3.166 \(\int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=181 \[ -\frac {(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a (5 B+6 i A) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[Out]

(-2-2*I)*a^(3/2)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+4/15*a*(9*A-10*I*B
)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/5*a*A*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)-2/15*a*(6*I*
A+5*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)

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Rubi [A]  time = 0.55, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3593, 3598, 12, 3544, 205} \[ -\frac {(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a (5 B+6 i A) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

((-2 - 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2
*a*A*Sqrt[a + I*a*Tan[c + d*x]])/(5*d*Tan[c + d*x]^(5/2)) - (2*a*((6*I)*A + 5*B)*Sqrt[a + I*a*Tan[c + d*x]])/(
15*d*Tan[c + d*x]^(3/2)) + (4*a*(9*A - (10*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*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 3593

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^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(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 - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

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} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (6 i A+5 B)-\frac {1}{2} a (4 A-5 i B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (6 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{2} a^2 (9 A-10 i B)-\frac {1}{2} a^2 (6 i A+5 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{15 a}\\ &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (6 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}+\frac {8 \int -\frac {15 a^3 (i A+B) \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {\tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (6 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-(2 a (i A+B)) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (6 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {\left (4 a^3 (A-i B)\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} (A-i B) \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 A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (6 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 9.85, size = 237, normalized size = 1.31 \[ \frac {a (A-i B) e^{-3 i (c+d x)} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right ) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {-1+e^{2 i (c+d x)}}}-\frac {a \csc ^2(c+d x) \sqrt {a+i a \tan (c+d x)} ((5 B+6 i A) \sin (2 (c+d x))+(21 A-20 i B) \cos (2 (c+d x))-15 A+20 i B)}{15 d \sqrt {\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

-1/15*(a*Csc[c + d*x]^2*(-15*A + (20*I)*B + (21*A - (20*I)*B)*Cos[2*(c + d*x)] + ((6*I)*A + 5*B)*Sin[2*(c + d*
x)])*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) + (a*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/
(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))^2*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]
*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^((3*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))])

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fricas [B]  time = 0.95, size = 572, normalized size = 3.16 \[ \frac {\sqrt {2} {\left ({\left (108 i \, A + 100 \, B\right )} a e^{\left (7 i \, d x + 7 i \, c\right )} + {\left (-12 i \, A - 60 \, B\right )} a e^{\left (5 i \, d x + 5 i \, c\right )} + {\left (-60 i \, A - 100 \, B\right )} a e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (60 i \, A + 60 \, B\right )} 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}} - 15 \, \sqrt {\frac {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (2 i \, A + 2 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (2 i \, A + 2 \, B\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 {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) + 15 \, \sqrt {\frac {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (2 i \, A + 2 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (2 i \, A + 2 \, B\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 {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right )}{30 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/30*(sqrt(2)*((108*I*A + 100*B)*a*e^(7*I*d*x + 7*I*c) + (-12*I*A - 60*B)*a*e^(5*I*d*x + 5*I*c) + (-60*I*A - 1
00*B)*a*e^(3*I*d*x + 3*I*c) + (60*I*A + 60*B)*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)) - 15*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*(d*e^(6*I*d*
x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x +
2*I*c) + (2*I*A + 2*B)*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^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)
) + 15*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*
I*d*x + 2*I*c) - d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (2*I*A + 2*B)*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^2 + 16*A*B - 8*I*B^2)*
a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)))/(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I
*c) + 3*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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(3/2)/tan(d*x + c)^(7/2), x)

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maple [B]  time = 0.34, size = 707, normalized size = 3.91 \[ -\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-72 A \left (\tan ^{2}\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}+60 i A \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 ^{3}\left (d x +c \right )\right ) a -15 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 ^{3}\left (d x +c \right )\right ) a +80 i B \left (\tan ^{2}\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}+60 B \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 ^{3}\left (d x +c \right )\right ) a +30 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 ^{3}\left (d x +c \right )\right ) a -15 \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 ^{3}\left (d x +c \right )\right ) a +24 i A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-30 \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 ^{3}\left (d x +c \right )\right ) a +20 B \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 )+12 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{30 d \tan \left (d x +c \right )^{\frac {5}{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)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x)

[Out]

-1/30/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(5/2)*(-72*A*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2
)*(I*a)^(1/2)*(-I*a)^(1/2)+60*I*A*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)*tan(d*x+c)^3*a-15*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)^3*a+80*I*B*tan(d*x+c)^2*(a*tan(d*x+c
)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+60*B*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)*tan(d*x+c)^3*a+30*I*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)*tan(d*x+c)^3*a-15*(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)^
3*a+24*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-30*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)*tan(d*x+c)^3*a+20*B*(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)+12*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c
)*(1+I*tan(d*x+c)))^(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 [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)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2))/tan(c + d*x)^(7/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2))/tan(c + d*x)^(7/2), 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)*(A+B*tan(d*x+c))/tan(d*x+c)**(7/2),x)

[Out]

Timed out

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