3.1023 \(\int \frac{1}{(a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}} \, dx\)

Optimal. Leaf size=186 \[ \frac{8 \tan (e+f x)}{35 a^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i}{35 a^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i}{35 a f (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}+\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}} \]

[Out]

(I/7)/(f*(a + I*a*Tan[e + f*x])^(7/2)*Sqrt[c - I*c*Tan[e + f*x]]) + ((4*I)/35)/(a*f*(a + I*a*Tan[e + f*x])^(5/
2)*Sqrt[c - I*c*Tan[e + f*x]]) + ((4*I)/35)/(a^2*f*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]]) +
(8*Tan[e + f*x])/(35*a^3*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.163004, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {3523, 45, 39} \[ \frac{8 \tan (e+f x)}{35 a^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i}{35 a^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i}{35 a f (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}+\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I/7)/(f*(a + I*a*Tan[e + f*x])^(7/2)*Sqrt[c - I*c*Tan[e + f*x]]) + ((4*I)/35)/(a*f*(a + I*a*Tan[e + f*x])^(5/
2)*Sqrt[c - I*c*Tan[e + f*x]]) + ((4*I)/35)/(a^2*f*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]]) +
(8*Tan[e + f*x])/(35*a^3*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 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} \sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{9/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}+\frac{(4 c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{7/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{7 f}\\ &=\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i}{35 a f (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}+\frac{(12 c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x)^{5/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{35 a f}\\ &=\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i}{35 a f (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i}{35 a^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{(8 c) \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 )}{35 a^2 f}\\ &=\frac{i}{7 f (a+i a \tan (e+f x))^{7/2} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i}{35 a f (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)}}+\frac{4 i}{35 a^2 f (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}+\frac{8 \tan (e+f x)}{35 a^3 f \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.5254, size = 115, normalized size = 0.62 \[ \frac{\sec ^2(e+f x) \sqrt{c-i c \tan (e+f x)} (56 \sin (2 (e+f x))-20 \sin (4 (e+f x))-84 i \cos (2 (e+f x))+15 i \cos (4 (e+f x))-35 i)}{280 a^3 c f (\tan (e+f x)-i)^2 \sqrt{a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*(-35*I - (84*I)*Cos[2*(e + f*x)] + (15*I)*Cos[4*(e + f*x)] + 56*Sin[2*(e + f*x)] - 20*Sin[4*(e
+ f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(280*a^3*c*f*(-I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]  time = 0.084, size = 130, normalized size = 0.7 \begin{align*}{\frac{24\,i \left ( \tan \left ( fx+e \right ) \right ) ^{5}-8\, \left ( \tan \left ( fx+e \right ) \right ) ^{6}+28\,i \left ( \tan \left ( fx+e \right ) \right ) ^{3}+12\, \left ( \tan \left ( fx+e \right ) \right ) ^{4}+4\,i\tan \left ( fx+e \right ) +33\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}+13}{35\,f{a}^{4}c \left ( \tan \left ( fx+e \right ) +i \right ) ^{2} \left ( -\tan \left ( fx+e \right ) +i \right ) ^{5}}\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(1/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(7/2),x)

[Out]

1/35/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^4/c*(24*I*tan(f*x+e)^5-8*tan(f*x+e)^6+28*I*ta
n(f*x+e)^3+12*tan(f*x+e)^4+4*I*tan(f*x+e)+33*tan(f*x+e)^2+13)/(tan(f*x+e)+I)^2/(-tan(f*x+e)+I)^5

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.448, size = 416, normalized size = 2.24 \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 (-35 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 208 i \, e^{\left (9 i \, f x + 9 i \, e\right )} + 105 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 208 i \, e^{\left (7 i \, f x + 7 i \, e\right )} + 210 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 98 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 33 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i\right )} e^{\left (-7 i \, f x - 7 i \, e\right )}}{560 \, a^{4} c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/560*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-35*I*e^(10*I*f*x + 10*I*e) - 208*I
*e^(9*I*f*x + 9*I*e) + 105*I*e^(8*I*f*x + 8*I*e) - 208*I*e^(7*I*f*x + 7*I*e) + 210*I*e^(6*I*f*x + 6*I*e) + 98*
I*e^(4*I*f*x + 4*I*e) + 33*I*e^(2*I*f*x + 2*I*e) + 5*I)*e^(-7*I*f*x - 7*I*e)/(a^4*c*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/(c-I*c*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(7/2),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}} \sqrt{-i \, c \tan \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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