)*a^2*sin(2*d*x + 2*c)^2 - (88200*I - 88200)*a^2*cos(2*d*x + 2*c) + (4410
0*I - 44100)*a^2)*cos(5*d*x + 5*c) + (-(102900*I - 102900)*a^2*cos(2*d*x + 2*c)^2 - (102900*I - 102900)*a^2*si
n(2*d*x + 2*c)^2 + (205800*I - 205800)*a^2*cos(2*d*x + 2*c) - (102900*I - 102900)*a^2)*cos(3*d*x + 3*c) + (-(1
27680*I - 127680)*a^2*cos(d*x + c) + (127680*I + 127680)*a^2*sin(d*x + c))*cos(2*d*x + 2*c) + (-(44100*I + 441
00)*a^2*cos(2*d*x + 2*c)^2 - (44100*I + 44100)*a^2*sin(2*d*x + 2*c)^2 + (88200*I + 88200)*a^2*cos(2*d*x + 2*c)
- (44100*I + 44100)*a^2)*sin(5*d*x + 5*c) + ((102900*I + 102900)*a^2*cos(2*d*x + 2*c)^2 + (102900*I + 102900)
*a^2*sin(2*d*x + 2*c)^2 - (205800*I + 205800)*a^2*cos(2*d*x + 2*c) + (102900*I + 102900)*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)) + ((-(48300*I - 48300)*a^2*cos(d*x + c) + (48300*I
+ 48300)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^4 + (-(48300*I - 48300)*a^2*cos(d*x + c) + (48300*I + 48300)*a^2*
sin(d*x + c))*sin(2*d*x + 2*c)^4 + ((193200*I - 193200)*a^2*cos(d*x + c) - (193200*I + 193200)*a^2*sin(d*x + c
))*cos(2*d*x + 2*c)^3 + (-(289800*I - 289800)*a^2*cos(d*x + c) + (289800*I + 289800)*a^2*sin(d*x + c))*cos(2*d
*x + 2*c)^2 - (48300*I - 48300)*a^2*cos(d*x + c) + ((-(96600*I - 96600)*a^2*cos(d*x + c) + (96600*I + 96600)*a
^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (96600*I - 96600)*a^2*cos(d*x + c) + (96600*I + 96600)*a^2*sin(d*x + c)
+ ((193200*I - 193200)*a^2*cos(d*x + c) - (193200*I + 193200)*a^2*sin(d*x + c))*cos(2*d*x + 2*c))*sin(2*d*x +
2*c)^2 + (48300*I + 48300)*a^2*sin(d*x + c) + ((193200*I - 193200)*a^2*cos(d*x + c) - (193200*I + 193200)*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*co
s(2*d*x + 2*c) + 1)^(1/4)*d)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{9/2}} \,d x \]
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)^(9/2),x)
[Out]
int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(9/2), x)
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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)**(9/2),x)
[Out]
Timed out
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