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

Optimal. Leaf size=130 \[ \frac{2 a (B+i A) \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 a (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a (B+i A) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{7}{2}}(c+d x)}{7 d} \]

[Out]

(-2*(-1)^(1/4)*a*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (2*a*(I*A + B)*Sqrt[Tan[c + d*x]])/d + (
2*a*(A - I*B)*Tan[c + d*x]^(3/2))/(3*d) + (2*a*(I*A + B)*Tan[c + d*x]^(5/2))/(5*d) + (((2*I)/7)*a*B*Tan[c + d*
x]^(7/2))/d

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Rubi [A]  time = 0.198178, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3592, 3528, 3533, 205} \[ \frac{2 a (B+i A) \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 a (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a (B+i A) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{7}{2}}(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-1)^(1/4)*a*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (2*a*(I*A + B)*Sqrt[Tan[c + d*x]])/d + (
2*a*(A - I*B)*Tan[c + d*x]^(3/2))/(3*d) + (2*a*(I*A + B)*Tan[c + d*x]^(5/2))/(5*d) + (((2*I)/7)*a*B*Tan[c + d*
x]^(7/2))/d

Rule 3592

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

Rule 3528

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

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 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]

Rubi steps

\begin{align*} \int \tan ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx &=\frac{2 i a B \tan ^{\frac{7}{2}}(c+d x)}{7 d}+\int \tan ^{\frac{5}{2}}(c+d x) (a (A-i B)+a (i A+B) \tan (c+d x)) \, dx\\ &=\frac{2 a (i A+B) \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 i a B \tan ^{\frac{7}{2}}(c+d x)}{7 d}+\int \tan ^{\frac{3}{2}}(c+d x) (-a (i A+B)+a (A-i B) \tan (c+d x)) \, dx\\ &=\frac{2 a (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 a (i A+B) \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 i a B \tan ^{\frac{7}{2}}(c+d x)}{7 d}+\int \sqrt{\tan (c+d x)} (-a (A-i B)-a (i A+B) \tan (c+d x)) \, dx\\ &=-\frac{2 a (i A+B) \sqrt{\tan (c+d x)}}{d}+\frac{2 a (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 a (i A+B) \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 i a B \tan ^{\frac{7}{2}}(c+d x)}{7 d}+\int \frac{a (i A+B)-a (A-i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a (i A+B) \sqrt{\tan (c+d x)}}{d}+\frac{2 a (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 a (i A+B) \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 i a B \tan ^{\frac{7}{2}}(c+d x)}{7 d}+\frac{\left (2 a^2 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a (i A+B)+a (A-i B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 \sqrt [4]{-1} a (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a (i A+B) \sqrt{\tan (c+d x)}}{d}+\frac{2 a (A-i B) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 a (i A+B) \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 i a B \tan ^{\frac{7}{2}}(c+d x)}{7 d}\\ \end{align*}

Mathematica [B]  time = 4.27937, size = 280, normalized size = 2.15 \[ \frac{\cos ^2(c+d x) (\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \left (\frac{2 e^{-i c} (B+i A) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac{1}{105} \cos (c) (\tan (c)+i) \sqrt{\tan (c+d x)} \sec ^2(c+d x) (5 (4 B+7 i A) \tan (c+d x)+\cos (2 (c+d x)) (5 (10 B+7 i A) \tan (c+d x)+126 (A-i B))+84 (A-i B))\right )}{d (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^2*(Cos[d*x] - I*Sin[d*x])*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x])*((2*(I*A + B)*Sqrt[((-I)*(
-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c
+ d*x)))]])/(E^(I*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]) - (Cos[c]*Sec[c + d*x]^2*(I
+ Tan[c])*Sqrt[Tan[c + d*x]]*(84*(A - I*B) + 5*((7*I)*A + 4*B)*Tan[c + d*x] + Cos[2*(c + d*x)]*(126*(A - I*B)
+ 5*((7*I)*A + 10*B)*Tan[c + d*x])))/105))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.015, size = 537, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I/d*a*A*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)))*2^(1/2)-2/3*I/
d*a*B*tan(d*x+c)^(3/2)+2/5/d*a*B*tan(d*x+c)^(5/2)+2/7*I*a*B*tan(d*x+c)^(7/2)/d+2/3/d*a*A*tan(d*x+c)^(3/2)+2/5*
I/d*a*A*tan(d*x+c)^(5/2)-2/d*a*B*tan(d*x+c)^(1/2)+1/2*I/d*a*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+1/4*
I/d*a*B*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)))*2^(1/2)+1/2*I/d*a*
B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+1/2/d*a*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+1/2/d*a*B*ar
ctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+1/4/d*a*B*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)))*2^(1/2)+1/2*I/d*a*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-2*I/d*a*A*tan(d*x+c)
^(1/2)+1/2*I/d*a*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/4/d*a*A*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)))*2^(1/2)-1/2/d*a*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/2
/d*a*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)

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Maxima [B]  time = 1.77148, size = 275, normalized size = 2.12 \begin{align*} -\frac{-120 i \, B a \tan \left (d x + c\right )^{\frac{7}{2}} + 168 \,{\left (-i \, A - B\right )} a \tan \left (d x + c\right )^{\frac{5}{2}} - 8 \,{\left (35 \, A - 35 i \, B\right )} a \tan \left (d x + c\right )^{\frac{3}{2}} + 840 \,{\left (i \, A + B\right )} a \sqrt{\tan \left (d x + c\right )} - 105 \,{\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}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*(-120*I*B*a*tan(d*x + c)^(7/2) + 168*(-I*A - B)*a*tan(d*x + c)^(5/2) - 8*(35*A - 35*I*B)*a*tan(d*x + c)
^(3/2) + 840*(I*A + B)*a*sqrt(tan(d*x + c)) - 105*(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)/d

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Fricas [B]  time = 2.31532, size = 1326, normalized size = 10.2 \begin{align*} \frac{105 \,{\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 )} \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 (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}}} \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 ) - 105 \,{\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 )} \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 (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}}} \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 ) +{\left ({\left (-1288 i \, A - 1408 \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (-2632 i \, A - 2272 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-2072 i \, A - 2432 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-728 i \, A - 608 \, 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}}}{420 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(105*(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)*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) + (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)*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)) - 105*(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)*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) - (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)*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)) + ((-1288*I*A - 1408*B)*a*e^(6*I*d*x + 6*I*c) + (-2632*I*A - 2272*B)*a*e^(4*I*d*x + 4*I*c)
+ (-2072*I*A - 2432*B)*a*e^(2*I*d*x + 2*I*c) + (-728*I*A - 608*B)*a)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I
*d*x + 2*I*c) + 1)))/(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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.26835, size = 193, normalized size = 1.48 \begin{align*} \frac{\left (i - 1\right ) \, \sqrt{2}{\left (4 \, A a - 4 i \, B a\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{4 \, d} - \frac{-30 i \, B a d^{6} \tan \left (d x + c\right )^{\frac{7}{2}} - 42 i \, A a d^{6} \tan \left (d x + c\right )^{\frac{5}{2}} - 42 \, B a d^{6} \tan \left (d x + c\right )^{\frac{5}{2}} - 70 \, A a d^{6} \tan \left (d x + c\right )^{\frac{3}{2}} + 70 i \, B a d^{6} \tan \left (d x + c\right )^{\frac{3}{2}} + 210 i \, A a d^{6} \sqrt{\tan \left (d x + c\right )} + 210 \, B a d^{6} \sqrt{\tan \left (d x + c\right )}}{105 \, d^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/4*I - 1/4)*sqrt(2)*(4*A*a - 4*I*B*a)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 1/105*(-30*I*B*a
*d^6*tan(d*x + c)^(7/2) - 42*I*A*a*d^6*tan(d*x + c)^(5/2) - 42*B*a*d^6*tan(d*x + c)^(5/2) - 70*A*a*d^6*tan(d*x
+ c)^(3/2) + 70*I*B*a*d^6*tan(d*x + c)^(3/2) + 210*I*A*a*d^6*sqrt(tan(d*x + c)) + 210*B*a*d^6*sqrt(tan(d*x +
c)))/d^7