3.1317 \(\int (c (d \tan (e+f x))^p)^n (a+i a \tan (e+f x))^m \, dx\)

Optimal. Leaf size=99 \[ \frac{\tan (e+f x) (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m F_1(n p+1;1-m,1;n p+2;-i \tan (e+f x),i \tan (e+f x)) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)} \]

[Out]

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

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Rubi [A]  time = 0.187287, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3578, 3564, 135, 133} \[ \frac{\tan (e+f x) (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m F_1(n p+1;1-m,1;n p+2;-i \tan (e+f x),i \tan (e+f x)) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3578

Int[((c_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(p_))^(n_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol]
:> Dist[(c^IntPart[n]*(c*(d*Tan[e + f*x])^p)^FracPart[n])/(d*Tan[e + f*x])^(p*FracPart[n]), Int[(a + b*Tan[e
+ f*x])^m*(d*Tan[e + f*x])^(n*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[n] &&  !Intege
rQ[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 \left (c (d \tan (e+f x))^p\right )^n (a+i a \tan (e+f x))^m \, dx &=\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \int (d \tan (e+f x))^{n p} (a+i a \tan (e+f x))^m \, dx\\ &=\frac{\left (i a^2 (d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i d x}{a}\right )^{n p} (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} (d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n (a+i a \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i d x}{a}\right )^{n p} \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 p;1-m,1;2+n p;-i \tan (e+f x),i \tan (e+f x)) (1+i \tan (e+f x))^{-m} \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n (a+i a \tan (e+f x))^m}{f (1+n p)}\\ \end{align*}

Mathematica [F]  time = 5.65535, size = 0, normalized size = 0. \[ \int \left (c (d \tan (e+f x))^p\right )^n (a+i a \tan (e+f x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.506, size = 0, normalized size = 0. \begin{align*} \int \left ( c \left ( d\tan \left ( fx+e \right ) \right ) ^{p} \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((c*(d*tan(f*x+e))^p)^n*(a+I*a*tan(f*x+e))^m,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (c \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 )^{p}\right )^{n} \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}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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