3.1186 \(\int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=123 \[ -\frac{(-d+i c) (a+i a \tan (e+f x))^m \sqrt{c+d \tan (e+f x)} F_1\left (m;-\frac{3}{2},1;m+1;-\frac{d (i \tan (e+f x)+1)}{i c-d},\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m \sqrt{\frac{c+d \tan (e+f x)}{c+i d}}} \]

[Out]

-((I*c - d)*AppellF1[m, -3/2, 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]])/(2*f*m*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)])

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Rubi [A]  time = 0.179219, antiderivative size = 123, 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} \[ -\frac{(-d+i c) (a+i a \tan (e+f x))^m \sqrt{c+d \tan (e+f x)} F_1\left (m;-\frac{3}{2},1;m+1;-\frac{d (i \tan (e+f x)+1)}{i c-d},\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m \sqrt{\frac{c+d \tan (e+f x)}{c+i d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((I*c - d)*AppellF1[m, -3/2, 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]])/(2*f*m*Sqrt[(c + d*Tan[e + f*x])/(c + I*d)])

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 (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m} \left (c-\frac{i d x}{a}\right )^{3/2}}{-a^2+a x} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac{\left (i a^2 (c+i d) \sqrt{c+d \tan (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m} \left (\frac{c}{c+i d}-\frac{i d x}{a (c+i d)}\right )^{3/2}}{-a^2+a x} \, dx,x,i a \tan (e+f x)\right )}{f \sqrt{\frac{c+d \tan (e+f x)}{c+i d}}}\\ &=-\frac{(i c-d) F_1\left (m;-\frac{3}{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{c+d \tan (e+f x)}}{2 f m \sqrt{\frac{c+d \tan (e+f x)}{c+i d}}}\\ \end{align*}

Mathematica [F]  time = 26.8655, size = 0, normalized size = 0. \[ \int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + I*a*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(3/2),x]

[Out]

Integrate[(a + I*a*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(3/2), x]

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Maple [F]  time = 0.375, 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{3}{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))^(3/2),x)

[Out]

int((a+I*a*tan(f*x+e))^m*(c+d*tan(f*x+e))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}\,{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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left ({\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d\right )} \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}}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, 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))^(3/2),x, algorithm="fricas")

[Out]

integral(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)*(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))/(e^(2*I*f*x + 2*I*e) + 1), 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))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}\,{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))^(3/2),x, algorithm="giac")

[Out]

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