3.1190 \(\int \frac{(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=125 \[ -\frac{i (a+i a \tan (e+f x))^m \sqrt{\frac{c+d \tan (e+f x)}{c+i d}} F_1\left (m;\frac{5}{2},1;m+1;-\frac{d (i \tan (e+f x)+1)}{i c-d},\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m (c+i d)^2 \sqrt{c+d \tan (e+f x)}} \]

[Out]

((-I/2)*AppellF1[m, 5/2, 1, 1 + m, -((d*(1 + I*Tan[e + f*x]))/(I*c - d)), (1 + I*Tan[e + f*x])/2]*(a + I*a*Tan
[e + f*x])^m*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)])/((c + I*d)^2*f*m*Sqrt[c + d*Tan[e + f*x]])

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Rubi [A]  time = 0.157546, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3564, 137, 136} \[ -\frac{i (a+i a \tan (e+f x))^m \sqrt{\frac{c+d \tan (e+f x)}{c+i d}} F_1\left (m;\frac{5}{2},1;m+1;-\frac{d (i \tan (e+f x)+1)}{i c-d},\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m (c+i d)^2 \sqrt{c+d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*AppellF1[m, 5/2, 1, 1 + m, -((d*(1 + I*Tan[e + f*x]))/(I*c - d)), (1 + I*Tan[e + f*x])/2]*(a + I*a*Tan
[e + f*x])^m*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)])/((c + I*d)^2*f*m*Sqrt[c + d*Tan[e + f*x]])

Rule 3564

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dis
t[(a*b)/f, Subst[Int[((a + x)^(m - 1)*(c + (d*x)/b)^n)/(b^2 + a*x), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b,
c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m}}{\left (-a^2+a x\right ) \left (c-\frac{i d x}{a}\right )^{5/2}} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac{\left (i a^2 \sqrt{\frac{c+d \tan (e+f x)}{c+i d}}\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m}}{\left (-a^2+a x\right ) \left (\frac{c}{c+i d}-\frac{i d x}{a (c+i d)}\right )^{5/2}} \, dx,x,i a \tan (e+f x)\right )}{(c+i d)^2 f \sqrt{c+d \tan (e+f x)}}\\ &=-\frac{i F_1\left (m;\frac{5}{2},1;1+m;-\frac{d (1+i \tan (e+f x))}{i c-d},\frac{1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m \sqrt{\frac{c+d \tan (e+f x)}{c+i d}}}{2 (c+i d)^2 f m \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [F]  time = 17.7164, size = 0, normalized size = 0. \[ \int \frac{(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + I*a*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

Integrate[(a + I*a*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^(5/2), x]

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Maple [F]  time = 0.325, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^(5/2),x)

[Out]

int((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \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((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (\frac{2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m} \sqrt{\frac{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (-i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - 3 i \, e^{\left (2 i \, f x + 2 i \, e\right )} - i\right )}}{-i \, c^{3} + 3 \, c^{2} d + 3 i \, c d^{2} - d^{3} +{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-3 i \, c^{3} - 3 \, c^{2} d - 3 i \, c d^{2} - 3 \, d^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-3 i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} + 3 \, d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/
(e^(2*I*f*x + 2*I*e) + 1))*(-I*e^(6*I*f*x + 6*I*e) - 3*I*e^(4*I*f*x + 4*I*e) - 3*I*e^(2*I*f*x + 2*I*e) - I)/(-
I*c^3 + 3*c^2*d + 3*I*c*d^2 - d^3 + (-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*e^(6*I*f*x + 6*I*e) + (-3*I*c^3 - 3*c
^2*d - 3*I*c*d^2 - 3*d^3)*e^(4*I*f*x + 4*I*e) + (-3*I*c^3 + 3*c^2*d - 3*I*c*d^2 + 3*d^3)*e^(2*I*f*x + 2*I*e)),
x)

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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+I*a*tan(f*x+e))**m/(c+d*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 \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \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((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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