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

Optimal. Leaf size=104 \[ \frac{(4 B+i A) (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(-B+i A) (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.228521, antiderivative size = 104, 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} \[ \frac{(4 B+i A) (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{(-B+i A) (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 7.77362, size = 92, normalized size = 0.88 \[ \frac{c (1-i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)} ((A-4 i B) \tan (e+f x)-4 i A-B)}{15 a^2 f (\tan (e+f x)-i)^2 \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(1 - I*Tan[e + f*x])*((-4*I)*A - B + (A - (4*I)*B)*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(15*a^2*f*(-I
+ Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]  time = 0.11, size = 92, normalized size = 0.9 \begin{align*}{\frac{{\frac{i}{15}}c \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( iA\tan \left ( fx+e \right ) -iB+4\,B\tan \left ( fx+e \right ) +4\,A \right ) }{f{a}^{3} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{4}}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.99084, size = 207, normalized size = 1.99 \begin{align*} \frac{{\left ({\left (150 \, A - 150 i \, B\right )} c \cos \left (4 \, f x + 4 \, e\right ) +{\left (240 \, A - 60 i \, B\right )} c \cos \left (2 \, f x + 2 \, e\right ) - 150 \,{\left (-i \, A - B\right )} c \sin \left (4 \, f x + 4 \, e\right ) - 60 \,{\left (-4 i \, A - B\right )} c \sin \left (2 \, f x + 2 \, e\right ) +{\left (90 \, A + 90 i \, B\right )} c\right )} \sqrt{a} \sqrt{c}}{{\left (-900 i \, a^{3} \cos \left (7 \, f x + 7 \, e\right ) - 900 i \, a^{3} \cos \left (5 \, f x + 5 \, e\right ) + 900 \, a^{3} \sin \left (7 \, f x + 7 \, e\right ) + 900 \, a^{3} \sin \left (5 \, f x + 5 \, e\right )\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.35945, size = 371, normalized size = 3.57 \begin{align*} \frac{{\left ({\left (-8 i \, A - 2 \, B\right )} c e^{\left (7 i \, f x + 7 i \, e\right )} +{\left (-8 i \, A - 2 \, B\right )} c e^{\left (5 i \, f x + 5 i \, e\right )} +{\left (5 i \, A + 5 \, B\right )} c e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (8 i \, A + 2 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (3 i \, A - 3 \, B\right )} c\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 (-5 i \, f x - 5 i \, e\right )}}{30 \, a^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*((-8*I*A - 2*B)*c*e^(7*I*f*x + 7*I*e) + (-8*I*A - 2*B)*c*e^(5*I*f*x + 5*I*e) + (5*I*A + 5*B)*c*e^(4*I*f*x
+ 4*I*e) + (8*I*A + 2*B)*c*e^(2*I*f*x + 2*I*e) + (3*I*A - 3*B)*c)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e
^(2*I*f*x + 2*I*e) + 1))*e^(-5*I*f*x - 5*I*e)/(a^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+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(3/2)/(a+I*a*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 \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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