3.200 \(\int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=235 \[ \frac {(2-2 i) a^{3/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 i a^2}{7 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {76 a \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {16 i a \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {268 i a \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}} \]
[Out]
(2-2*I)*a^(3/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+268/105*I*a*(a+I*a*tan(d*x+
c))^(1/2)/d/tan(d*x+c)^(1/2)-2/7*a^2/d/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2)-2/7*I*a^2/d/(a+I*a*tan(d*x+c)
)^(1/2)/tan(d*x+c)^(5/2)-16/35*I*a*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)+76/105*a*(a+I*a*tan(d*x+c))^(1/
2)/d/tan(d*x+c)^(3/2)
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Rubi [A] time = 0.74, antiderivative size = 235, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used =
{3553, 3596, 3598, 12, 3544, 205} \[ -\frac {2 i a^2}{7 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {(2-2 i) a^{3/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {76 a \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {16 i a \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {268 i a \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(9/2),x]
[Out]
((2 - 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*a^2)/(7*d*
Tan[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/7)*a^2)/(d*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*
x]]) - (((16*I)/35)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(5/2)) + (76*a*Sqrt[a + I*a*Tan[c + d*x]])/(
105*d*Tan[c + d*x]^(3/2)) + (((268*I)/105)*a*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 3596
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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n,
0]
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))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx &=-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2}{7} \int \frac {-\frac {13 i a^2}{2}+\frac {15}{2} a^2 \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\\ &=-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{7 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-4 i a^3+3 a^3 \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx}{7 a^2}\\ &=-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{7 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {19 a^4}{2}+8 i a^4 \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{35 a^3}\\ &=-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{7 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {76 a \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {67 i a^5}{4}-\frac {19}{2} a^5 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{105 a^4}\\ &=-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{7 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {76 a \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {268 i a \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}-\frac {16 \int -\frac {105 a^6 \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{105 a^5}\\ &=-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{7 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {76 a \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {268 i a \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}+(2 a) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{7 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {76 a \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {268 i a \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}-\frac {\left (4 i a^3\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 {(2-2 i) a^{3/2} \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}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{7 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {76 a \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {268 i a \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 3.15, size = 224, normalized size = 0.95 \[ -\frac {2 i \sqrt {2} a e^{-i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-\frac {a \csc ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (-378 i \sin (c+d x)+158 i \sin (3 (c+d x))+7 \cos (c+d x)+53 \cos (3 (c+d x)))}{210 d \sqrt {\tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(9/2),x]
[Out]
((-2*I)*Sqrt[2]*a*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*ArcTa
nh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/(d*E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/
(1 + E^((2*I)*(c + d*x)))]) - (a*Csc[c + d*x]^3*(7*Cos[c + d*x] + 53*Cos[3*(c + d*x)] - (378*I)*Sin[c + d*x] +
(158*I)*Sin[3*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(210*d*Sqrt[Tan[c + d*x]])
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fricas [B] time = 0.48, size = 491, normalized size = 2.09 \[ -\frac {4 \, \sqrt {2} {\left (211 \, a e^{\left (9 i \, d x + 9 i \, c\right )} - 160 \, a e^{\left (7 i \, d x + 7 i \, c\right )} + 14 \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 280 \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 105 \, a 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}} - 105 \, {\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 )} \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\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 {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) + 105 \, {\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 )} \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\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 {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{210 \, {\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))^(3/2)/tan(d*x+c)^(9/2),x, algorithm="fricas")
[Out]
-1/210*(4*sqrt(2)*(211*a*e^(9*I*d*x + 9*I*c) - 160*a*e^(7*I*d*x + 7*I*c) + 14*a*e^(5*I*d*x + 5*I*c) + 280*a*e^
(3*I*d*x + 3*I*c) - 105*a*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)) - 105*(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)*sqrt(-8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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(-8*I*a^3/d^2)*d*e^(I*d
*x + I*c))*e^(-I*d*x - I*c)/a) + 105*(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)*sqrt(-8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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(-8*I*a^3/d^2)*d*e^(
I*d*x + I*c))*e^(-I*d*x - I*c)/a))/(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 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^(3/2)/tan(d*x + c)^(9/2), x)
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maple [B] time = 0.21, size = 457, normalized size = 1.94 \[ \frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (105 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{4}\left (d x +c \right )\right ) a +420 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, \left (\tan ^{4}\left (d x +c \right )\right ) a -105 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{4}\left (d x +c \right )\right ) a +152 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )+536 i \left (\tan ^{3}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-96 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )-60 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{210 d \tan \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),x)
[Out]
1/210/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(7/2)*(105*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+420*I*ln(1/2*(2*I*a*t
an(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-105*
(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))/(ta
n(d*x+c)+I))*tan(d*x+c)^4*a+152*(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+53
6*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)-96*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)-60*(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 = 1.02, size = 2840, normalized size = 12.09 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),x, algorithm="maxima")
[Out]
1/176400*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(352800*I + 352800)*a*cos(
7*d*x + 7*c) + (176400*I + 176400)*a*cos(5*d*x + 5*c) - (167580*I + 167580)*a*cos(3*d*x + 3*c) - (59220*I + 59
220)*a*cos(d*x + c) - (352800*I - 352800)*a*sin(7*d*x + 7*c) + (176400*I - 176400)*a*sin(5*d*x + 5*c) - (16758
0*I - 167580)*a*sin(3*d*x + 3*c) - (59220*I - 59220)*a*sin(d*x + c))*cos(7/2*arctan2(sin(2*d*x + 2*c), -cos(2*
d*x + 2*c) + 1)) + (((236880*I + 236880)*a*cos(d*x + c) + (236880*I - 236880)*a*sin(d*x + c))*cos(2*d*x + 2*c)
^2 + ((236880*I + 236880)*a*cos(d*x + c) + (236880*I - 236880)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + (-(352800*
I + 352800)*a*cos(2*d*x + 2*c)^2 - (352800*I + 352800)*a*sin(2*d*x + 2*c)^2 + (705600*I + 705600)*a*cos(2*d*x
+ 2*c) - (352800*I + 352800)*a)*cos(3*d*x + 3*c) + (-(473760*I + 473760)*a*cos(d*x + c) - (473760*I - 473760)*
a*sin(d*x + c))*cos(2*d*x + 2*c) + (236880*I + 236880)*a*cos(d*x + c) + (-(352800*I - 352800)*a*cos(2*d*x + 2*
c)^2 - (352800*I - 352800)*a*sin(2*d*x + 2*c)^2 + (705600*I - 705600)*a*cos(2*d*x + 2*c) - (352800*I - 352800)
*a)*sin(3*d*x + 3*c) + (236880*I - 236880)*a*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c)
+ 1)) + (-(352800*I - 352800)*a*cos(7*d*x + 7*c) + (176400*I - 176400)*a*cos(5*d*x + 5*c) - (167580*I - 16758
0)*a*cos(3*d*x + 3*c) - (59220*I - 59220)*a*cos(d*x + c) + (352800*I + 352800)*a*sin(7*d*x + 7*c) - (176400*I
+ 176400)*a*sin(5*d*x + 5*c) + (167580*I + 167580)*a*sin(3*d*x + 3*c) + (59220*I + 59220)*a*sin(d*x + c))*sin(
7/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (((236880*I - 236880)*a*cos(d*x + c) - (236880*I + 236
880)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((236880*I - 236880)*a*cos(d*x + c) - (236880*I + 236880)*a*sin(d*x
+ c))*sin(2*d*x + 2*c)^2 + (-(352800*I - 352800)*a*cos(2*d*x + 2*c)^2 - (352800*I - 352800)*a*sin(2*d*x + 2*c)
^2 + (705600*I - 705600)*a*cos(2*d*x + 2*c) - (352800*I - 352800)*a)*cos(3*d*x + 3*c) + (-(473760*I - 473760)*
a*cos(d*x + c) + (473760*I + 473760)*a*sin(d*x + c))*cos(2*d*x + 2*c) + (236880*I - 236880)*a*cos(d*x + c) + (
(352800*I + 352800)*a*cos(2*d*x + 2*c)^2 + (352800*I + 352800)*a*sin(2*d*x + 2*c)^2 - (705600*I + 705600)*a*co
s(2*d*x + 2*c) + (352800*I + 352800)*a)*sin(3*d*x + 3*c) - (236880*I + 236880)*a*sin(d*x + c))*sin(3/2*arctan2
(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a) + (((352800*I + 352800)*a*cos(2*d*x + 2*c)^4 + (352800*I +
352800)*a*sin(2*d*x + 2*c)^4 - (1411200*I + 1411200)*a*cos(2*d*x + 2*c)^3 + (2116800*I + 2116800)*a*cos(2*d*x
+ 2*c)^2 + ((705600*I + 705600)*a*cos(2*d*x + 2*c)^2 - (1411200*I + 1411200)*a*cos(2*d*x + 2*c) + (705600*I +
705600)*a)*sin(2*d*x + 2*c)^2 - (1411200*I + 1411200)*a*cos(2*d*x + 2*c) + (352800*I + 352800)*a)*arctan2((co
s(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)) + ((176400*I - 176400)*a*cos(2*d*x + 2*
c)^4 + (176400*I - 176400)*a*sin(2*d*x + 2*c)^4 - (705600*I - 705600)*a*cos(2*d*x + 2*c)^3 + (1058400*I - 1058
400)*a*cos(2*d*x + 2*c)^2 + ((352800*I - 352800)*a*cos(2*d*x + 2*c)^2 - (705600*I - 705600)*a*cos(2*d*x + 2*c)
+ (352800*I - 352800)*a)*sin(2*d*x + 2*c)^2 - (705600*I - 705600)*a*cos(2*d*x + 2*c) + (176400*I - 176400)*a)
*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(si
n(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) + ((((5245
80*I + 524580)*a*cos(d*x + c) + (524580*I - 524580)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((524580*I + 524580)*
a*cos(d*x + c) + (524580*I - 524580)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + ((352800*I + 352800)*a*cos(2*d*x + 2
*c)^2 + (352800*I + 352800)*a*sin(2*d*x + 2*c)^2 - (705600*I + 705600)*a*cos(2*d*x + 2*c) + (352800*I + 352800
)*a)*cos(5*d*x + 5*c) + (-(823200*I + 823200)*a*cos(2*d*x + 2*c)^2 - (823200*I + 823200)*a*sin(2*d*x + 2*c)^2
+ (1646400*I + 1646400)*a*cos(2*d*x + 2*c) - (823200*I + 823200)*a)*cos(3*d*x + 3*c) + (-(1049160*I + 1049160)
*a*cos(d*x + c) - (1049160*I - 1049160)*a*sin(d*x + c))*cos(2*d*x + 2*c) + (524580*I + 524580)*a*cos(d*x + c)
+ ((352800*I - 352800)*a*cos(2*d*x + 2*c)^2 + (352800*I - 352800)*a*sin(2*d*x + 2*c)^2 - (705600*I - 705600)*a
*cos(2*d*x + 2*c) + (352800*I - 352800)*a)*sin(5*d*x + 5*c) + (-(823200*I - 823200)*a*cos(2*d*x + 2*c)^2 - (82
3200*I - 823200)*a*sin(2*d*x + 2*c)^2 + (1646400*I - 1646400)*a*cos(2*d*x + 2*c) - (823200*I - 823200)*a)*sin(
3*d*x + 3*c) + (524580*I - 524580)*a*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) +
((-(349440*I + 349440)*a*cos(d*x + c) - (349440*I - 349440)*a*sin(d*x + c))*cos(2*d*x + 2*c)^4 + (-(349440*I
+ 349440)*a*cos(d*x + c) - (349440*I - 349440)*a*sin(d*x + c))*sin(2*d*x + 2*c)^4 + ((1397760*I + 1397760)*a*c
os(d*x + c) + (1397760*I - 1397760)*a*sin(d*x + c))*cos(2*d*x + 2*c)^3 + (-(2096640*I + 2096640)*a*cos(d*x + c
) - (2096640*I - 2096640)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((-(698880*I + 698880)*a*cos(d*x + c) - (698880
*I - 698880)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((1397760*I + 1397760)*a*cos(d*x + c) + (1397760*I - 1397760
)*a*sin(d*x + c))*cos(2*d*x + 2*c) - (698880*I + 698880)*a*cos(d*x + c) - (698880*I - 698880)*a*sin(d*x + c))*
sin(2*d*x + 2*c)^2 + ((1397760*I + 1397760)*a*cos(d*x + c) + (1397760*I - 1397760)*a*sin(d*x + c))*cos(2*d*x +
2*c) - (349440*I + 349440)*a*cos(d*x + c) - (349440*I - 349440)*a*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2
*c), -cos(2*d*x + 2*c) + 1)) + (((524580*I - 524580)*a*cos(d*x + c) - (524580*I + 524580)*a*sin(d*x + c))*cos(
2*d*x + 2*c)^2 + ((524580*I - 524580)*a*cos(d*x + c) - (524580*I + 524580)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2
+ ((352800*I - 352800)*a*cos(2*d*x + 2*c)^2 + (352800*I - 352800)*a*sin(2*d*x + 2*c)^2 - (705600*I - 705600)*a
*cos(2*d*x + 2*c) + (352800*I - 352800)*a)*cos(5*d*x + 5*c) + (-(823200*I - 823200)*a*cos(2*d*x + 2*c)^2 - (82
3200*I - 823200)*a*sin(2*d*x + 2*c)^2 + (1646400*I - 1646400)*a*cos(2*d*x + 2*c) - (823200*I - 823200)*a)*cos(
3*d*x + 3*c) + (-(1049160*I - 1049160)*a*cos(d*x + c) + (1049160*I + 1049160)*a*sin(d*x + c))*cos(2*d*x + 2*c)
+ (524580*I - 524580)*a*cos(d*x + c) + (-(352800*I + 352800)*a*cos(2*d*x + 2*c)^2 - (352800*I + 352800)*a*sin
(2*d*x + 2*c)^2 + (705600*I + 705600)*a*cos(2*d*x + 2*c) - (352800*I + 352800)*a)*sin(5*d*x + 5*c) + ((823200*
I + 823200)*a*cos(2*d*x + 2*c)^2 + (823200*I + 823200)*a*sin(2*d*x + 2*c)^2 - (1646400*I + 1646400)*a*cos(2*d*
x + 2*c) + (823200*I + 823200)*a)*sin(3*d*x + 3*c) - (524580*I + 524580)*a*sin(d*x + c))*sin(5/2*arctan2(sin(2
*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + ((-(349440*I - 349440)*a*cos(d*x + c) + (349440*I + 349440)*a*sin(d*x +
c))*cos(2*d*x + 2*c)^4 + (-(349440*I - 349440)*a*cos(d*x + c) + (349440*I + 349440)*a*sin(d*x + c))*sin(2*d*x
+ 2*c)^4 + ((1397760*I - 1397760)*a*cos(d*x + c) - (1397760*I + 1397760)*a*sin(d*x + c))*cos(2*d*x + 2*c)^3 +
(-(2096640*I - 2096640)*a*cos(d*x + c) + (2096640*I + 2096640)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((-(69888
0*I - 698880)*a*cos(d*x + c) + (698880*I + 698880)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((1397760*I - 1397760)
*a*cos(d*x + c) - (1397760*I + 1397760)*a*sin(d*x + c))*cos(2*d*x + 2*c) - (698880*I - 698880)*a*cos(d*x + c)
+ (698880*I + 698880)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + ((1397760*I - 1397760)*a*cos(d*x + c) - (1397760*I
+ 1397760)*a*sin(d*x + c))*cos(2*d*x + 2*c) - (349440*I - 349440)*a*cos(d*x + c) + (349440*I + 349440)*a*sin(d
*x + 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)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{9/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(9/2),x)
[Out]
int((a + a*tan(c + d*x)*1i)^(3/2)/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))**(3/2)/tan(d*x+c)**(9/2),x)
[Out]
Timed out
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