3.1004 \(\int \frac{(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=90 \[ \frac{i (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{i (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.126306, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {3523, 45, 37} \[ \frac{i (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac{i (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3523

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((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), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

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{(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{\sqrt{c-i c x}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{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 (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac{i (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 = 2.72341, size = 79, normalized size = 0.88 \[ \frac{c (1-i \tan (e+f x)) (\tan (e+f x)-4 i) \sqrt{c-i c \tan (e+f x)}}{15 a^2 f (\tan (e+f x)-i)^2 \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(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 + 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 = 0.034, size = 75, normalized size = 0.8 \begin{align*}{\frac{c \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 4\,i-\tan \left ( fx+e \right ) \right ) }{15\,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((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x)

[Out]

1/15/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)*(4*I-tan(f*x+e))/(-tan(f
*x+e)+I)^4

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Maxima [A]  time = 1.93798, size = 116, normalized size = 1.29 \begin{align*} \frac{{\left (3 i \, c \cos \left (5 \, f x + 5 \, e\right ) + 5 i \, c \cos \left (\frac{3}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right ) + 3 \, c \sin \left (5 \, f x + 5 \, e\right ) + 5 \, c \sin \left (\frac{3}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right )\right )} \sqrt{c}}{30 \, a^{\frac{5}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(3*I*c*cos(5*f*x + 5*e) + 5*I*c*cos(3/5*arctan2(sin(5*f*x + 5*e), cos(5*f*x + 5*e))) + 3*c*sin(5*f*x + 5*
e) + 5*c*sin(3/5*arctan2(sin(5*f*x + 5*e), cos(5*f*x + 5*e))))*sqrt(c)/(a^(5/2)*f)

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Fricas [A]  time = 1.44953, size = 302, normalized size = 3.36 \begin{align*} \frac{{\left (-8 i \, c e^{\left (7 i \, f x + 7 i \, e\right )} - 8 i \, c e^{\left (5 i \, f x + 5 i \, e\right )} + 5 i \, c e^{\left (4 i \, f x + 4 i \, e\right )} + 8 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, 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((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*c*e^(7*I*f*x + 7*I*e) - 8*I*c*e^(5*I*f*x + 5*I*e) + 5*I*c*e^(4*I*f*x + 4*I*e) + 8*I*c*e^(2*I*f*x +
2*I*e) + 3*I*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((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 (-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((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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