3.827 \(\int \frac{(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx\)

Optimal. Leaf size=208 \[ -\frac{2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{9009 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac{2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac{(-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}} \]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(13*f*(c - I*c*Tan[e + f*x])^(13/2)) - (((3*I)*A - 10*B)*(a + I*a*Ta
n[e + f*x])^(7/2))/(143*c*f*(c - I*c*Tan[e + f*x])^(11/2)) - (2*((3*I)*A - 10*B)*(a + I*a*Tan[e + f*x])^(7/2))
/(1287*c^2*f*(c - I*c*Tan[e + f*x])^(9/2)) - (2*((3*I)*A - 10*B)*(a + I*a*Tan[e + f*x])^(7/2))/(9009*c^3*f*(c
- I*c*Tan[e + f*x])^(7/2))

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Rubi [A]  time = 0.285863, 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} \[ -\frac{2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{9009 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac{2 (-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac{(-10 B+3 i A) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(13/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(13*f*(c - I*c*Tan[e + f*x])^(13/2)) - (((3*I)*A - 10*B)*(a + I*a*Ta
n[e + f*x])^(7/2))/(143*c*f*(c - I*c*Tan[e + f*x])^(11/2)) - (2*((3*I)*A - 10*B)*(a + I*a*Tan[e + f*x])^(7/2))
/(1287*c^2*f*(c - I*c*Tan[e + f*x])^(9/2)) - (2*((3*I)*A - 10*B)*(a + I*a*Tan[e + f*x])^(7/2))/(9009*c^3*f*(c
- I*c*Tan[e + f*x])^(7/2))

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))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{15/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}+\frac{(a (3 A+10 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{13/2}} \, dx,x,\tan (e+f x)\right )}{13 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac{(3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}+\frac{(2 a (3 A+10 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{143 c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac{(3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{2 (3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}+\frac{(2 a (3 A+10 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{1287 c^2 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac{(3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{2 (3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{1287 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac{2 (3 i A-10 B) (a+i a \tan (e+f x))^{7/2}}{9009 c^3 f (c-i c \tan (e+f x))^{7/2}}\\ \end{align*}

Mathematica [B]  time = 17.0674, size = 495, normalized size = 2.38 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((B-i A) \cos (6 f x) \left (\frac{\cos (3 e)}{112 c^7}+\frac{i \sin (3 e)}{112 c^7}\right )+(A+i B) \sin (6 f x) \left (\frac{\cos (3 e)}{112 c^7}+\frac{i \sin (3 e)}{112 c^7}\right )+(8 B-15 i A) \cos (8 f x) \left (\frac{\cos (5 e)}{504 c^7}+\frac{i \sin (5 e)}{504 c^7}\right )+(B-30 i A) \cos (10 f x) \left (\frac{\cos (7 e)}{792 c^7}+\frac{i \sin (7 e)}{792 c^7}\right )+(25 A-12 i B) \cos (12 f x) \left (\frac{\sin (9 e)}{1144 c^7}-\frac{i \cos (9 e)}{1144 c^7}\right )+(A-i B) \cos (14 f x) \left (\frac{\sin (11 e)}{208 c^7}-\frac{i \cos (11 e)}{208 c^7}\right )+(15 A+8 i B) \sin (8 f x) \left (\frac{\cos (5 e)}{504 c^7}+\frac{i \sin (5 e)}{504 c^7}\right )+(30 A+i B) \sin (10 f x) \left (\frac{\cos (7 e)}{792 c^7}+\frac{i \sin (7 e)}{792 c^7}\right )+(25 A-12 i B) \sin (12 f x) \left (\frac{\cos (9 e)}{1144 c^7}+\frac{i \sin (9 e)}{1144 c^7}\right )+(A-i B) \sin (14 f x) \left (\frac{\cos (11 e)}{208 c^7}+\frac{i \sin (11 e)}{208 c^7}\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(13/2),x]

[Out]

(Cos[e + f*x]^4*(((-I)*A + B)*Cos[6*f*x]*(Cos[3*e]/(112*c^7) + ((I/112)*Sin[3*e])/c^7) + ((-15*I)*A + 8*B)*Cos
[8*f*x]*(Cos[5*e]/(504*c^7) + ((I/504)*Sin[5*e])/c^7) + ((-30*I)*A + B)*Cos[10*f*x]*(Cos[7*e]/(792*c^7) + ((I/
792)*Sin[7*e])/c^7) + (25*A - (12*I)*B)*Cos[12*f*x]*(((-I/1144)*Cos[9*e])/c^7 + Sin[9*e]/(1144*c^7)) + (A - I*
B)*Cos[14*f*x]*(((-I/208)*Cos[11*e])/c^7 + Sin[11*e]/(208*c^7)) + (A + I*B)*(Cos[3*e]/(112*c^7) + ((I/112)*Sin
[3*e])/c^7)*Sin[6*f*x] + (15*A + (8*I)*B)*(Cos[5*e]/(504*c^7) + ((I/504)*Sin[5*e])/c^7)*Sin[8*f*x] + (30*A + I
*B)*(Cos[7*e]/(792*c^7) + ((I/792)*Sin[7*e])/c^7)*Sin[10*f*x] + (25*A - (12*I)*B)*(Cos[9*e]/(1144*c^7) + ((I/1
144)*Sin[9*e])/c^7)*Sin[12*f*x] + (A - I*B)*(Cos[11*e]/(208*c^7) + ((I/208)*Sin[11*e])/c^7)*Sin[14*f*x])*Sqrt[
Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(f*(Cos[f
*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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Maple [A]  time = 0.114, size = 184, normalized size = 0.9 \begin{align*}{\frac{{\frac{i}{9009}}{a}^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 6\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{5}-160\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{4}-20\,B \left ( \tan \left ( fx+e \right ) \right ) ^{5}-177\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{3}-48\,A \left ( \tan \left ( fx+e \right ) \right ) ^{4}-1643\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{2}+590\,B \left ( \tan \left ( fx+e \right ) \right ) ^{3}-1569\,iA\tan \left ( fx+e \right ) +408\,A \left ( \tan \left ( fx+e \right ) \right ) ^{2}-97\,iB-776\,B\tan \left ( fx+e \right ) -930\,A \right ) }{f{c}^{7} \left ( \tan \left ( fx+e \right ) +i \right ) ^{8}}\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))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x)

[Out]

1/9009*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^7*(1+tan(f*x+e)^2)*(6*I*A*tan(f*x+e)^
5-160*I*B*tan(f*x+e)^4-20*B*tan(f*x+e)^5-177*I*A*tan(f*x+e)^3-48*A*tan(f*x+e)^4-1643*I*B*tan(f*x+e)^2+590*B*ta
n(f*x+e)^3-1569*I*A*tan(f*x+e)+408*A*tan(f*x+e)^2-97*I*B-776*B*tan(f*x+e)-930*A)/(tan(f*x+e)+I)^8

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Maxima [A]  time = 3.11868, size = 373, normalized size = 1.79 \begin{align*} \frac{{\left (693 \,{\left (-i \, A - B\right )} a^{3} \cos \left (\frac{13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 819 \,{\left (-3 i \, A - B\right )} a^{3} \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 ) + 1001 \,{\left (-3 i \, A + B\right )} a^{3} \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 ) + 1287 \,{\left (-i \, A + B\right )} a^{3} \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 ) +{\left (693 \, A - 693 i \, B\right )} a^{3} \sin \left (\frac{13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (2457 \, A - 819 i \, B\right )} a^{3} \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 (3003 \, A + 1001 i \, B\right )} a^{3} \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 (1287 \, A + 1287 i \, B\right )} a^{3} \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 )\right )} \sqrt{a}}{72072 \, c^{\frac{13}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x, algorithm="maxima")

[Out]

1/72072*(693*(-I*A - B)*a^3*cos(13/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 819*(-3*I*A - B)*a^3*cos(1
1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1001*(-3*I*A + B)*a^3*cos(9/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))) + 1287*(-I*A + B)*a^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (693*A - 693*I*B)
*a^3*sin(13/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (2457*A - 819*I*B)*a^3*sin(11/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))) + (3003*A + 1001*I*B)*a^3*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (
1287*A + 1287*I*B)*a^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(13/2)*f)

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Fricas [A]  time = 1.30005, size = 458, normalized size = 2.2 \begin{align*} \frac{{\left ({\left (-693 i \, A - 693 \, B\right )} a^{3} e^{\left (14 i \, f x + 14 i \, e\right )} +{\left (-3150 i \, A - 1512 \, B\right )} a^{3} e^{\left (12 i \, f x + 12 i \, e\right )} +{\left (-5460 i \, A + 182 \, B\right )} a^{3} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-4290 i \, A + 2288 \, B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-1287 i \, A + 1287 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 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 )}}{72072 \, c^{7} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x, algorithm="fricas")

[Out]

1/72072*((-693*I*A - 693*B)*a^3*e^(14*I*f*x + 14*I*e) + (-3150*I*A - 1512*B)*a^3*e^(12*I*f*x + 12*I*e) + (-546
0*I*A + 182*B)*a^3*e^(10*I*f*x + 10*I*e) + (-4290*I*A + 2288*B)*a^3*e^(8*I*f*x + 8*I*e) + (-1287*I*A + 1287*B)
*a^3*e^(6*I*f*x + 6*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^7*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))**(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(13/2),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{7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(13/2), x)