3.1318 \(\int (c (d \tan (e+f x))^p)^n (a+i a \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=132 \[ \frac{4 a^3 \tan (e+f x) \, _2F_1(1,n p+1;n p+2;i \tan (e+f x)) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)}-\frac{i a^3 \tan ^2(e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+2)}-\frac{3 a^3 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)} \]

[Out]

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

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Rubi [A]  time = 0.310687, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1586, 6677, 88, 64, 43} \[ \frac{4 a^3 \tan (e+f x) \, _2F_1(1,n p+1;n p+2;i \tan (e+f x)) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)}-\frac{i a^3 \tan ^2(e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+2)}-\frac{3 a^3 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 6677

Int[(u_)*((c_.)*((a_.) + (b_.)*(x_))^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c*(a + b*x)^n)^FracPart[p])/
(a + b*x)^(n*FracPart[p]), Int[u*(a + b*x)^(n*p), x], x] /; FreeQ[{a, b, c, n, p}, x] &&  !IntegerQ[p] &&  !Ma
tchQ[u, x^(n1_.)*(v_.) /; EqQ[n, n1 + 1]]

Rule 88

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

Rule 64

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \left (c (d \tan (e+f x))^p\right )^n (a+i a \tan (e+f x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (c (d x)^p\right )^n (a+i a x)^3}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (c (d x)^p\right )^n (a+i a x)^2}{\frac{1}{a}-\frac{i x}{a}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int \frac{(d x)^{n p} (a+i a x)^2}{\frac{1}{a}-\frac{i x}{a}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int \left (-2 a^3 (d x)^{n p}+\frac{4 a^2 (d x)^{n p}}{\frac{1}{a}-\frac{i x}{a}}-a^2 (d x)^{n p} (a+i a x)\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{2 a^3 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}-\frac{\left (a^2 (d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int (d x)^{n p} (a+i a x) \, dx,x,\tan (e+f x)\right )}{f}+\frac{\left (4 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{(d x)^{n p}}{\frac{1}{a}-\frac{i x}{a}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{2 a^3 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac{4 a^3 \, _2F_1(1,1+n p;2+n p;i \tan (e+f x)) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}-\frac{\left (a^2 (d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \operatorname{Subst}\left (\int \left (a (d x)^{n p}+\frac{i a (d x)^{1+n p}}{d}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{3 a^3 \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}+\frac{4 a^3 \, _2F_1(1,1+n p;2+n p;i \tan (e+f x)) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (1+n p)}-\frac{i a^3 \tan ^2(e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (2+n p)}\\ \end{align*}

Mathematica [A]  time = 4.36246, size = 240, normalized size = 1.82 \[ \frac{e^{-3 i e} 2^{-n p} \cos ^3(e+f x) (a+i a \tan (e+f x))^3 \left (-\frac{i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )^{n p+1} \left (2 (n p+2) \left (1+e^{2 i (e+f x)}\right )^{n p+2} \, _2F_1\left (n p+1,n p+1;n p+2;\frac{1}{2} \left (1-e^{2 i (e+f x)}\right )\right )-2^{n p} \left ((4 n p+7) e^{2 i (e+f x)}+2 n p+5\right )\right ) \tan ^{-n p}(e+f x) \left (c (d \tan (e+f x))^p\right )^n}{f (n p+1) (n p+2) \left (1+e^{2 i (e+f x)}\right ) (\cos (f x)+i \sin (f x))^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((((-I)*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x))))^(1 + n*p)*Cos[e + f*x]^3*(-(2^(n*p)*(5 + 2*n*p
+ E^((2*I)*(e + f*x))*(7 + 4*n*p))) + 2*(1 + E^((2*I)*(e + f*x)))^(2 + n*p)*(2 + n*p)*Hypergeometric2F1[1 + n*
p, 1 + n*p, 2 + n*p, (1 - E^((2*I)*(e + f*x)))/2])*(c*(d*Tan[e + f*x])^p)^n*(a + I*a*Tan[e + f*x])^3)/(2^(n*p)
*E^((3*I)*e)*(1 + E^((2*I)*(e + f*x)))*f*(1 + n*p)*(2 + n*p)*(Cos[f*x] + I*Sin[f*x])^3*Tan[e + f*x]^(n*p))

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Maple [F]  time = 0.769, 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 ) ^{3}\, 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))^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} \left (\left (d \tan \left (f x + e\right )\right )^{p} c\right )^{n}\,{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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{8 \, \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} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )}}{e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, 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((c*(d*tan(f*x+e))^p)^n*(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \left (c \left (d \tan{\left (e + f x \right )}\right )^{p}\right )^{n}\, dx + \int - 3 \left (c \left (d \tan{\left (e + f x \right )}\right )^{p}\right )^{n} \tan ^{2}{\left (e + f x \right )}\, dx + \int 3 i \left (c \left (d \tan{\left (e + f x \right )}\right )^{p}\right )^{n} \tan{\left (e + f x \right )}\, dx + \int - i \left (c \left (d \tan{\left (e + f x \right )}\right )^{p}\right )^{n} \tan ^{3}{\left (e + f x \right )}\, dx\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))**3,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} \left (\left (d \tan \left (f x + e\right )\right )^{p} c\right )^{n}\,{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))^3,x, algorithm="giac")

[Out]

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