\(\int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx\) [1174]

Optimal result

Integrand size = 28, antiderivative size = 157 \[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx=\frac {a^3 (i c-d (5+2 n)) (c+d \tan (e+f x))^{1+n}}{d^2 f (1+n) (2+n)}+\frac {4 a^3 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d \tan (e+f x)}{c-i d}\right ) (c+d \tan (e+f x))^{1+n}}{(i c+d) f (1+n)}-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{1+n}}{d f (2+n)} \]

[Out]

a^3*(I*c-d*(5+2*n))*(c+d*tan(f*x+e))^(1+n)/d^2/f/(1+n)/(2+n)+4*a^3*hypergeom([1, 1+n],[2+n],(c+d*tan(f*x+e))/(
c-I*d))*(c+d*tan(f*x+e))^(1+n)/(I*c+d)/f/(1+n)-(a^3+I*a^3*tan(f*x+e))*(c+d*tan(f*x+e))^(1+n)/d/f/(2+n)

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3637, 3673, 3618, 70} \[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx=\frac {a^3 (-d (2 n+5)+i c) (c+d \tan (e+f x))^{n+1}}{d^2 f (n+1) (n+2)}+\frac {4 a^3 (c+d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {c+d \tan (e+f x)}{c-i d}\right )}{f (n+1) (d+i c)}-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{n+1}}{d f (n+2)} \]

[In]

Int[(a + I*a*Tan[e + f*x])^3*(c + d*Tan[e + f*x])^n,x]

[Out]

(a^3*(I*c - d*(5 + 2*n))*(c + d*Tan[e + f*x])^(1 + n))/(d^2*f*(1 + n)*(2 + n)) + (4*a^3*Hypergeometric2F1[1, 1
+ n, 2 + n, (c + d*Tan[e + f*x])/(c - I*d)]*(c + d*Tan[e + f*x])^(1 + n))/((I*c + d)*f*(1 + n)) - ((a^3 + I*a
^3*Tan[e + f*x])*(c + d*Tan[e + f*x])^(1 + n))/(d*f*(2 + n))

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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3673

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{1+n}}{d f (2+n)}+\frac {a \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^n (a (i c+d (3+2 n))+a (c+i d (5+2 n)) \tan (e+f x)) \, dx}{d (2+n)} \\ & = \frac {a^3 (i c-d (5+2 n)) (c+d \tan (e+f x))^{1+n}}{d^2 f (1+n) (2+n)}-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{1+n}}{d f (2+n)}+\frac {a \int (c+d \tan (e+f x))^n \left (4 a^2 d (2+n)+4 i a^2 d (2+n) \tan (e+f x)\right ) \, dx}{d (2+n)} \\ & = \frac {a^3 (i c-d (5+2 n)) (c+d \tan (e+f x))^{1+n}}{d^2 f (1+n) (2+n)}-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{1+n}}{d f (2+n)}+\frac {\left (16 i a^5 d (2+n)\right ) \text {Subst}\left (\int \frac {\left (c-\frac {i x}{4 a^2 (2+n)}\right )^n}{-16 a^4 d^2 (2+n)^2+4 a^2 d (2+n) x} \, dx,x,4 i a^2 d (2+n) \tan (e+f x)\right )}{f} \\ & = \frac {a^3 (i c-d (5+2 n)) (c+d \tan (e+f x))^{1+n}}{d^2 f (1+n) (2+n)}+\frac {4 a^3 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d \tan (e+f x)}{c-i d}\right ) (c+d \tan (e+f x))^{1+n}}{(i c+d) f (1+n)}-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{1+n}}{d f (2+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.76 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.71 \[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx=-\frac {a^3 (c+d \tan (e+f x))^{1+n} \left (-4 d^2 (2+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d \tan (e+f x)}{c-i d}\right )+(c-i d) (c+3 i d (2+n)-d (1+n) \tan (e+f x))\right )}{d^2 (i c+d) f (1+n) (2+n)} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])^3*(c + d*Tan[e + f*x])^n,x]

[Out]

-((a^3*(c + d*Tan[e + f*x])^(1 + n)*(-4*d^2*(2 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (c + d*Tan[e + f*x])/(c
- I*d)] + (c - I*d)*(c + (3*I)*d*(2 + n) - d*(1 + n)*Tan[e + f*x])))/(d^2*(I*c + d)*f*(1 + n)*(2 + n)))

Maple [F]

\[\int \left (a +i a \tan \left (f x +e \right )\right )^{3} \left (c +d \tan \left (f x +e \right )\right )^{n}d x\]

[In]

int((a+I*a*tan(f*x+e))^3*(c+d*tan(f*x+e))^n,x)

[Out]

int((a+I*a*tan(f*x+e))^3*(c+d*tan(f*x+e))^n,x)

Fricas [F]

\[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (d \tan \left (f x + e\right ) + c\right )}^{n} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(c+d*tan(f*x+e))^n,x, algorithm="fricas")

[Out]

integral(8*a^3*(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))^n*e^(6*I*f*x + 6*I*e)/(e^
(6*I*f*x + 6*I*e) + 3*e^(4*I*f*x + 4*I*e) + 3*e^(2*I*f*x + 2*I*e) + 1), x)

Sympy [F]

\[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx=- i a^{3} \left (\int i \left (c + d \tan {\left (e + f x \right )}\right )^{n}\, dx + \int \left (- 3 \left (c + d \tan {\left (e + f x \right )}\right )^{n} \tan {\left (e + f x \right )}\right )\, dx + \int \left (c + d \tan {\left (e + f x \right )}\right )^{n} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 3 i \left (c + d \tan {\left (e + f x \right )}\right )^{n} \tan ^{2}{\left (e + f x \right )}\right )\, dx\right ) \]

[In]

integrate((a+I*a*tan(f*x+e))**3*(c+d*tan(f*x+e))**n,x)

[Out]

-I*a**3*(Integral(I*(c + d*tan(e + f*x))**n, x) + Integral(-3*(c + d*tan(e + f*x))**n*tan(e + f*x), x) + Integ
ral((c + d*tan(e + f*x))**n*tan(e + f*x)**3, x) + Integral(-3*I*(c + d*tan(e + f*x))**n*tan(e + f*x)**2, x))

Maxima [F]

\[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (d \tan \left (f x + e\right ) + c\right )}^{n} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(c+d*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^3*(d*tan(f*x + e) + c)^n, x)

Giac [F]

\[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (d \tan \left (f x + e\right ) + c\right )}^{n} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^3*(c+d*tan(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^3*(d*tan(f*x + e) + c)^n, x)

Mupad [F(-1)]

Timed out. \[ \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx=\int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \]

[In]

int((a + a*tan(e + f*x)*1i)^3*(c + d*tan(e + f*x))^n,x)

[Out]

int((a + a*tan(e + f*x)*1i)^3*(c + d*tan(e + f*x))^n, x)