\(\int \frac {(c+d x^n)^3}{a+b x^n} \, dx\) [299]

Optimal result

Integrand size = 19, antiderivative size = 173 \[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\frac {d \left (a^2 d^2 \left (1+3 n+2 n^2\right )+b^2 c^2 \left (1+4 n+6 n^2\right )-a b c d \left (2+7 n+6 n^2\right )\right ) x}{b^3 (1+n) (1+2 n)}-\frac {d (a d (1+2 n)-b (c+4 c n)) x \left (c+d x^n\right )}{b^2 (1+n) (1+2 n)}+\frac {d x \left (c+d x^n\right )^2}{b (1+2 n)}+\frac {(b c-a d)^3 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {427, 542, 396, 251} \[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\frac {d x \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (6 n^2+7 n+2\right )+b^2 c^2 \left (6 n^2+4 n+1\right )\right )}{b^3 (n+1) (2 n+1)}+\frac {x (b c-a d)^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a b^3}-\frac {d x \left (c+d x^n\right ) (a d (2 n+1)-b (4 c n+c))}{b^2 (n+1) (2 n+1)}+\frac {d x \left (c+d x^n\right )^2}{b (2 n+1)} \]

[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 251

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])

Rule 396

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 427

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 542

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 1.49 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.60 \[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\frac {x \left (3 c^2 d x^n \Phi \left (-\frac {b x^n}{a},1,1+\frac {1}{n}\right )+3 c d^2 x^{2 n} \Phi \left (-\frac {b x^n}{a},1,2+\frac {1}{n}\right )+d^3 x^{3 n} \Phi \left (-\frac {b x^n}{a},1,3+\frac {1}{n}\right )+c^3 \Phi \left (-\frac {b x^n}{a},1,\frac {1}{n}\right )\right )}{a n} \]

[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)

Maple [F]

\[\int \frac {\left (c +d \,x^{n}\right )^{3}}{a +b \,x^{n}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{3}}{b x^{n} + a} \,d x } \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.38 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.08 \[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\frac {a^{\frac {1}{n}} a^{-1 - \frac {1}{n}} c^{3} x \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, \frac {1}{n}\right ) \Gamma \left (\frac {1}{n}\right )}{n^{2} \Gamma \left (1 + \frac {1}{n}\right )} + \frac {3 a^{-4 - \frac {1}{n}} a^{3 + \frac {1}{n}} d^{3} x^{3 n + 1} \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, 3 + \frac {1}{n}\right ) \Gamma \left (3 + \frac {1}{n}\right )}{n \Gamma \left (4 + \frac {1}{n}\right )} + \frac {a^{-4 - \frac {1}{n}} a^{3 + \frac {1}{n}} d^{3} x^{3 n + 1} \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, 3 + \frac {1}{n}\right ) \Gamma \left (3 + \frac {1}{n}\right )}{n^{2} \Gamma \left (4 + \frac {1}{n}\right )} + \frac {6 a^{-3 - \frac {1}{n}} a^{2 + \frac {1}{n}} c d^{2} x^{2 n + 1} \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, 2 + \frac {1}{n}\right ) \Gamma \left (2 + \frac {1}{n}\right )}{n \Gamma \left (3 + \frac {1}{n}\right )} + \frac {3 a^{-3 - \frac {1}{n}} a^{2 + \frac {1}{n}} c d^{2} x^{2 n + 1} \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, 2 + \frac {1}{n}\right ) \Gamma \left (2 + \frac {1}{n}\right )}{n^{2} \Gamma \left (3 + \frac {1}{n}\right )} - \frac {3 a^{- \frac {1}{n}} a^{1 + \frac {1}{n}} b^{\frac {1}{n}} b^{-1 - \frac {1}{n}} c^{2} d x \Phi \left (\frac {a x^{- n} e^{i \pi }}{b}, 1, \frac {e^{i \pi }}{n}\right ) \Gamma \left (\frac {1}{n}\right )}{a n^{2} \Gamma \left (1 + \frac {1}{n}\right )} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{3}}{b x^{n} + a} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{3}}{b x^{n} + a} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^n\right )^3}{a+b x^n} \, dx=\int \frac {{\left (c+d\,x^n\right )}^3}{a+b\,x^n} \,d x \]

[In]

int((c + d*x^n)^3/(a + b*x^n),x)

[Out]

int((c + d*x^n)^3/(a + b*x^n), x)