3.826 \(\int \frac{(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=155 \[ -\frac{(-9 B+2 i A) (a+i a \tan (e+f x))^{7/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}-\frac{(-9 B+2 i A) (a+i a \tan (e+f x))^{7/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}} \]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(11*f*(c - I*c*Tan[e + f*x])^(11/2)) - (((2*I)*A - 9*B)*(a + I*a*Tan
[e + f*x])^(7/2))/(99*c*f*(c - I*c*Tan[e + f*x])^(9/2)) - (((2*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(7/2))/(693*
c^2*f*(c - I*c*Tan[e + f*x])^(7/2))

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Rubi [A]  time = 0.263789, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ -\frac{(-9 B+2 i A) (a+i a \tan (e+f x))^{7/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}-\frac{(-9 B+2 i A) (a+i a \tan (e+f x))^{7/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(7/2)*(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])^(7/2))/(11*f*(c - I*c*Tan[e + f*x])^(11/2)) - (((2*I)*A - 9*B)*(a + I*a*Tan
[e + f*x])^(7/2))/(99*c*f*(c - I*c*Tan[e + f*x])^(9/2)) - (((2*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(7/2))/(693*
c^2*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))^{11/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2} (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))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}}+\frac{(a (2 A+9 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 )}{11 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac{(2 i A-9 B) (a+i a \tan (e+f x))^{7/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}+\frac{(a (2 A+9 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 )}{99 c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{11 f (c-i c \tan (e+f x))^{11/2}}-\frac{(2 i A-9 B) (a+i a \tan (e+f x))^{7/2}}{99 c f (c-i c \tan (e+f x))^{9/2}}-\frac{(2 i A-9 B) (a+i a \tan (e+f x))^{7/2}}{693 c^2 f (c-i c \tan (e+f x))^{7/2}}\\ \end{align*}

Mathematica [B]  time = 15.7224, size = 417, normalized size = 2.69 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((B-i A) \cos (6 f x) \left (\frac{\cos (3 e)}{56 c^6}+\frac{i \sin (3 e)}{56 c^6}\right )+(A+i B) \sin (6 f x) \left (\frac{\cos (3 e)}{56 c^6}+\frac{i \sin (3 e)}{56 c^6}\right )+(9 B-23 i A) \cos (8 f x) \left (\frac{\cos (5 e)}{504 c^6}+\frac{i \sin (5 e)}{504 c^6}\right )+(31 A-9 i B) \cos (10 f x) \left (\frac{\sin (7 e)}{792 c^6}-\frac{i \cos (7 e)}{792 c^6}\right )+(A-i B) \cos (12 f x) \left (\frac{\sin (9 e)}{88 c^6}-\frac{i \cos (9 e)}{88 c^6}\right )+(23 A+9 i B) \sin (8 f x) \left (\frac{\cos (5 e)}{504 c^6}+\frac{i \sin (5 e)}{504 c^6}\right )+(31 A-9 i B) \sin (10 f x) \left (\frac{\cos (7 e)}{792 c^6}+\frac{i \sin (7 e)}{792 c^6}\right )+(A-i B) \sin (12 f x) \left (\frac{\cos (9 e)}{88 c^6}+\frac{i \sin (9 e)}{88 c^6}\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

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])^(11/2),x]

[Out]

(Cos[e + f*x]^4*(((-I)*A + B)*Cos[6*f*x]*(Cos[3*e]/(56*c^6) + ((I/56)*Sin[3*e])/c^6) + ((-23*I)*A + 9*B)*Cos[8
*f*x]*(Cos[5*e]/(504*c^6) + ((I/504)*Sin[5*e])/c^6) + (31*A - (9*I)*B)*Cos[10*f*x]*(((-I/792)*Cos[7*e])/c^6 +
Sin[7*e]/(792*c^6)) + (A - I*B)*Cos[12*f*x]*(((-I/88)*Cos[9*e])/c^6 + Sin[9*e]/(88*c^6)) + (A + I*B)*(Cos[3*e]
/(56*c^6) + ((I/56)*Sin[3*e])/c^6)*Sin[6*f*x] + (23*A + (9*I)*B)*(Cos[5*e]/(504*c^6) + ((I/504)*Sin[5*e])/c^6)
*Sin[8*f*x] + (31*A - (9*I)*B)*(Cos[7*e]/(792*c^6) + ((I/792)*Sin[7*e])/c^6)*Sin[10*f*x] + (A - I*B)*(Cos[9*e]
/(88*c^6) + ((I/88)*Sin[9*e])/c^6)*Sin[12*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 = 0.11, size = 161, normalized size = 1. \begin{align*}{\frac{{\frac{i}{693}}{a}^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 2\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{4}-63\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}-9\,B \left ( \tan \left ( fx+e \right ) \right ) ^{4}-45\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}-14\,A \left ( \tan \left ( fx+e \right ) \right ) ^{3}+63\,iB\tan \left ( fx+e \right ) -144\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}+79\,iA-140\,A\tan \left ( fx+e \right ) -9\,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))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x)

[Out]

1/693*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^6*(1+tan(f*x+e)^2)*(2*I*A*tan(f*x+e)^4
-63*I*B*tan(f*x+e)^3-9*B*tan(f*x+e)^4-45*I*A*tan(f*x+e)^2-14*A*tan(f*x+e)^3+63*I*B*tan(f*x+e)-144*B*tan(f*x+e)
^2+79*I*A-140*A*tan(f*x+e)-9*B)/(tan(f*x+e)+I)^7

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Maxima [A]  time = 3.05426, size = 270, normalized size = 1.74 \begin{align*} \frac{{\left (63 \,{\left (-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 ) - 154 i \, A 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 ) + 99 \,{\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 (63 \, A - 63 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 ) + 154 \, A 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 (99 \, A + 99 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}}{2772 \, 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))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

1/2772*(63*(-I*A - B)*a^3*cos(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 154*I*A*a^3*cos(9/2*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 99*(-I*A + B)*a^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ (63*A - 63*I*B)*a^3*sin(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 154*A*a^3*sin(9/2*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e))) + (99*A + 99*I*B)*a^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqr
t(a)/(c^(11/2)*f)

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Fricas [A]  time = 1.41822, size = 375, normalized size = 2.42 \begin{align*} \frac{{\left ({\left (-63 i \, A - 63 \, B\right )} a^{3} e^{\left (12 i \, f x + 12 i \, e\right )} +{\left (-217 i \, A - 63 \, B\right )} a^{3} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-253 i \, A + 99 \, B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-99 i \, A + 99 \, 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 )}}{2772 \, c^{6} 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))^(11/2),x, algorithm="fricas")

[Out]

1/2772*((-63*I*A - 63*B)*a^3*e^(12*I*f*x + 12*I*e) + (-217*I*A - 63*B)*a^3*e^(10*I*f*x + 10*I*e) + (-253*I*A +
99*B)*a^3*e^(8*I*f*x + 8*I*e) + (-99*I*A + 99*B)*a^3*e^(6*I*f*x + 6*I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*s
qrt(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))**(7/2)*(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{7}{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))^(7/2)*(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)^(7/2)/(-I*c*tan(f*x + e) + c)^(11/2), x)