3.1040 \(\int \frac{1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=234 \[ -\frac{8 i \sqrt{a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{8 i \sqrt{a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{4 i \sqrt{a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{4 i}{3 a f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \]

[Out]

(I/3)/(f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2)) + ((4*I)/3)/(a*f*Sqrt[a + I*a*Tan[e + f*x]
]*(c - I*c*Tan[e + f*x])^(5/2)) - (((4*I)/5)*Sqrt[a + I*a*Tan[e + f*x]])/(a^2*f*(c - I*c*Tan[e + f*x])^(5/2))
- (((8*I)/15)*Sqrt[a + I*a*Tan[e + f*x]])/(a^2*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (((8*I)/15)*Sqrt[a + I*a*Ta
n[e + f*x]])/(a^2*c^2*f*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.190929, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {3523, 45, 37} \[ -\frac{8 i \sqrt{a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}-\frac{8 i \sqrt{a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{4 i \sqrt{a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{4 i}{3 a f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(I/3)/(f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2)) + ((4*I)/3)/(a*f*Sqrt[a + I*a*Tan[e + f*x]
]*(c - I*c*Tan[e + f*x])^(5/2)) - (((4*I)/5)*Sqrt[a + I*a*Tan[e + f*x]])/(a^2*f*(c - I*c*Tan[e + f*x])^(5/2))
- (((8*I)/15)*Sqrt[a + I*a*Tan[e + f*x]])/(a^2*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (((8*I)/15)*Sqrt[a + I*a*Ta
n[e + f*x]])/(a^2*c^2*f*Sqrt[c - I*c*Tan[e + f*x]])

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{1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{5/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac{(4 c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{3/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac{4 i}{3 a f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac{(4 c) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac{4 i}{3 a f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac{4 i \sqrt{a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac{8 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac{4 i}{3 a f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac{4 i \sqrt{a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac{8 i \sqrt{a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}+\frac{8 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a c f}\\ &=\frac{i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac{4 i}{3 a f \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac{4 i \sqrt{a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac{8 i \sqrt{a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac{8 i \sqrt{a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 5.63567, size = 130, normalized size = 0.56 \[ \frac{\sec (e+f x) \sqrt{c-i c \tan (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x))) (-40 i \sin (2 (e+f x))-4 i \sin (4 (e+f x))+20 \cos (2 (e+f x))+\cos (4 (e+f x))-45)}{120 a c^3 f (\tan (e+f x)-i) \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]

[Out]

(Sec[e + f*x]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*(-45 + 20*Cos[2*(e + f*x)] + Cos[4*(e + f*x)] - (40*I)*S
in[2*(e + f*x)] - (4*I)*Sin[4*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(120*a*c^3*f*(-I + Tan[e + f*x])*Sqrt[a
+ I*a*Tan[e + f*x]])

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Maple [A]  time = 0.044, size = 130, normalized size = 0.6 \begin{align*} -{\frac{8\,i \left ( \tan \left ( fx+e \right ) \right ) ^{5}+8\, \left ( \tan \left ( fx+e \right ) \right ) ^{6}+20\,i \left ( \tan \left ( fx+e \right ) \right ) ^{3}+20\, \left ( \tan \left ( fx+e \right ) \right ) ^{4}+12\,i\tan \left ( fx+e \right ) +15\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}+3}{15\,f{a}^{2}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{4} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{3}}\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(1/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

-1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/a^2/c^3*(8*I*tan(f*x+e)^5+8*tan(f*x+e)^6+20*I*
tan(f*x+e)^3+20*tan(f*x+e)^4+12*I*tan(f*x+e)+15*tan(f*x+e)^2+3)/(tan(f*x+e)+I)^4/(-tan(f*x+e)+I)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.42636, size = 413, normalized size = 1.76 \begin{align*} \frac{\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}}{\left (-3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 23 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 110 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 48 i \, e^{\left (5 i \, f x + 5 i \, e\right )} - 30 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 48 i \, e^{\left (3 i \, f x + 3 i \, e\right )} + 65 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{240 \, a^{2} c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/240*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-3*I*e^(10*I*f*x + 10*I*e) - 23*I*e
^(8*I*f*x + 8*I*e) - 110*I*e^(6*I*f*x + 6*I*e) + 48*I*e^(5*I*f*x + 5*I*e) - 30*I*e^(4*I*f*x + 4*I*e) + 48*I*e^
(3*I*f*x + 3*I*e) + 65*I*e^(2*I*f*x + 2*I*e) + 5*I)*e^(-3*I*f*x - 3*I*e)/(a^2*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(1/(a+I*a*tan(f*x+e))**(3/2)/(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{1}{{\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(1/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(f*x + e) + a)^(3/2)*(-I*c*tan(f*x + e) + c)^(5/2)), x)