3.801 \(\int \frac{(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=102 \[ -\frac{(-4 B+i A) (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(3/2))/(5*f*(c - I*c*Tan[e + f*x])^(5/2)) - ((I*A - 4*B)*(a + I*a*Tan[e + f
*x])^(3/2))/(15*c*f*(c - I*c*Tan[e + f*x])^(3/2))

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Rubi [A]  time = 0.233717, 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} \[ -\frac{(-4 B+i A) (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(3/2))/(5*f*(c - I*c*Tan[e + f*x])^(5/2)) - ((I*A - 4*B)*(a + I*a*Tan[e + f
*x])^(3/2))/(15*c*f*(c - I*c*Tan[e + f*x])^(3/2))

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))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x} (A+B x)}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac{(a (A+4 i B)) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac{(i A-4 B) (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 11.6563, size = 117, normalized size = 1.15 \[ \frac{a \cos (e+f x) (\cos (f x)-i \sin (f x)) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)} (\cos (4 e+5 f x)+i \sin (4 e+5 f x)) ((B-4 i A) \cos (e+f x)-(A+4 i B) \sin (e+f x))}{15 c^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(a*Cos[e + f*x]*(Cos[f*x] - I*Sin[f*x])*(((-4*I)*A + B)*Cos[e + f*x] - (A + (4*I)*B)*Sin[e + f*x])*(Cos[4*e +
5*f*x] + I*Sin[4*e + 5*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(15*c^3*f)

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Maple [A]  time = 0.106, size = 90, normalized size = 0.9 \begin{align*}{\frac{{\frac{i}{15}}a \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( -4\,A+iA\tan \left ( fx+e \right ) -iB-4\,B\tan \left ( fx+e \right ) \right ) }{f{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4}}\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))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

1/15*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a/c^3*(1+tan(f*x+e)^2)*(-4*A+I*A*tan(f*x+e)-I
*B-4*B*tan(f*x+e))/(tan(f*x+e)+I)^4

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Maxima [A]  time = 2.28958, size = 209, normalized size = 2.05 \begin{align*} -\frac{{\left ({\left (90 \, A - 90 i \, B\right )} a \cos \left (7 \, f x + 7 \, e\right ) +{\left (240 \, A + 60 i \, B\right )} a \cos \left (5 \, f x + 5 \, e\right ) +{\left (150 \, A + 150 i \, B\right )} a \cos \left (3 \, f x + 3 \, e\right ) - 90 \,{\left (-i \, A - B\right )} a \sin \left (7 \, f x + 7 \, e\right ) - 60 \,{\left (-4 i \, A + B\right )} a \sin \left (5 \, f x + 5 \, e\right ) - 150 \,{\left (-i \, A + B\right )} a \sin \left (3 \, f x + 3 \, e\right )\right )} \sqrt{a} \sqrt{c}}{{\left (-900 i \, c^{3} \cos \left (2 \, f x + 2 \, e\right ) + 900 \, c^{3} \sin \left (2 \, f x + 2 \, e\right ) - 900 i \, c^{3}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-((90*A - 90*I*B)*a*cos(7*f*x + 7*e) + (240*A + 60*I*B)*a*cos(5*f*x + 5*e) + (150*A + 150*I*B)*a*cos(3*f*x + 3
*e) - 90*(-I*A - B)*a*sin(7*f*x + 7*e) - 60*(-4*I*A + B)*a*sin(5*f*x + 5*e) - 150*(-I*A + B)*a*sin(3*f*x + 3*e
))*sqrt(a)*sqrt(c)/((-900*I*c^3*cos(2*f*x + 2*e) + 900*c^3*sin(2*f*x + 2*e) - 900*I*c^3)*f)

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Fricas [A]  time = 1.44042, size = 290, normalized size = 2.84 \begin{align*} \frac{{\left ({\left (-3 i \, A - 3 \, B\right )} a e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-8 i \, A + 2 \, B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-5 i \, A + 5 \, B\right )} a e^{\left (2 i \, f x + 2 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 )}}{30 \, c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/30*((-3*I*A - 3*B)*a*e^(6*I*f*x + 6*I*e) + (-8*I*A + 2*B)*a*e^(4*I*f*x + 4*I*e) + (-5*I*A + 5*B)*a*e^(2*I*f*
x + 2*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^3*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))**(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(5/2),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{3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(3/2)/(-I*c*tan(f*x + e) + c)^(5/2), x)