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

Optimal. Leaf size=127 \[ \frac{4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (5 B+7 i A)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (A-i B)}{d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

[Out]

(4*(-1)^(1/4)*a^2*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (2*a^2*((7*I)*A + 5*B))/(15*d*Tan[c + d
*x]^(3/2)) + (4*a^2*(A - I*B))/(d*Sqrt[Tan[c + d*x]]) - (2*A*(a^2 + I*a^2*Tan[c + d*x]))/(5*d*Tan[c + d*x]^(5/
2))

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Rubi [A]  time = 0.257492, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {3593, 3591, 3529, 3533, 205} \[ \frac{4 \sqrt [4]{-1} a^2 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (5 B+7 i A)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (A-i B)}{d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1)^(1/4)*a^2*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (2*a^2*((7*I)*A + 5*B))/(15*d*Tan[c + d
*x]^(3/2)) + (4*a^2*(A - I*B))/(d*Sqrt[Tan[c + d*x]]) - (2*A*(a^2 + I*a^2*Tan[c + d*x]))/(5*d*Tan[c + d*x]^(5/
2))

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 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]

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 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 \frac{(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+i a \tan (c+d x)) \left (\frac{1}{2} a (7 i A+5 B)-\frac{1}{2} a (3 A-5 i B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (7 i A+5 B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{-5 a^2 (A-i B)-5 a^2 (i A+B) \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (7 i A+5 B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (A-i B)}{d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{-5 a^2 (i A+B)+5 a^2 (A-i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 (7 i A+5 B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (A-i B)}{d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{\left (20 a^4 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-5 a^2 (i A+B)-5 a^2 (A-i B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{4 \sqrt [4]{-1} a^2 (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 a^2 (7 i A+5 B)}{15 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (A-i B)}{d \sqrt{\tan (c+d x)}}-\frac{2 A \left (a^2+i a^2 \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [B]  time = 4.94616, size = 272, normalized size = 2.14 \[ \frac{\cos ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (-\frac{4 i e^{-2 i c} (A-i B) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac{(\cos (2 c)-i \sin (2 c)) \csc ^2(c+d x) (5 (B+2 i A) \sin (2 (c+d x))+(33 A-30 i B) \cos (2 (c+d x))-27 A+30 i B)}{15 \sqrt{\tan (c+d x)}}\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^3*(((-4*I)*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[S
qrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])/(E^((2*I)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 +
E^((2*I)*(c + d*x)))]) - (Csc[c + d*x]^2*(Cos[2*c] - I*Sin[2*c])*(-27*A + (30*I)*B + (33*A - (30*I)*B)*Cos[2*(
c + d*x)] + 5*((2*I)*A + B)*Sin[2*(c + d*x)]))/(15*Sqrt[Tan[c + d*x]]))*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c
+ d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.018, size = 537, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.82405, size = 265, normalized size = 2.09 \begin{align*} -\frac{15 \,{\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^{2} - \frac{4 \,{\left ({\left (30 \, A - 30 i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + 5 \,{\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - 3 \, A a^{2}\right )}}{\tan \left (d x + c\right )^{\frac{5}{2}}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*(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^2 - 4*((30*A - 30*I*B)*a^2*tan(d*x + c)^2 + 5*(-2*I*A - B)*a^2*tan(d*x + c)
- 3*A*a^2)/tan(d*x + c)^(5/2))/d

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Fricas [B]  time = 1.92264, size = 1359, normalized size = 10.7 \begin{align*} -\frac{15 \, \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{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 (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 15 \, \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{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 (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) -{\left ({\left (344 i \, A + 280 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (-88 i \, A - 200 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-248 i \, A - 280 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (184 i \, A + 200 \, B\right )} a^{2}\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}}}{60 \,{\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((a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/60*(15*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/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((4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^
2)*(d*e^(2*I*d*x + 2*I*c) + d)*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)/((2*I*A + 2*B)*a^2)) - 15*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/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((4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32*A*
B - 16*I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*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)/((2*I*A + 2*B)*a^2)) - ((344*I*A + 280*B)*a^2*e^(6*I*d*x + 6*I*c) + (-88*I*A - 200*B)*
a^2*e^(4*I*d*x + 4*I*c) + (-248*I*A - 280*B)*a^2*e^(2*I*d*x + 2*I*c) + (184*I*A + 200*B)*a^2)*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((a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c))/tan(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.25522, size = 146, normalized size = 1.15 \begin{align*} \frac{\left (i + 1\right ) \, \sqrt{2}{\left (8 i \, A a^{2} + 8 \, B a^{2}\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{60 \, A a^{2} \tan \left (d x + c\right )^{2} - 60 i \, B a^{2} \tan \left (d x + c\right )^{2} - 20 i \, A a^{2} \tan \left (d x + c\right ) - 10 \, B a^{2} \tan \left (d x + c\right ) - 6 \, A a^{2}}{15 \, d \tan \left (d x + c\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/4*I + 1/4)*sqrt(2)*(8*I*A*a^2 + 8*B*a^2)*arctan((1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d + 1/15*(60*A*a^
2*tan(d*x + c)^2 - 60*I*B*a^2*tan(d*x + c)^2 - 20*I*A*a^2*tan(d*x + c) - 10*B*a^2*tan(d*x + c) - 6*A*a^2)/(d*t
an(d*x + c)^(5/2))