3.851 \(\int (a+i a \tan (e+f x))^m (A+B \tan (e+f x)) (c-i c \tan (e+f x))^n \, dx\)

Optimal. Leaf size=150 \[ \frac{(B+i A) (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^n}{2 f n}-\frac{2^{n-1} (B (m-n)+i A (m+n)) (1-i \tan (e+f x))^{-n} (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^n \text{Hypergeometric2F1}\left (m,-n,m+1,\frac{1}{2} (1+i \tan (e+f x))\right )}{f m n} \]

[Out]

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

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Rubi [A]  time = 0.227203, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {3588, 79, 70, 69} \[ \frac{(B+i A) (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^n}{2 f n}-\frac{2^{n-1} (B (m-n)+i A (m+n)) (1-i \tan (e+f x))^{-n} (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^n \, _2F_1\left (m,-n;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{f m n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 33.303, size = 197, normalized size = 1.31 \[ \frac{2^{m+n-1} \left (e^{i f x}\right )^m \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m \left (\frac{c}{1+e^{2 i (e+f x)}}\right )^n \sec ^{-m}(e+f x) (\cos (f x)+i \sin (f x))^{-m} (a+i a \tan (e+f x))^m \left ((m+1) (B-i A) \text{Hypergeometric2F1}\left (1,-n,m+1,-e^{2 i (e+f x)}\right )-i m (A-i B) e^{2 i (e+f x)} \text{Hypergeometric2F1}\left (1,1-n,m+2,-e^{2 i (e+f x)}\right )\right )}{f m (m+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(-1 + m + n)*(E^(I*f*x))^m*(c/(1 + E^((2*I)*(e + f*x))))^n*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^m*((
-I)*(A - I*B)*E^((2*I)*(e + f*x))*m*Hypergeometric2F1[1, 1 - n, 2 + m, -E^((2*I)*(e + f*x))] + ((-I)*A + B)*(1
+ m)*Hypergeometric2F1[1, -n, 1 + m, -E^((2*I)*(e + f*x))])*(a + I*a*Tan[e + f*x])^m)/(f*m*(1 + m)*Sec[e + f*
x]^m*(Cos[f*x] + I*Sin[f*x])^m)

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Maple [F]  time = 1.095, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m} \left ( A+B\tan \left ( fx+e \right ) \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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} \left (\frac{2 \, c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n}}{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*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^n,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*(2*c/
(e^(2*I*f*x + 2*I*e) + 1))^n/(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*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**n,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}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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