3.299 \(\int \frac{(c+d x^n)^3}{a+b x^n} \, dx\)

Optimal. Leaf size=173 \[ \frac{d x \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (6 n^2+7 n+2\right )+b^2 c^2 \left (6 n^2+4 n+1\right )\right )}{b^3 (n+1) (2 n+1)}+\frac{x (b c-a d)^3 \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a b^3}-\frac{d x \left (c+d x^n\right ) (a d (2 n+1)-b (4 c n+c))}{b^2 (n+1) (2 n+1)}+\frac{d x \left (c+d x^n\right )^2}{b (2 n+1)} \]

[Out]

(d*(a^2*d^2*(1 + 3*n + 2*n^2) + b^2*c^2*(1 + 4*n + 6*n^2) - a*b*c*d*(2 + 7*n + 6*n^2))*x)/(b^3*(1 + n)*(1 + 2*
n)) - (d*(a*d*(1 + 2*n) - b*(c + 4*c*n))*x*(c + d*x^n))/(b^2*(1 + n)*(1 + 2*n)) + (d*x*(c + d*x^n)^2)/(b*(1 +
2*n)) + ((b*c - a*d)^3*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*b^3)

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Rubi [A]  time = 0.267249, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {416, 528, 388, 245} \[ \frac{d x \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (6 n^2+7 n+2\right )+b^2 c^2 \left (6 n^2+4 n+1\right )\right )}{b^3 (n+1) (2 n+1)}+\frac{x (b c-a d)^3 \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a b^3}-\frac{d x \left (c+d x^n\right ) (a d (2 n+1)-b (4 c n+c))}{b^2 (n+1) (2 n+1)}+\frac{d x \left (c+d x^n\right )^2}{b (2 n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^n)^3/(a + b*x^n),x]

[Out]

(d*(a^2*d^2*(1 + 3*n + 2*n^2) + b^2*c^2*(1 + 4*n + 6*n^2) - a*b*c*d*(2 + 7*n + 6*n^2))*x)/(b^3*(1 + n)*(1 + 2*
n)) - (d*(a*d*(1 + 2*n) - b*(c + 4*c*n))*x*(c + d*x^n))/(b^2*(1 + n)*(1 + 2*n)) + (d*x*(c + d*x^n)^2)/(b*(1 +
2*n)) + ((b*c - a*d)^3*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*b^3)

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 528

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]

Rule 388

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

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\left (c+d x^n\right )^3}{a+b x^n} \, dx &=\frac{d x \left (c+d x^n\right )^2}{b (1+2 n)}+\frac{\int \frac{\left (c+d x^n\right ) \left (-c (a d-b (c+2 c n))-d (a d (1+2 n)-b (c+4 c n)) x^n\right )}{a+b x^n} \, dx}{b (1+2 n)}\\ &=-\frac{d (a d (1+2 n)-b (c+4 c n)) x \left (c+d x^n\right )}{b^2 (1+n) (1+2 n)}+\frac{d x \left (c+d x^n\right )^2}{b (1+2 n)}+\frac{\int \frac{c \left (a^2 d^2 (1+2 n)-a b c d (2+5 n)+b^2 c^2 \left (1+3 n+2 n^2\right )\right )+d \left (a^2 d^2 \left (1+3 n+2 n^2\right )+b^2 c^2 \left (1+4 n+6 n^2\right )-a b c d \left (2+7 n+6 n^2\right )\right ) x^n}{a+b x^n} \, dx}{b^2 (1+n) (1+2 n)}\\ &=\frac{d \left (a^2 d^2 \left (1+3 n+2 n^2\right )+b^2 c^2 \left (1+4 n+6 n^2\right )-a b c d \left (2+7 n+6 n^2\right )\right ) x}{b^3 (1+n) (1+2 n)}-\frac{d (a d (1+2 n)-b (c+4 c n)) x \left (c+d x^n\right )}{b^2 (1+n) (1+2 n)}+\frac{d x \left (c+d x^n\right )^2}{b (1+2 n)}+\frac{(b c-a d)^3 \int \frac{1}{a+b x^n} \, dx}{b^3}\\ &=\frac{d \left (a^2 d^2 \left (1+3 n+2 n^2\right )+b^2 c^2 \left (1+4 n+6 n^2\right )-a b c d \left (2+7 n+6 n^2\right )\right ) x}{b^3 (1+n) (1+2 n)}-\frac{d (a d (1+2 n)-b (c+4 c n)) x \left (c+d x^n\right )}{b^2 (1+n) (1+2 n)}+\frac{d x \left (c+d x^n\right )^2}{b (1+2 n)}+\frac{(b c-a d)^3 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a b^3}\\ \end{align*}

Mathematica [C]  time = 0.982742, size = 104, normalized size = 0.6 \[ \frac{x \left (3 c^2 d x^n \Phi \left (-\frac{b x^n}{a},1,1+\frac{1}{n}\right )+c^3 \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+3 c d^2 x^{2 n} \Phi \left (-\frac{b x^n}{a},1,2+\frac{1}{n}\right )+d^3 x^{3 n} \Phi \left (-\frac{b x^n}{a},1,3+\frac{1}{n}\right )\right )}{a n} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^n)^3/(a + b*x^n),x]

[Out]

(x*(3*c^2*d*x^n*HurwitzLerchPhi[-((b*x^n)/a), 1, 1 + n^(-1)] + 3*c*d^2*x^(2*n)*HurwitzLerchPhi[-((b*x^n)/a), 1
, 2 + n^(-1)] + d^3*x^(3*n)*HurwitzLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + c^3*HurwitzLerchPhi[-((b*x^n)/a), 1
, n^(-1)]))/(a*n)

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Maple [F]  time = 0.382, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c+d{x}^{n} \right ) ^{3}}{a+b{x}^{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*x^n)^3/(a+b*x^n),x)

[Out]

int((c+d*x^n)^3/(a+b*x^n),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \int \frac{1}{b^{4} x^{n} + a b^{3}}\,{d x} + \frac{b^{2} d^{3}{\left (n + 1\right )} x x^{2 \, n} +{\left (3 \, b^{2} c d^{2}{\left (2 \, n + 1\right )} - a b d^{3}{\left (2 \, n + 1\right )}\right )} x x^{n} +{\left (3 \,{\left (2 \, n^{2} + 3 \, n + 1\right )} b^{2} c^{2} d - 3 \,{\left (2 \, n^{2} + 3 \, n + 1\right )} a b c d^{2} +{\left (2 \, n^{2} + 3 \, n + 1\right )} a^{2} d^{3}\right )} x}{{\left (2 \, n^{2} + 3 \, n + 1\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^n)^3/(a+b*x^n),x, algorithm="maxima")

[Out]

(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*integrate(1/(b^4*x^n + a*b^3), x) + (b^2*d^3*(n + 1)*x*x^(
2*n) + (3*b^2*c*d^2*(2*n + 1) - a*b*d^3*(2*n + 1))*x*x^n + (3*(2*n^2 + 3*n + 1)*b^2*c^2*d - 3*(2*n^2 + 3*n + 1
)*a*b*c*d^2 + (2*n^2 + 3*n + 1)*a^2*d^3)*x)/((2*n^2 + 3*n + 1)*b^3)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d^{3} x^{3 \, n} + 3 \, c d^{2} x^{2 \, n} + 3 \, c^{2} d x^{n} + c^{3}}{b x^{n} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^n)^3/(a+b*x^n),x, algorithm="fricas")

[Out]

integral((d^3*x^(3*n) + 3*c*d^2*x^(2*n) + 3*c^2*d*x^n + c^3)/(b*x^n + a), x)

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Sympy [C]  time = 4.7349, size = 269, normalized size = 1.55 \begin{align*} - \frac{3 c^{2} d x \Phi \left (\frac{a x^{- n} e^{i \pi }}{b}, 1, \frac{e^{i \pi }}{n}\right ) \Gamma \left (\frac{1}{n}\right )}{b n^{2} \Gamma \left (1 + \frac{1}{n}\right )} + \frac{c^{3} x \Phi \left (\frac{b x^{n} e^{i \pi }}{a}, 1, \frac{1}{n}\right ) \Gamma \left (\frac{1}{n}\right )}{a n^{2} \Gamma \left (1 + \frac{1}{n}\right )} + \frac{6 c d^{2} x x^{2 n} \Phi \left (\frac{b x^{n} e^{i \pi }}{a}, 1, 2 + \frac{1}{n}\right ) \Gamma \left (2 + \frac{1}{n}\right )}{a n \Gamma \left (3 + \frac{1}{n}\right )} + \frac{3 c d^{2} x x^{2 n} \Phi \left (\frac{b x^{n} e^{i \pi }}{a}, 1, 2 + \frac{1}{n}\right ) \Gamma \left (2 + \frac{1}{n}\right )}{a n^{2} \Gamma \left (3 + \frac{1}{n}\right )} + \frac{3 d^{3} x x^{3 n} \Phi \left (\frac{b x^{n} e^{i \pi }}{a}, 1, 3 + \frac{1}{n}\right ) \Gamma \left (3 + \frac{1}{n}\right )}{a n \Gamma \left (4 + \frac{1}{n}\right )} + \frac{d^{3} x x^{3 n} \Phi \left (\frac{b x^{n} e^{i \pi }}{a}, 1, 3 + \frac{1}{n}\right ) \Gamma \left (3 + \frac{1}{n}\right )}{a n^{2} \Gamma \left (4 + \frac{1}{n}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x**n)**3/(a+b*x**n),x)

[Out]

-3*c**2*d*x*lerchphi(a*x**(-n)*exp_polar(I*pi)/b, 1, exp_polar(I*pi)/n)*gamma(1/n)/(b*n**2*gamma(1 + 1/n)) + c
**3*x*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, 1/n)*gamma(1/n)/(a*n**2*gamma(1 + 1/n)) + 6*c*d**2*x*x**(2*n)*lerc
hphi(b*x**n*exp_polar(I*pi)/a, 1, 2 + 1/n)*gamma(2 + 1/n)/(a*n*gamma(3 + 1/n)) + 3*c*d**2*x*x**(2*n)*lerchphi(
b*x**n*exp_polar(I*pi)/a, 1, 2 + 1/n)*gamma(2 + 1/n)/(a*n**2*gamma(3 + 1/n)) + 3*d**3*x*x**(3*n)*lerchphi(b*x*
*n*exp_polar(I*pi)/a, 1, 3 + 1/n)*gamma(3 + 1/n)/(a*n*gamma(4 + 1/n)) + d**3*x*x**(3*n)*lerchphi(b*x**n*exp_po
lar(I*pi)/a, 1, 3 + 1/n)*gamma(3 + 1/n)/(a*n**2*gamma(4 + 1/n))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{n} + c\right )}^{3}}{b x^{n} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^n)^3/(a+b*x^n),x, algorithm="giac")

[Out]

integrate((d*x^n + c)^3/(b*x^n + a), x)