3.1062 \(\int \frac{(a+i a \tan (e+f x))^m}{\sqrt{c-i c \tan (e+f x)}} \, dx\)

Optimal. Leaf size=65 \[ -\frac{i (a+i a \tan (e+f x))^m \, _2F_1\left (1,m-\frac{1}{2};\frac{1}{2};\frac{1}{2} (1-i \tan (e+f x))\right )}{f \sqrt{c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.10828, antiderivative size = 86, normalized size of antiderivative = 1.32, 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} \[ -\frac{i 2^m (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m \, _2F_1\left (-\frac{1}{2},1-m;\frac{1}{2};\frac{1}{2} (1-i \tan (e+f x))\right )}{f \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*2^m*Hypergeometric2F1[-1/2, 1 - m, 1/2, (1 - I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(f*(1 + I*Tan[
e + f*x])^m*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 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 \frac{(a+i a \tan (e+f x))^m}{\sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(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 \frac{\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{1}{2},1-m;\frac{1}{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}{f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 6.21993, size = 141, normalized size = 2.17 \[ -\frac{i c 2^{m-\frac{3}{2}} \left (e^{i f x}\right )^m \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m \, _2F_1\left (1,\frac{3}{2};m+1;-e^{2 i (e+f x)}\right ) \sec ^{-m}(e+f x) (\cos (f x)+i \sin (f x))^{-m} (a+i a \tan (e+f x))^m}{f m \left (\frac{c}{1+e^{2 i (e+f x)}}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.476, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}\, 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))^(1/2),x)

[Out]

int((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e))^(1/2),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}}{\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((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \left (i \tan{\left (e + f x \right )} + 1\right )\right )^{m}}{\sqrt{- c \left (i \tan{\left (e + f x \right )} - 1\right )}}\, dx \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))**(1/2),x)

[Out]

Integral((a*(I*tan(e + f*x) + 1))**m/sqrt(-c*(I*tan(e + f*x) - 1)), x)

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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}}{\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((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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