3.174 \(\int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=185 \[ -\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (5 B+8 i A) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (38 A-35 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[Out]

(-4-4*I)*a^(5/2)*(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+2/15*a^2*(38*A-35*
I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/15*a^2*(8*I*A+5*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3
/2)-2/5*a*A*(a+I*a*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(5/2)

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Rubi [A]  time = 0.57, antiderivative size = 185, 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} \[ -\frac {2 a^2 (5 B+8 i A) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (38 A-35 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

((-4 - 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2
*a^2*((8*I)*A + 5*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*Tan[c + d*x]^(3/2)) + (2*a^2*(38*A - (35*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(15*d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^(3/2))/(5*d*Tan[c + d*x]^(5/2))

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))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {1}{2} a (8 i A+5 B)-\frac {1}{2} a (2 A-5 i B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (8 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4}{15} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (38 A-35 i B)-\frac {1}{4} a^2 (22 i A+25 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (8 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (38 A-35 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 \int -\frac {15 a^3 (i A+B) \sqrt {a+i a \tan (c+d x)}}{2 \sqrt {\tan (c+d x)}} \, dx}{15 a}\\ &=-\frac {2 a^2 (8 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (38 A-35 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\left (4 a^2 (i A+B)\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 (8 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (38 A-35 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {\left (8 a^4 (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 {(4+4 i) a^{5/2} (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^2 (8 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (38 A-35 i B) \sqrt {a+i a \tan (c+d x)}}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 11.87, size = 323, normalized size = 1.75 \[ -\frac {4 \sqrt {2} e^{-2 i c} \sqrt {e^{i d x}} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} (a+i a \tan (c+d x))^{5/2} \left (e^{i (c+d x)} \sqrt {1-e^{2 i (c+d x)}} \left (i A \left (-35 e^{2 i (c+d x)}+26 e^{4 i (c+d x)}+15\right )+5 B \left (-7 e^{2 i (c+d x)}+4 e^{4 i (c+d x)}+3\right )\right )+15 (B+i A) \left (-1+e^{2 i (c+d x)}\right )^3 \sin ^{-1}\left (e^{i (c+d x)}\right )\right ) (A+B \tan (c+d x))}{15 d \left (1-e^{2 i (c+d x)}\right )^{7/2} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac {7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(-4*Sqrt[2]*Sqrt[E^(I*d*x)]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(E^(I*(c + d*x))
*Sqrt[1 - E^((2*I)*(c + d*x))]*(5*B*(3 - 7*E^((2*I)*(c + d*x)) + 4*E^((4*I)*(c + d*x))) + I*A*(15 - 35*E^((2*I
)*(c + d*x)) + 26*E^((4*I)*(c + d*x)))) + 15*(I*A + B)*(-1 + E^((2*I)*(c + d*x)))^3*ArcSin[E^(I*(c + d*x))])*(
a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(15*d*E^((2*I)*c)*(1 - E^((2*I)*(c + d*x)))^(7/2)*Sqrt[E^(I*
(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sec[c + d*x]^(7/2)*(Cos[d*x] + I*Sin[d*x])^(5/2)*(A*Cos[c + d*x] + B*Sin
[c + d*x]))

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fricas [B]  time = 1.05, size = 588, normalized size = 3.18 \[ \frac {\sqrt {2} {\left ({\left (208 i \, A + 160 \, B\right )} a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + {\left (-72 i \, A - 120 \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + {\left (-160 i \, A - 160 \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (120 i \, A + 120 \, B\right )} 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}} - 15 \, \sqrt {\frac {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, B^{2}\right )} a^{5}}{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 (4 i \, A + 4 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (4 i \, A + 4 \, B\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 {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{2}}\right ) + 15 \, \sqrt {\frac {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, B^{2}\right )} a^{5}}{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 (4 i \, A + 4 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (4 i \, A + 4 \, B\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 {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{2}}\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))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/30*(sqrt(2)*((208*I*A + 160*B)*a^2*e^(7*I*d*x + 7*I*c) + (-72*I*A - 120*B)*a^2*e^(5*I*d*x + 5*I*c) + (-160*I
*A - 160*B)*a^2*e^(3*I*d*x + 3*I*c) + (120*I*A + 120*B)*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)) - 15*sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/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)*((4*I*A + 4*B)*
a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*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^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*
c)/((4*I*A + 4*B)*a^2)) + 15*sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/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)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B
)*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^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((4*I*A + 4*B)*a^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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(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)^(5/2)/tan(d*x + c)^(7/2), x)

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maple [B]  time = 0.33, size = 709, normalized size = 3.83 \[ -\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-76 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 +70 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 +22 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 +10 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 )+6 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 )}{15 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))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x)

[Out]

-1/15/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(5/2)*(-76*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+70*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+22*I*A*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)-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+10*B*(a*t
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+6*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))^(5/2)*(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 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]

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)^(5/2))/tan(c + d*x)^(7/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/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))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(7/2),x)

[Out]

Timed out

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