3.852 \(\int (a+i a \tan (e+f x))^{1+m} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{-1-m} \, dx\)

Optimal. Leaf size=147 \[ \frac{a B 2^m (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{-m} \text{Hypergeometric2F1}\left (-m,-m,1-m,\frac{1}{2} (1-i \tan (e+f x))\right )}{c f m}-\frac{(B+i A) (a+i a \tan (e+f x))^{m+1} (c-i c \tan (e+f x))^{-m-1}}{2 f (m+1)} \]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(1 + m)*(c - I*c*Tan[e + f*x])^(-1 - m))/(2*f*(1 + m)) + (2^m*a*B*Hypergeom
etric2F1[-m, -m, 1 - m, (1 - I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(c*f*m*(1 + I*Tan[e + f*x])^m*(c - I
*c*Tan[e + f*x])^m)

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Rubi [A]  time = 0.224891, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.085, Rules used = {3588, 79, 70, 69} \[ \frac{a B 2^m (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{-m} \, _2F_1\left (-m,-m;1-m;\frac{1}{2} (1-i \tan (e+f x))\right )}{c f m}-\frac{(B+i A) (a+i a \tan (e+f x))^{m+1} (c-i c \tan (e+f x))^{-m-1}}{2 f (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(1 + m)*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(-1 - m),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(1 + m)*(c - I*c*Tan[e + f*x])^(-1 - m))/(2*f*(1 + m)) + (2^m*a*B*Hypergeom
etric2F1[-m, -m, 1 - m, (1 - I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(c*f*m*(1 + I*Tan[e + f*x])^m*(c - I
*c*Tan[e + f*x])^m)

Rule 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((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)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

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

Mathematica [A]  time = 84.1838, size = 177, normalized size = 1.2 \[ \frac{a e^{i (e+2 f x)} \left (e^{i f x}\right )^m \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m (\tan (e+f x)-i) \left (\frac{c}{1+e^{2 i (e+f x)}}\right )^{-m} \sec ^{-m-1}(e+f x) (\cos (f x)+i \sin (f x))^{-m-1} (a+i a \tan (e+f x))^m \left (2 i B \text{Hypergeometric2F1}\left (1,m+1,m+2,-e^{2 i (e+f x)}\right )+A-i B\right )}{2 c f (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(1 + m)*(A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(-1 - m),x]

[Out]

(a*E^(I*(e + 2*f*x))*(E^(I*f*x))^m*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^m*(A - I*B + (2*I)*B*Hypergeome
tric2F1[1, 1 + m, 2 + m, -E^((2*I)*(e + f*x))])*Sec[e + f*x]^(-1 - m)*(Cos[f*x] + I*Sin[f*x])^(-1 - m)*(-I + T
an[e + f*x])*(a + I*a*Tan[e + f*x])^m)/(2*c*(c/(1 + E^((2*I)*(e + f*x))))^m*f*(1 + m))

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Maple [F]  time = 1.369, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{1+m} \left ( A+B\tan \left ( fx+e \right ) \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-1-m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(-1-m),x)

[Out]

int((a+I*a*tan(f*x+e))^(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(-1-m),x)

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Maxima [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))^(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(-1-m),x, algorithm="maxima")

[Out]

Timed out

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

[Out]

integral(((A - I*B)*e^(2*I*f*x + 2*I*e) + A + I*B)*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^(m + 1)
*(2*c/(e^(2*I*f*x + 2*I*e) + 1))^(-m - 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))**(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(-1-m),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m + 1}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{-m - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1+m)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(-1-m),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(m + 1)*(-I*c*tan(f*x + e) + c)^(-m - 1), x)