3.1059 \(\int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=67 \[ \frac{i (c-i c \tan (e+f x))^{5/2} (a+i a \tan (e+f x))^m \, _2F_1\left (1,m+\frac{5}{2};\frac{7}{2};\frac{1}{2} (1-i \tan (e+f x))\right )}{5 f} \]

[Out]

((I/5)*Hypergeometric2F1[1, 5/2 + m, 7/2, (1 - I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m*(c - I*c*Tan[e + f*
x])^(5/2))/f

________________________________________________________________________________________

Rubi [A]  time = 0.112948, antiderivative size = 88, normalized size of antiderivative = 1.31, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3523, 70, 69} \[ \frac{i 2^m (c-i c \tan (e+f x))^{5/2} (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m \, _2F_1\left (\frac{5}{2},1-m;\frac{7}{2};\frac{1}{2} (1-i \tan (e+f x))\right )}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/5)*2^m*Hypergeometric2F1[5/2, 1 - m, 7/2, (1 - I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m*(c - I*c*Tan[e
+ f*x])^(5/2))/(f*(1 + I*Tan[e + f*x])^m)

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 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int (a+i a x)^{-1+m} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left (2^{-1+m} c (a+i a \tan (e+f x))^m \left (\frac{a+i a \tan (e+f x)}{a}\right )^{-m}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{i x}{2}\right )^{-1+m} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i 2^m \, _2F_1\left (\frac{5}{2},1-m;\frac{7}{2};\frac{1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2}}{5 f}\\ \end{align*}

Mathematica [B]  time = 63.1625, size = 141, normalized size = 2.1 \[ -\frac{i c 2^{m+\frac{3}{2}} \left (e^{i f x}\right )^m \left (\frac{c}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m \, _2F_1\left (-\frac{3}{2},1;m+1;-e^{2 i (e+f x)}\right ) \sec ^{-m}(e+f x) (\cos (f x)+i \sin (f x))^{-m} (a+i a \tan (e+f x))^m}{f m} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(3/2 + m)*c*(E^(I*f*x))^m*(c/(1 + E^((2*I)*(e + f*x))))^(3/2)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)
)))^m*Hypergeometric2F1[-3/2, 1, 1 + m, -E^((2*I)*(e + f*x))]*(a + I*a*Tan[e + f*x])^m)/(f*m*Sec[e + f*x]^m*(C
os[f*x] + I*Sin[f*x])^m)

________________________________________________________________________________________

Maple [F]  time = 0.441, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m} \left ( c-ic\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-I*c*tan(f*x+e))^(5/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{4 \, \sqrt{2} c^{2} \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{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, 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-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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-I*c*tan(f*x+e))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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