3.812 \(\int \frac{(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=102 \[ -\frac{(-6 B+i A) (a+i a \tan (e+f x))^{5/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{7 f (c-i c \tan (e+f x))^{7/2}} \]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) - ((I*A - 6*B)*(a + I*a*Tan[e + f
*x])^(5/2))/(35*c*f*(c - I*c*Tan[e + f*x])^(5/2))

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Rubi [A]  time = 0.230742, 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} \[ -\frac{(-6 B+i A) (a+i a \tan (e+f x))^{5/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{7 f (c-i c \tan (e+f x))^{7/2}} \]

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])^(7/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) - ((I*A - 6*B)*(a + I*a*Tan[e + f
*x])^(5/2))/(35*c*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 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))^{7/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{(a (A+6 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 )}{7 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac{(i A-6 B) (a+i a \tan (e+f x))^{5/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 12.1307, size = 121, normalized size = 1.19 \[ \frac{a^2 \cos (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)} (\cos (6 e+8 f x)+i \sin (6 e+8 f x)) ((B-6 i A) \cos (e+f x)-(A+6 i B) \sin (e+f x))}{35 c^4 f (\cos (f x)+i \sin (f x))^2} \]

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])^(7/2),x]

[Out]

(a^2*Cos[e + f*x]*(((-6*I)*A + B)*Cos[e + f*x] - (A + (6*I)*B)*Sin[e + f*x])*(Cos[6*e + 8*f*x] + I*Sin[6*e + 8
*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(35*c^4*f*(Cos[f*x] + I*Sin[f*x])^2)

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Maple [A]  time = 0.108, size = 115, normalized size = 1.1 \begin{align*}{\frac{-{\frac{i}{35}}{a}^{2} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}+5\,iB\tan \left ( fx+e \right ) -6\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}+6\,iA-5\,A\tan \left ( fx+e \right ) -B \right ) }{f{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{5}}\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))^(7/2),x)

[Out]

-1/35*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2/c^4*(1+tan(f*x+e)^2)*(I*A*tan(f*x+e)^2+5
*I*B*tan(f*x+e)-6*B*tan(f*x+e)^2+6*I*A-5*A*tan(f*x+e)-B)/(tan(f*x+e)+I)^5

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Maxima [B]  time = 2.45494, size = 225, normalized size = 2.21 \begin{align*} -\frac{{\left ({\left (350 \, A - 350 i \, B\right )} a^{2} \cos \left (9 \, f x + 9 \, e\right ) +{\left (840 \, A + 140 i \, B\right )} a^{2} \cos \left (7 \, f x + 7 \, e\right ) +{\left (490 \, A + 490 i \, B\right )} a^{2} \cos \left (5 \, f x + 5 \, e\right ) - 350 \,{\left (-i \, A - B\right )} a^{2} \sin \left (9 \, f x + 9 \, e\right ) - 140 \,{\left (-6 i \, A + B\right )} a^{2} \sin \left (7 \, f x + 7 \, e\right ) - 490 \,{\left (-i \, A + B\right )} a^{2} \sin \left (5 \, f x + 5 \, e\right )\right )} \sqrt{a} \sqrt{c}}{{\left (-4900 i \, c^{4} \cos \left (2 \, f x + 2 \, e\right ) + 4900 \, c^{4} \sin \left (2 \, f x + 2 \, e\right ) - 4900 i \, c^{4}\right )} 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))^(7/2),x, algorithm="maxima")

[Out]

-((350*A - 350*I*B)*a^2*cos(9*f*x + 9*e) + (840*A + 140*I*B)*a^2*cos(7*f*x + 7*e) + (490*A + 490*I*B)*a^2*cos(
5*f*x + 5*e) - 350*(-I*A - B)*a^2*sin(9*f*x + 9*e) - 140*(-6*I*A + B)*a^2*sin(7*f*x + 7*e) - 490*(-I*A + B)*a^
2*sin(5*f*x + 5*e))*sqrt(a)*sqrt(c)/((-4900*I*c^4*cos(2*f*x + 2*e) + 4900*c^4*sin(2*f*x + 2*e) - 4900*I*c^4)*f
)

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Fricas [A]  time = 1.39234, size = 300, normalized size = 2.94 \begin{align*} \frac{{\left ({\left (-5 i \, A - 5 \, B\right )} a^{2} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-12 i \, A + 2 \, B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-7 i \, A + 7 \, 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 )}}{70 \, c^{4} 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))^(7/2),x, algorithm="fricas")

[Out]

1/70*((-5*I*A - 5*B)*a^2*e^(8*I*f*x + 8*I*e) + (-12*I*A + 2*B)*a^2*e^(6*I*f*x + 6*I*e) + (-7*I*A + 7*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^4*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))**(7/2),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{7}{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))^(7/2),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)^(7/2), x)