3.1032 \(\int \frac{1}{(a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=193 \[ \frac{8 \tan (e+f x)}{21 a^3 c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}+\frac{4 \tan (e+f x)}{21 a^2 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i}{7 a f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}} \]

[Out]

(I/7)/(f*(a + I*a*Tan[e + f*x])^(7/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (I/7)/(a*f*(a + I*a*Tan[e + f*x])^(5/2)*
(c - I*c*Tan[e + f*x])^(3/2)) + (4*Tan[e + f*x])/(21*a^2*f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])
^(3/2)) + (8*Tan[e + f*x])/(21*a^3*c*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.169957, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {3523, 45, 40, 39} \[ \frac{8 \tan (e+f x)}{21 a^3 c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}+\frac{4 \tan (e+f x)}{21 a^2 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i}{7 a f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(7/2)*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

(I/7)/(f*(a + I*a*Tan[e + f*x])^(7/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (I/7)/(a*f*(a + I*a*Tan[e + f*x])^(5/2)*
(c - I*c*Tan[e + f*x])^(3/2)) + (4*Tan[e + f*x])/(21*a^2*f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])
^(3/2)) + (8*Tan[e + f*x])/(21*a^3*c*f*Sqrt[a + I*a*Tan[e + f*x]]*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 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(x*(a + b*x)^(m + 1)*(c + d*x)^(m +
1))/(2*a*c*(m + 1)), x] + Dist[(2*m + 3)/(2*a*c*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /; F
reeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 0]

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{9/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}+\frac{(5 c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{7/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{7 f}\\ &=\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i}{7 a f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{(4 c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{7 a f}\\ &=\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i}{7 a f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{4 \tan (e+f x)}{21 a^2 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{8 \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{21 a^2 f}\\ &=\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}+\frac{i}{7 a f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac{4 \tan (e+f x)}{21 a^2 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac{8 \tan (e+f x)}{21 a^3 c f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 6.4826, size = 151, normalized size = 0.78 \[ \frac{\sec ^3(e+f x) \sqrt{c-i c \tan (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) (-70 i \sin (e+f x)+63 i \sin (3 (e+f x))+5 i \sin (5 (e+f x))-140 \cos (e+f x)+42 \cos (3 (e+f x))+2 \cos (5 (e+f x)))}{336 a^3 c^2 f (\tan (e+f x)-i)^3 \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(7/2)*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

(Sec[e + f*x]^3*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*(-140*Cos[e + f*x] + 42*Cos[3*(e + f*x)] + 2*Cos[5*(e
+ f*x)] - (70*I)*Sin[e + f*x] + (63*I)*Sin[3*(e + f*x)] + (5*I)*Sin[5*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/
(336*a^3*c^2*f*(-I + Tan[e + f*x])^3*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]  time = 0.045, size = 141, normalized size = 0.7 \begin{align*}{\frac{16\,i \left ( \tan \left ( fx+e \right ) \right ) ^{6}-8\, \left ( \tan \left ( fx+e \right ) \right ) ^{7}+40\,i \left ( \tan \left ( fx+e \right ) \right ) ^{4}-12\, \left ( \tan \left ( fx+e \right ) \right ) ^{5}+30\,i \left ( \tan \left ( fx+e \right ) \right ) ^{2}+5\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}+6\,i+9\,\tan \left ( fx+e \right ) }{21\,f{a}^{4}{c}^{2} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{5} \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))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

1/21/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/a^4/c^2*(16*I*tan(f*x+e)^6-8*tan(f*x+e)^7+40*I*
tan(f*x+e)^4-12*tan(f*x+e)^5+30*I*tan(f*x+e)^2+5*tan(f*x+e)^3+6*I+9*tan(f*x+e))/(-tan(f*x+e)+I)^5/(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))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.47561, size = 458, normalized size = 2.37 \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 (-7 i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 112 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 192 i \, e^{\left (9 i \, f x + 9 i \, e\right )} + 105 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 192 i \, e^{\left (7 i \, f x + 7 i \, e\right )} + 280 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 91 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 24 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (-7 i \, f x - 7 i \, e\right )}}{672 \, a^{4} c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/672*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-7*I*e^(12*I*f*x + 12*I*e) - 112*I*
e^(10*I*f*x + 10*I*e) - 192*I*e^(9*I*f*x + 9*I*e) + 105*I*e^(8*I*f*x + 8*I*e) - 192*I*e^(7*I*f*x + 7*I*e) + 28
0*I*e^(6*I*f*x + 6*I*e) + 91*I*e^(4*I*f*x + 4*I*e) + 24*I*e^(2*I*f*x + 2*I*e) + 3*I)*e^(-7*I*f*x - 7*I*e)/(a^4
*c^2*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))**(7/2)/(c-I*c*tan(f*x+e))**(3/2),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{7}{2}}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{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))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(f*x + e) + a)^(7/2)*(-I*c*tan(f*x + e) + c)^(3/2)), x)