\(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx\) [715]

Optimal result

Integrand size = 41, antiderivative size = 115 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=\frac {(i A (2-n)+B (2+n)) \operatorname {Hypergeometric2F1}\left (2,n,1+n,\frac {1}{2} (1-i \tan (e+f x))\right ) (c-i c \tan (e+f x))^n}{16 a^2 f n}+\frac {(i A-B) (c-i c \tan (e+f x))^n}{4 a^2 f (1+i \tan (e+f x))^2} \]

[Out]

1/16*(I*A*(2-n)+B*(2+n))*hypergeom([2, n],[1+n],1/2-1/2*I*tan(f*x+e))*(c-I*c*tan(f*x+e))^n/a^2/f/n+1/4*(I*A-B)
*(c-I*c*tan(f*x+e))^n/a^2/f/(1+I*tan(f*x+e))^2

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {3669, 79, 70} \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=\frac {(B (n+2)+i A (2-n)) (c-i c \tan (e+f x))^n \operatorname {Hypergeometric2F1}\left (2,n,n+1,\frac {1}{2} (1-i \tan (e+f x))\right )}{16 a^2 f n}+\frac {(-B+i A) (c-i c \tan (e+f x))^n}{4 a^2 f (1+i \tan (e+f x))^2} \]

[In]

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

[Out]

((I*A*(2 - n) + B*(2 + n))*Hypergeometric2F1[2, n, 1 + n, (1 - I*Tan[e + f*x])/2]*(c - I*c*Tan[e + f*x])^n)/(1
6*a^2*f*n) + ((I*A - B)*(c - I*c*Tan[e + f*x])^n)/(4*a^2*f*(1 + I*Tan[e + f*x])^2)

Rule 70

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

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)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 3669

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {(A+B x) (c-i c x)^{-1+n}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(i A-B) (c-i c \tan (e+f x))^n}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(c (A (2-n)-i B (2+n))) \text {Subst}\left (\int \frac {(c-i c x)^{-1+n}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{4 f} \\ & = \frac {(i A (2-n)+B (2+n)) \operatorname {Hypergeometric2F1}\left (2,n,1+n,\frac {1}{2} (1-i \tan (e+f x))\right ) (c-i c \tan (e+f x))^n}{16 a^2 f n}+\frac {(i A-B) (c-i c \tan (e+f x))^n}{4 a^2 f (1+i \tan (e+f x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.48 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.32 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=\frac {(c-i c \tan (e+f x))^n \left (-i A (-3+n)+B (1+n)+\frac {i (-1+n) (A (-2+n)+i B (2+n)) \operatorname {Hypergeometric2F1}\left (1,n,1+n,-\frac {1}{2} i (i+\tan (e+f x))\right )}{n}+(-i A (-3+n)+B (1+n)) (\cos (2 (e+f x))-i \sin (2 (e+f x)))-\frac {2 (A+i B) (i+\tan (e+f x))}{(-i+\tan (e+f x))^2}\right )}{16 a^2 f} \]

[In]

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

[Out]

((c - I*c*Tan[e + f*x])^n*((-I)*A*(-3 + n) + B*(1 + n) + (I*(-1 + n)*(A*(-2 + n) + I*B*(2 + n))*Hypergeometric
2F1[1, n, 1 + n, (-1/2*I)*(I + Tan[e + f*x])])/n + ((-I)*A*(-3 + n) + B*(1 + n))*(Cos[2*(e + f*x)] - I*Sin[2*(
e + f*x)]) - (2*(A + I*B)*(I + Tan[e + f*x]))/(-I + Tan[e + f*x])^2))/(16*a^2*f)

Maple [F]

\[\int \frac {\left (A +B \tan \left (f x +e \right )\right ) \left (c -i c \tan \left (f x +e \right )\right )^{n}}{\left (a +i a \tan \left (f x +e \right )\right )^{2}}d x\]

[In]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,x)

[Out]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,x)

Fricas [F]

\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{n}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

integral(1/4*((A - I*B)*e^(4*I*f*x + 4*I*e) + 2*A*e^(2*I*f*x + 2*I*e) + A + I*B)*(2*c/(e^(2*I*f*x + 2*I*e) + 1
))^n*e^(-4*I*f*x - 4*I*e)/a^2, x)

Sympy [F]

\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {A \left (- i c \tan {\left (e + f x \right )} + c\right )^{n}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \frac {B \left (- i c \tan {\left (e + f x \right )} + c\right )^{n} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**n/(a+I*a*tan(f*x+e))**2,x)

[Out]

-(Integral(A*(-I*c*tan(e + f*x) + c)**n/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(B*(-I*c*tan(e
+ f*x) + c)**n*tan(e + f*x)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x))/a**2

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{n}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^n/(I*a*tan(f*x + e) + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^n}{(a+i a \tan (e+f x))^2} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^n}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]

[In]

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^n)/(a + a*tan(e + f*x)*1i)^2,x)

[Out]

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^n)/(a + a*tan(e + f*x)*1i)^2, x)