3.814 \(\int \frac{(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=208 \[ -\frac{2 (-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{3465 c^3 f (c-i c \tan (e+f x))^{5/2}}-\frac{2 (-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}-\frac{(-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}} \]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(11*f*(c - I*c*Tan[e + f*x])^(11/2)) - (((3*I)*A - 8*B)*(a + I*a*Tan
[e + f*x])^(5/2))/(99*c*f*(c - I*c*Tan[e + f*x])^(9/2)) - (2*((3*I)*A - 8*B)*(a + I*a*Tan[e + f*x])^(5/2))/(69
3*c^2*f*(c - I*c*Tan[e + f*x])^(7/2)) - (2*((3*I)*A - 8*B)*(a + I*a*Tan[e + f*x])^(5/2))/(3465*c^3*f*(c - I*c*
Tan[e + f*x])^(5/2))

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Rubi [A]  time = 0.291019, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ -\frac{2 (-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{3465 c^3 f (c-i c \tan (e+f x))^{5/2}}-\frac{2 (-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}-\frac{(-8 B+3 i A) (a+i a \tan (e+f x))^{5/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(11/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(11*f*(c - I*c*Tan[e + f*x])^(11/2)) - (((3*I)*A - 8*B)*(a + I*a*Tan
[e + f*x])^(5/2))/(99*c*f*(c - I*c*Tan[e + f*x])^(9/2)) - (2*((3*I)*A - 8*B)*(a + I*a*Tan[e + f*x])^(5/2))/(69
3*c^2*f*(c - I*c*Tan[e + f*x])^(7/2)) - (2*((3*I)*A - 8*B)*(a + I*a*Tan[e + f*x])^(5/2))/(3465*c^3*f*(c - I*c*
Tan[e + f*x])^(5/2))

Rule 3588

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{13/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}}+\frac{(a (3 A+8 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{11 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac{(3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}+\frac{(2 a (3 A+8 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{99 c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac{(3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac{2 (3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}+\frac{(2 a (3 A+8 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{693 c^2 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac{(3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac{2 (3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}-\frac{2 (3 i A-8 B) (a+i a \tan (e+f x))^{5/2}}{3465 c^3 f (c-i c \tan (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 13.4844, size = 156, normalized size = 0.75 \[ \frac{a^2 \cos (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)} (\cos (8 e+10 f x)+i \sin (8 e+10 f x)) (-(3 A+8 i B) (55 \sin (e+f x)+63 \sin (3 (e+f x)))+55 (B-24 i A) \cos (e+f x)+63 (3 B-8 i A) \cos (3 (e+f x)))}{13860 c^6 f (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(11/2),x]

[Out]

(a^2*Cos[e + f*x]*(55*((-24*I)*A + B)*Cos[e + f*x] + 63*((-8*I)*A + 3*B)*Cos[3*(e + f*x)] - (3*A + (8*I)*B)*(5
5*Sin[e + f*x] + 63*Sin[3*(e + f*x)]))*(Cos[8*e + 10*f*x] + I*Sin[8*e + 10*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sq
rt[c - I*c*Tan[e + f*x]])/(13860*c^6*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]  time = 0.158, size = 161, normalized size = 0.8 \begin{align*}{\frac{-{\frac{i}{3465}}{a}^{2} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 6\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{4}-112\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}-16\,B \left ( \tan \left ( fx+e \right ) \right ) ^{4}-135\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}-42\,A \left ( \tan \left ( fx+e \right ) \right ) ^{3}-427\,iB\tan \left ( fx+e \right ) +360\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}-456\,iA+273\,A\tan \left ( fx+e \right ) +61\,B \right ) }{f{c}^{6} \left ( \tan \left ( fx+e \right ) +i \right ) ^{7}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x)

[Out]

-1/3465*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2/c^6*(1+tan(f*x+e)^2)*(6*I*A*tan(f*x+e)
^4-112*I*B*tan(f*x+e)^3-16*B*tan(f*x+e)^4-135*I*A*tan(f*x+e)^2-42*A*tan(f*x+e)^3-427*I*B*tan(f*x+e)+360*B*tan(
f*x+e)^2-456*I*A+273*A*tan(f*x+e)+61*B)/(tan(f*x+e)+I)^7

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Maxima [A]  time = 2.46048, size = 373, normalized size = 1.79 \begin{align*} \frac{{\left (315 \,{\left (-i \, A - B\right )} a^{2} \cos \left (\frac{11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 385 \,{\left (-3 i \, A - B\right )} a^{2} \cos \left (\frac{9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 495 \,{\left (-3 i \, A + B\right )} a^{2} \cos \left (\frac{7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 693 \,{\left (-i \, A + B\right )} a^{2} \cos \left (\frac{5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (315 \, A - 315 i \, B\right )} a^{2} \sin \left (\frac{11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (1155 \, A - 385 i \, B\right )} a^{2} \sin \left (\frac{9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (1485 \, A + 495 i \, B\right )} a^{2} \sin \left (\frac{7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (693 \, A + 693 i \, B\right )} a^{2} \sin \left (\frac{5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt{a}}{27720 \, c^{\frac{11}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

1/27720*(315*(-I*A - B)*a^2*cos(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 385*(-3*I*A - B)*a^2*cos(9
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 495*(-3*I*A + B)*a^2*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2
*f*x + 2*e))) + 693*(-I*A + B)*a^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (315*A - 315*I*B)*a^
2*sin(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (1155*A - 385*I*B)*a^2*sin(9/2*arctan2(sin(2*f*x + 2
*e), cos(2*f*x + 2*e))) + (1485*A + 495*I*B)*a^2*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (693*A
+ 693*I*B)*a^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(11/2)*f)

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Fricas [A]  time = 1.41673, size = 451, normalized size = 2.17 \begin{align*} \frac{{\left ({\left (-315 i \, A - 315 \, B\right )} a^{2} e^{\left (12 i \, f x + 12 i \, e\right )} +{\left (-1470 i \, A - 700 \, B\right )} a^{2} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-2640 i \, A + 110 \, B\right )} a^{2} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-2178 i \, A + 1188 \, B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-693 i \, A + 693 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )}}{27720 \, c^{6} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

1/27720*((-315*I*A - 315*B)*a^2*e^(12*I*f*x + 12*I*e) + (-1470*I*A - 700*B)*a^2*e^(10*I*f*x + 10*I*e) + (-2640
*I*A + 110*B)*a^2*e^(8*I*f*x + 8*I*e) + (-2178*I*A + 1188*B)*a^2*e^(6*I*f*x + 6*I*e) + (-693*I*A + 693*B)*a^2*
e^(4*I*f*x + 4*I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e)/(c^6*
f)

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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(f*x+e))**(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(5/2)/(-I*c*tan(f*x + e) + c)^(11/2), x)