3.298 \(\int \frac{(c+d x^n)^4}{a+b x^n} \, dx\)

Optimal. Leaf size=310 \[ -\frac{d x \left (c+d x^n\right ) \left (-a^2 d^2 \left (6 n^2+5 n+1\right )+2 a b c d (3 n+1)^2-b^2 c^2 \left (18 n^2+7 n+1\right )\right )}{b^3 (n+1) (2 n+1) (3 n+1)}-\frac{d x \left (-a^2 b c d^2 \left (24 n^3+38 n^2+19 n+3\right )+a^3 d^3 \left (6 n^3+11 n^2+6 n+1\right )+a b^2 c^2 d \left (36 n^3+45 n^2+20 n+3\right )-b^3 c^3 \left (24 n^3+18 n^2+7 n+1\right )\right )}{b^4 (n+1) (2 n+1) (3 n+1)}+\frac{x (b c-a d)^4 \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a b^4}-\frac{d x \left (c+d x^n\right )^2 (a d (3 n+1)-b (6 c n+c))}{b^2 \left (6 n^2+5 n+1\right )}+\frac{d x \left (c+d x^n\right )^3}{b (3 n+1)} \]

[Out]

-((d*(a^3*d^3*(1 + 6*n + 11*n^2 + 6*n^3) - b^3*c^3*(1 + 7*n + 18*n^2 + 24*n^3) - a^2*b*c*d^2*(3 + 19*n + 38*n^
2 + 24*n^3) + a*b^2*c^2*d*(3 + 20*n + 45*n^2 + 36*n^3))*x)/(b^4*(1 + n)*(1 + 2*n)*(1 + 3*n))) - (d*(2*a*b*c*d*
(1 + 3*n)^2 - a^2*d^2*(1 + 5*n + 6*n^2) - b^2*c^2*(1 + 7*n + 18*n^2))*x*(c + d*x^n))/(b^3*(1 + n)*(1 + 2*n)*(1
+ 3*n)) - (d*(a*d*(1 + 3*n) - b*(c + 6*c*n))*x*(c + d*x^n)^2)/(b^2*(1 + 5*n + 6*n^2)) + (d*x*(c + d*x^n)^3)/(
b*(1 + 3*n)) + ((b*c - a*d)^4*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*b^4)

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Rubi [A]  time = 0.501013, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {416, 528, 388, 245} \[ -\frac{d x \left (c+d x^n\right ) \left (-a^2 d^2 \left (6 n^2+5 n+1\right )+2 a b c d (3 n+1)^2-b^2 c^2 \left (18 n^2+7 n+1\right )\right )}{b^3 (n+1) (2 n+1) (3 n+1)}-\frac{d x \left (-a^2 b c d^2 \left (24 n^3+38 n^2+19 n+3\right )+a^3 d^3 \left (6 n^3+11 n^2+6 n+1\right )+a b^2 c^2 d \left (36 n^3+45 n^2+20 n+3\right )-b^3 c^3 \left (24 n^3+18 n^2+7 n+1\right )\right )}{b^4 (n+1) (2 n+1) (3 n+1)}+\frac{x (b c-a d)^4 \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a b^4}-\frac{d x \left (c+d x^n\right )^2 (a d (3 n+1)-b (6 c n+c))}{b^2 \left (6 n^2+5 n+1\right )}+\frac{d x \left (c+d x^n\right )^3}{b (3 n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^n)^4/(a + b*x^n),x]

[Out]

-((d*(a^3*d^3*(1 + 6*n + 11*n^2 + 6*n^3) - b^3*c^3*(1 + 7*n + 18*n^2 + 24*n^3) - a^2*b*c*d^2*(3 + 19*n + 38*n^
2 + 24*n^3) + a*b^2*c^2*d*(3 + 20*n + 45*n^2 + 36*n^3))*x)/(b^4*(1 + n)*(1 + 2*n)*(1 + 3*n))) - (d*(2*a*b*c*d*
(1 + 3*n)^2 - a^2*d^2*(1 + 5*n + 6*n^2) - b^2*c^2*(1 + 7*n + 18*n^2))*x*(c + d*x^n))/(b^3*(1 + n)*(1 + 2*n)*(1
+ 3*n)) - (d*(a*d*(1 + 3*n) - b*(c + 6*c*n))*x*(c + d*x^n)^2)/(b^2*(1 + 5*n + 6*n^2)) + (d*x*(c + d*x^n)^3)/(
b*(1 + 3*n)) + ((b*c - a*d)^4*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*b^4)

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 )^4}{a+b x^n} \, dx &=\frac{d x \left (c+d x^n\right )^3}{b (1+3 n)}+\frac{\int \frac{\left (c+d x^n\right )^2 \left (-c (a d-b (c+3 c n))-d (a d (1+3 n)-b (c+6 c n)) x^n\right )}{a+b x^n} \, dx}{b (1+3 n)}\\ &=-\frac{d (a d (1+3 n)-b (c+6 c n)) x \left (c+d x^n\right )^2}{b^2 \left (1+5 n+6 n^2\right )}+\frac{d x \left (c+d x^n\right )^3}{b (1+3 n)}+\frac{\int \frac{\left (c+d x^n\right ) \left (c \left (a^2 d^2 (1+3 n)-2 a b c d (1+4 n)+b^2 c^2 \left (1+5 n+6 n^2\right )\right )-d \left (2 a b c d (1+3 n)^2-a^2 d^2 \left (1+5 n+6 n^2\right )-b^2 c^2 \left (1+7 n+18 n^2\right )\right ) x^n\right )}{a+b x^n} \, dx}{b^2 \left (1+5 n+6 n^2\right )}\\ &=-\frac{d \left (2 a b c d (1+3 n)^2-a^2 d^2 \left (1+5 n+6 n^2\right )-b^2 c^2 \left (1+7 n+18 n^2\right )\right ) x \left (c+d x^n\right )}{b^3 (1+n) \left (1+5 n+6 n^2\right )}-\frac{d (a d (1+3 n)-b (c+6 c n)) x \left (c+d x^n\right )^2}{b^2 \left (1+5 n+6 n^2\right )}+\frac{d x \left (c+d x^n\right )^3}{b (1+3 n)}+\frac{\int \frac{-c \left (a^3 d^3 \left (1+5 n+6 n^2\right )-a^2 b c d^2 \left (3+16 n+21 n^2\right )+a b^2 c^2 d \left (3+17 n+26 n^2\right )-b^3 c^3 \left (1+6 n+11 n^2+6 n^3\right )\right )-d \left (a^3 d^3 \left (1+6 n+11 n^2+6 n^3\right )-b^3 c^3 \left (1+7 n+18 n^2+24 n^3\right )-a^2 b c d^2 \left (3+19 n+38 n^2+24 n^3\right )+a b^2 c^2 d \left (3+20 n+45 n^2+36 n^3\right )\right ) x^n}{a+b x^n} \, dx}{b^3 (1+n) \left (1+5 n+6 n^2\right )}\\ &=-\frac{d \left (a^3 d^3 \left (1+6 n+11 n^2+6 n^3\right )-b^3 c^3 \left (1+7 n+18 n^2+24 n^3\right )-a^2 b c d^2 \left (3+19 n+38 n^2+24 n^3\right )+a b^2 c^2 d \left (3+20 n+45 n^2+36 n^3\right )\right ) x}{b^4 (1+n) \left (1+5 n+6 n^2\right )}-\frac{d \left (2 a b c d (1+3 n)^2-a^2 d^2 \left (1+5 n+6 n^2\right )-b^2 c^2 \left (1+7 n+18 n^2\right )\right ) x \left (c+d x^n\right )}{b^3 (1+n) \left (1+5 n+6 n^2\right )}-\frac{d (a d (1+3 n)-b (c+6 c n)) x \left (c+d x^n\right )^2}{b^2 \left (1+5 n+6 n^2\right )}+\frac{d x \left (c+d x^n\right )^3}{b (1+3 n)}+\frac{(b c-a d)^4 \int \frac{1}{a+b x^n} \, dx}{b^4}\\ &=-\frac{d \left (a^3 d^3 \left (1+6 n+11 n^2+6 n^3\right )-b^3 c^3 \left (1+7 n+18 n^2+24 n^3\right )-a^2 b c d^2 \left (3+19 n+38 n^2+24 n^3\right )+a b^2 c^2 d \left (3+20 n+45 n^2+36 n^3\right )\right ) x}{b^4 (1+n) \left (1+5 n+6 n^2\right )}-\frac{d \left (2 a b c d (1+3 n)^2-a^2 d^2 \left (1+5 n+6 n^2\right )-b^2 c^2 \left (1+7 n+18 n^2\right )\right ) x \left (c+d x^n\right )}{b^3 (1+n) \left (1+5 n+6 n^2\right )}-\frac{d (a d (1+3 n)-b (c+6 c n)) x \left (c+d x^n\right )^2}{b^2 \left (1+5 n+6 n^2\right )}+\frac{d x \left (c+d x^n\right )^3}{b (1+3 n)}+\frac{(b c-a d)^4 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a b^4}\\ \end{align*}

Mathematica [C]  time = 2.47083, size = 133, normalized size = 0.43 \[ \frac{x \left (6 c^2 d^2 x^{2 n} \Phi \left (-\frac{b x^n}{a},1,2+\frac{1}{n}\right )+4 c^3 d x^n \Phi \left (-\frac{b x^n}{a},1,1+\frac{1}{n}\right )+c^4 \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+4 c d^3 x^{3 n} \Phi \left (-\frac{b x^n}{a},1,3+\frac{1}{n}\right )+d^4 x^{4 n} \Phi \left (-\frac{b x^n}{a},1,4+\frac{1}{n}\right )\right )}{a n} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^n)^4/(a + b*x^n),x]

[Out]

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

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Maple [F]  time = 0.364, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c+d{x}^{n} \right ) ^{4}}{a+b{x}^{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*x^n)^4/(a+b*x^n),x)

[Out]

int((c+d*x^n)^4/(a+b*x^n),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \int \frac{1}{b^{5} x^{n} + a b^{4}}\,{d x} + \frac{{\left (2 \, n^{2} + 3 \, n + 1\right )} b^{3} d^{4} x x^{3 \, n} +{\left (4 \,{\left (3 \, n^{2} + 4 \, n + 1\right )} b^{3} c d^{3} -{\left (3 \, n^{2} + 4 \, n + 1\right )} a b^{2} d^{4}\right )} x x^{2 \, n} +{\left (6 \,{\left (6 \, n^{2} + 5 \, n + 1\right )} b^{3} c^{2} d^{2} - 4 \,{\left (6 \, n^{2} + 5 \, n + 1\right )} a b^{2} c d^{3} +{\left (6 \, n^{2} + 5 \, n + 1\right )} a^{2} b d^{4}\right )} x x^{n} +{\left (4 \,{\left (6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1\right )} b^{3} c^{3} d - 6 \,{\left (6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1\right )} a b^{2} c^{2} d^{2} + 4 \,{\left (6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1\right )} a^{2} b c d^{3} -{\left (6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1\right )} a^{3} d^{4}\right )} x}{{\left (6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1\right )} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^n)^4/(a+b*x^n),x, algorithm="maxima")

[Out]

(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*integrate(1/(b^5*x^n + a*b^4), x) + ((
2*n^2 + 3*n + 1)*b^3*d^4*x*x^(3*n) + (4*(3*n^2 + 4*n + 1)*b^3*c*d^3 - (3*n^2 + 4*n + 1)*a*b^2*d^4)*x*x^(2*n) +
(6*(6*n^2 + 5*n + 1)*b^3*c^2*d^2 - 4*(6*n^2 + 5*n + 1)*a*b^2*c*d^3 + (6*n^2 + 5*n + 1)*a^2*b*d^4)*x*x^n + (4*
(6*n^3 + 11*n^2 + 6*n + 1)*b^3*c^3*d - 6*(6*n^3 + 11*n^2 + 6*n + 1)*a*b^2*c^2*d^2 + 4*(6*n^3 + 11*n^2 + 6*n +
1)*a^2*b*c*d^3 - (6*n^3 + 11*n^2 + 6*n + 1)*a^3*d^4)*x)/((6*n^3 + 11*n^2 + 6*n + 1)*b^4)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d^{4} x^{4 \, n} + 4 \, c d^{3} x^{3 \, n} + 6 \, c^{2} d^{2} x^{2 \, n} + 4 \, c^{3} d x^{n} + c^{4}}{b x^{n} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^n)^4/(a+b*x^n),x, algorithm="fricas")

[Out]

integral((d^4*x^(4*n) + 4*c*d^3*x^(3*n) + 6*c^2*d^2*x^(2*n) + 4*c^3*d*x^n + c^4)/(b*x^n + a), x)

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Sympy [C]  time = 6.64649, size = 369, normalized size = 1.19 \begin{align*} - \frac{4 c^{3} 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^{4} 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{12 c^{2} 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{6 c^{2} 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{12 c 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{4 c 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 )} + \frac{4 d^{4} x x^{4 n} \Phi \left (\frac{b x^{n} e^{i \pi }}{a}, 1, 4 + \frac{1}{n}\right ) \Gamma \left (4 + \frac{1}{n}\right )}{a n \Gamma \left (5 + \frac{1}{n}\right )} + \frac{d^{4} x x^{4 n} \Phi \left (\frac{b x^{n} e^{i \pi }}{a}, 1, 4 + \frac{1}{n}\right ) \Gamma \left (4 + \frac{1}{n}\right )}{a n^{2} \Gamma \left (5 + \frac{1}{n}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x**n)**4/(a+b*x**n),x)

[Out]

-4*c**3*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
**4*x*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, 1/n)*gamma(1/n)/(a*n**2*gamma(1 + 1/n)) + 12*c**2*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*gamma(3 + 1/n)) + 6*c**2*d**2*x*x**(2*n)*le
rchphi(b*x**n*exp_polar(I*pi)/a, 1, 2 + 1/n)*gamma(2 + 1/n)/(a*n**2*gamma(3 + 1/n)) + 12*c*d**3*x*x**(3*n)*ler
chphi(b*x**n*exp_polar(I*pi)/a, 1, 3 + 1/n)*gamma(3 + 1/n)/(a*n*gamma(4 + 1/n)) + 4*c*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**2*gamma(4 + 1/n)) + 4*d**4*x*x**(4*n)*lerchphi(b*x
**n*exp_polar(I*pi)/a, 1, 4 + 1/n)*gamma(4 + 1/n)/(a*n*gamma(5 + 1/n)) + d**4*x*x**(4*n)*lerchphi(b*x**n*exp_p
olar(I*pi)/a, 1, 4 + 1/n)*gamma(4 + 1/n)/(a*n**2*gamma(5 + 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 )}^{4}}{b x^{n} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^n)^4/(a+b*x^n),x, algorithm="giac")

[Out]

integrate((d*x^n + c)^4/(b*x^n + a), x)