3.825 \(\int \frac{(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=102 \[ -\frac{(-8 B+i A) (a+i a \tan (e+f x))^{7/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{9 f (c-i c \tan (e+f x))^{9/2}} \]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(9*f*(c - I*c*Tan[e + f*x])^(9/2)) - ((I*A - 8*B)*(a + I*a*Tan[e + f
*x])^(7/2))/(63*c*f*(c - I*c*Tan[e + f*x])^(7/2))

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Rubi [A]  time = 0.228282, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3588, 78, 37} \[ -\frac{(-8 B+i A) (a+i a \tan (e+f x))^{7/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{9 f (c-i c \tan (e+f x))^{9/2}} \]

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])^(9/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(9*f*(c - I*c*Tan[e + f*x])^(9/2)) - ((I*A - 8*B)*(a + I*a*Tan[e + f
*x])^(7/2))/(63*c*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 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))^{9/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{9 f (c-i c \tan (e+f x))^{9/2}}+\frac{(a (A+8 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 )}{9 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{9 f (c-i c \tan (e+f x))^{9/2}}-\frac{(i A-8 B) (a+i a \tan (e+f x))^{7/2}}{63 c f (c-i c \tan (e+f x))^{7/2}}\\ \end{align*}

Mathematica [B]  time = 13.7657, size = 335, normalized size = 3.28 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((B-i A) \cos (6 f x) \left (\frac{\cos (3 e)}{28 c^5}+\frac{i \sin (3 e)}{28 c^5}\right )+(A+i B) \sin (6 f x) \left (\frac{\cos (3 e)}{28 c^5}+\frac{i \sin (3 e)}{28 c^5}\right )+(B-8 i A) \cos (8 f x) \left (\frac{\cos (5 e)}{126 c^5}+\frac{i \sin (5 e)}{126 c^5}\right )+(A-i B) \cos (10 f x) \left (\frac{\sin (7 e)}{36 c^5}-\frac{i \cos (7 e)}{36 c^5}\right )+(8 A+i B) \sin (8 f x) \left (\frac{\cos (5 e)}{126 c^5}+\frac{i \sin (5 e)}{126 c^5}\right )+(A-i B) \sin (10 f x) \left (\frac{\cos (7 e)}{36 c^5}+\frac{i \sin (7 e)}{36 c^5}\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

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])^(9/2),x]

[Out]

(Cos[e + f*x]^4*(((-I)*A + B)*Cos[6*f*x]*(Cos[3*e]/(28*c^5) + ((I/28)*Sin[3*e])/c^5) + ((-8*I)*A + B)*Cos[8*f*
x]*(Cos[5*e]/(126*c^5) + ((I/126)*Sin[5*e])/c^5) + (A - I*B)*Cos[10*f*x]*(((-I/36)*Cos[7*e])/c^5 + Sin[7*e]/(3
6*c^5)) + (A + I*B)*(Cos[3*e]/(28*c^5) + ((I/28)*Sin[3*e])/c^5)*Sin[6*f*x] + (8*A + I*B)*(Cos[5*e]/(126*c^5) +
((I/126)*Sin[5*e])/c^5)*Sin[8*f*x] + (A - I*B)*(Cos[7*e]/(36*c^5) + ((I/36)*Sin[7*e])/c^5)*Sin[10*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 = 134, normalized size = 1.3 \begin{align*} -{\frac{{a}^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 8\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}+6\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}+A \left ( \tan \left ( fx+e \right ) \right ) ^{3}-6\,iB\tan \left ( fx+e \right ) +15\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}-8\,iA+15\,A\tan \left ( fx+e \right ) +B \right ) }{63\,f{c}^{5} \left ( \tan \left ( fx+e \right ) +i \right ) ^{6}}\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))^(9/2),x)

[Out]

-1/63/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^5*(1+tan(f*x+e)^2)*(8*I*B*tan(f*x+e)^3+6
*I*A*tan(f*x+e)^2+A*tan(f*x+e)^3-6*I*B*tan(f*x+e)+15*B*tan(f*x+e)^2-8*I*A+15*A*tan(f*x+e)+B)/(tan(f*x+e)+I)^6

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Maxima [B]  time = 2.76656, size = 225, normalized size = 2.21 \begin{align*} -\frac{{\left ({\left (882 \, A - 882 i \, B\right )} a^{3} \cos \left (11 \, f x + 11 \, e\right ) +{\left (2016 \, A + 252 i \, B\right )} a^{3} \cos \left (9 \, f x + 9 \, e\right ) +{\left (1134 \, A + 1134 i \, B\right )} a^{3} \cos \left (7 \, f x + 7 \, e\right ) - 882 \,{\left (-i \, A - B\right )} a^{3} \sin \left (11 \, f x + 11 \, e\right ) - 252 \,{\left (-8 i \, A + B\right )} a^{3} \sin \left (9 \, f x + 9 \, e\right ) - 1134 \,{\left (-i \, A + B\right )} a^{3} \sin \left (7 \, f x + 7 \, e\right )\right )} \sqrt{a} \sqrt{c}}{{\left (-15876 i \, c^{5} \cos \left (2 \, f x + 2 \, e\right ) + 15876 \, c^{5} \sin \left (2 \, f x + 2 \, e\right ) - 15876 i \, c^{5}\right )} 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))^(9/2),x, algorithm="maxima")

[Out]

-((882*A - 882*I*B)*a^3*cos(11*f*x + 11*e) + (2016*A + 252*I*B)*a^3*cos(9*f*x + 9*e) + (1134*A + 1134*I*B)*a^3
*cos(7*f*x + 7*e) - 882*(-I*A - B)*a^3*sin(11*f*x + 11*e) - 252*(-8*I*A + B)*a^3*sin(9*f*x + 9*e) - 1134*(-I*A
+ B)*a^3*sin(7*f*x + 7*e))*sqrt(a)*sqrt(c)/((-15876*I*c^5*cos(2*f*x + 2*e) + 15876*c^5*sin(2*f*x + 2*e) - 158
76*I*c^5)*f)

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Fricas [A]  time = 1.41371, size = 304, normalized size = 2.98 \begin{align*} \frac{{\left ({\left (-7 i \, A - 7 \, B\right )} a^{3} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-16 i \, A + 2 \, B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-9 i \, A + 9 \, 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 )}}{126 \, c^{5} 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))^(9/2),x, algorithm="fricas")

[Out]

1/126*((-7*I*A - 7*B)*a^3*e^(10*I*f*x + 10*I*e) + (-16*I*A + 2*B)*a^3*e^(8*I*f*x + 8*I*e) + (-9*I*A + 9*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^5
*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))**(9/2),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{9}{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))^(9/2),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)^(9/2), x)