3.324 \(\int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx\)

Optimal. Leaf size=88 \[ \frac{(1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m (d \tan (e+f x))^{n+1} F_1(n+1;1-m,1;n+2;-i \tan (e+f x),i \tan (e+f x))}{d f (n+1)} \]

[Out]

(AppellF1[1 + n, 1 - m, 1, 2 + n, (-I)*Tan[e + f*x], I*Tan[e + f*x]]*(d*Tan[e + f*x])^(1 + n)*(a + I*a*Tan[e +
f*x])^m)/(d*f*(1 + n)*(1 + I*Tan[e + f*x])^m)

________________________________________________________________________________________

Rubi [A]  time = 0.0970577, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3564, 135, 133} \[ \frac{(1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m (d \tan (e+f x))^{n+1} F_1(n+1;1-m,1;n+2;-i \tan (e+f x),i \tan (e+f x))}{d f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(AppellF1[1 + n, 1 - m, 1, 2 + n, (-I)*Tan[e + f*x], I*Tan[e + f*x]]*(d*Tan[e + f*x])^(1 + n)*(a + I*a*Tan[e +
f*x])^m)/(d*f*(1 + n)*(1 + I*Tan[e + f*x])^m)

Rule 3564

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dis
t[(a*b)/f, Subst[Int[((a + x)^(m - 1)*(c + (d*x)/b)^n)/(b^2 + a*x), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b,
c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i d x}{a}\right )^n (a+x)^{-1+m}}{-a^2+a x} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac{\left (i a (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i d x}{a}\right )^n \left (1+\frac{x}{a}\right )^{-1+m}}{-a^2+a x} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac{F_1(1+n;1-m,1;2+n;-i \tan (e+f x),i \tan (e+f x)) (1+i \tan (e+f x))^{-m} (d \tan (e+f x))^{1+n} (a+i a \tan (e+f x))^m}{d f (1+n)}\\ \end{align*}

Mathematica [F]  time = 5.27207, size = 0, normalized size = 0. \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.835, size = 0, normalized size = 0. \begin{align*} \int \left ( d\tan \left ( fx+e \right ) \right ) ^{n} \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\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{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x +
2*I*e) + 1))^n, x)

________________________________________________________________________________________

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((d*tan(f*x+e))**n*(a+I*a*tan(f*x+e))**m,x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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