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

Optimal. Leaf size=78 \[ \frac {2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (B+i A)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[Out]

2*(-1)^(1/4)*a*(A-I*B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d-2*a*(I*A+B)/d/tan(d*x+c)^(1/2)-2/3*a*A/d/tan(d*x+
c)^(3/2)

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Rubi [A]  time = 0.13, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3591, 3529, 3533, 205} \[ \frac {2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (B+i A)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-1)^(1/4)*a*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (2*a*A)/(3*d*Tan[c + d*x]^(3/2)) - (2*a*(
I*A + B))/(d*Sqrt[Tan[c + d*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 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3591

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2
+ b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx &=-\frac {2 a A}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\int \frac {a (i A+B)-a (A-i B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a (i A+B)}{d \sqrt {\tan (c+d x)}}+\int \frac {-a (A-i B)-a (i A+B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a A}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a (i A+B)}{d \sqrt {\tan (c+d x)}}+\frac {\left (2 a^2 (A-i B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a (A-i B)+a (i A+B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a A}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a (i A+B)}{d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 2.06, size = 94, normalized size = 1.21 \[ -\frac {2 a \left (-3 i (A-i B) \sqrt {i \tan (c+d x)} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+A \cot (c+d x)+3 i A+3 B\right )}{3 d \sqrt {\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*((3*I)*A + 3*B + A*Cot[c + d*x] - (3*I)*(A - I*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*
(c + d*x)))]]*Sqrt[I*Tan[c + d*x]]))/(3*d*Sqrt[Tan[c + d*x]])

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fricas [B]  time = 0.74, size = 423, normalized size = 5.42 \[ -\frac {3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 8 \, {\left ({\left (4 \, A - 3 i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, A a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (2 \, A - 3 i \, B\right )} a\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.10, size = 474, normalized size = 6.08 \[ -\frac {2 a A}{3 d \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 i a A}{d \sqrt {\tan \left (d x +c \right )}}-\frac {2 a B}{d \sqrt {\tan \left (d x +c \right )}}+\frac {i a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {i a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {i a B \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}-\frac {a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a A \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}-\frac {i a A \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}-\frac {i a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {i a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a B \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}-\frac {a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a*A/d/tan(d*x+c)^(3/2)-2*I/d*a/tan(d*x+c)^(1/2)*A-2/d*a/tan(d*x+c)^(1/2)*B+1/2*I/d*a*B*2^(1/2)*arctan(1+2
^(1/2)*tan(d*x+c)^(1/2))+1/2*I/d*a*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))+1/4*I/d*a*B*2^(1/2)*ln((1+2^(
1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))-1/2/d*a*A*2^(1/2)*arctan(1+2^(1/2)*
tan(d*x+c)^(1/2))-1/2/d*a*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))-1/4/d*a*A*2^(1/2)*ln((1+2^(1/2)*tan(d*
x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))-1/4*I/d*a*A*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(
1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))-1/2*I/d*a*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2
))-1/2*I/d*a*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))-1/4/d*a*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+ta
n(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))-1/2/d*a*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-1/2/d*
a*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))

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maxima [B]  time = 1.13, size = 169, normalized size = 2.17 \[ \frac {3 \, {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a - \frac {8 \, {\left ({\left (3 i \, A + 3 \, B\right )} a \tan \left (d x + c\right ) + A a\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/12*(3*(2*sqrt(2)*(-(I + 1)*A + (I - 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(
-(I + 1)*A + (I - 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*((I - 1)*A + (I + 1)*B
)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*((I - 1)*A + (I + 1)*B)*log(-sqrt(2)*sqrt(tan(d
*x + c)) + tan(d*x + c) + 1))*a - 8*((3*I*A + 3*B)*a*tan(d*x + c) + A*a)/tan(d*x + c)^(3/2))/d

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mupad [B]  time = 7.53, size = 99, normalized size = 1.27 \[ \frac {2\,{\left (-1\right )}^{1/4}\,B\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}-\frac {2\,B\,a}{d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}-\frac {\frac {2\,A\,a}{3\,d}+\frac {A\,a\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {\sqrt {2}\,A\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )}{d} \]

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))/tan(c + d*x)^(5/2),x)

[Out]

(2*(-1)^(1/4)*B*a*atanh((-1)^(1/4)*tan(c + d*x)^(1/2)))/d - (2*B*a)/(d*tan(c + d*x)^(1/2)) - (2^(1/2)*A*a*atan
(2^(1/2)*tan(c + d*x)^(1/2)*(1/2 - 1i/2))*(1 + 1i))/d - ((2*A*a)/(3*d) + (A*a*tan(c + d*x)*2i)/d)/tan(c + d*x)
^(3/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \frac {A}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B}{\sqrt {\tan {\left (c + d x \right )}}}\, dx + \int \left (- \frac {i A}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\right )\, dx + \int \left (- \frac {i B}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)

[Out]

I*a*(Integral(A/tan(c + d*x)**(3/2), x) + Integral(B/sqrt(tan(c + d*x)), x) + Integral(-I*A/tan(c + d*x)**(5/2
), x) + Integral(-I*B/tan(c + d*x)**(3/2), x))

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