3.305 \(\int \frac {(c+d x^n)^4}{(a+b x^n)^2} \, dx\)

Optimal. Leaf size=341 \[ -\frac {x (b c-a d)^3 (b c (1-n)-a d (3 n+1)) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2 b^4 n}-\frac {d x \left (c+d x^n\right ) \left (a^2 d^2 \left (6 n^2+5 n+1\right )-2 a b c d \left (5 n^2+4 n+1\right )+b^2 c^2 \left (2 n^2+3 n+1\right )\right )}{a b^3 n (n+1) (2 n+1)}-\frac {d x \left (-a^3 d^3 \left (6 n^3+11 n^2+6 n+1\right )+a^2 b c d^2 \left (16 n^3+26 n^2+15 n+3\right )-a b^2 c^2 d \left (12 n^3+17 n^2+12 n+3\right )+b^3 c^3 \left (2 n^2+3 n+1\right )\right )}{a b^4 n (n+1) (2 n+1)}+\frac {d x \left (c+d x^n\right )^2 (a d (3 n+1)-b (2 c n+c))}{a b^2 n (2 n+1)}+\frac {x (b c-a d) \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )} \]

[Out]

-d*(b^3*c^3*(2*n^2+3*n+1)-a^3*d^3*(6*n^3+11*n^2+6*n+1)-a*b^2*c^2*d*(12*n^3+17*n^2+12*n+3)+a^2*b*c*d^2*(16*n^3+
26*n^2+15*n+3))*x/a/b^4/n/(2*n^2+3*n+1)-d*(b^2*c^2*(2*n^2+3*n+1)-2*a*b*c*d*(5*n^2+4*n+1)+a^2*d^2*(6*n^2+5*n+1)
)*x*(c+d*x^n)/a/b^3/n/(2*n^2+3*n+1)-d*(-3*a*d*n+2*b*c*n-a*d+b*c)*x*(c+d*x^n)^2/a/b^2/n/(1+2*n)+(-a*d+b*c)*x*(c
+d*x^n)^3/a/b/n/(a+b*x^n)-(-a*d+b*c)^3*(b*c*(1-n)-a*d*(1+3*n))*x*hypergeom([1, 1/n],[1+1/n],-b*x^n/a)/a^2/b^4/
n

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Rubi [A]  time = 0.55, antiderivative size = 341, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {413, 528, 388, 245} \[ -\frac {d x \left (c+d x^n\right ) \left (a^2 d^2 \left (6 n^2+5 n+1\right )-2 a b c d \left (5 n^2+4 n+1\right )+b^2 c^2 \left (2 n^2+3 n+1\right )\right )}{a b^3 n (n+1) (2 n+1)}-\frac {d x \left (a^2 b c d^2 \left (16 n^3+26 n^2+15 n+3\right )-a^3 d^3 \left (6 n^3+11 n^2+6 n+1\right )-a b^2 c^2 d \left (12 n^3+17 n^2+12 n+3\right )+b^3 c^3 \left (2 n^2+3 n+1\right )\right )}{a b^4 n (n+1) (2 n+1)}-\frac {x (b c-a d)^3 (b c (1-n)-a d (3 n+1)) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2 b^4 n}+\frac {d x \left (c+d x^n\right )^2 (a d (3 n+1)-b (2 c n+c))}{a b^2 n (2 n+1)}+\frac {x (b c-a d) \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^n)^4/(a + b*x^n)^2,x]

[Out]

-((d*(b^3*c^3*(1 + 3*n + 2*n^2) - a^3*d^3*(1 + 6*n + 11*n^2 + 6*n^3) - a*b^2*c^2*d*(3 + 12*n + 17*n^2 + 12*n^3
) + a^2*b*c*d^2*(3 + 15*n + 26*n^2 + 16*n^3))*x)/(a*b^4*n*(1 + n)*(1 + 2*n))) - (d*(b^2*c^2*(1 + 3*n + 2*n^2)
- 2*a*b*c*d*(1 + 4*n + 5*n^2) + a^2*d^2*(1 + 5*n + 6*n^2))*x*(c + d*x^n))/(a*b^3*n*(1 + n)*(1 + 2*n)) + (d*(a*
d*(1 + 3*n) - b*(c + 2*c*n))*x*(c + d*x^n)^2)/(a*b^2*n*(1 + 2*n)) + ((b*c - a*d)*x*(c + d*x^n)^3)/(a*b*n*(a +
b*x^n)) - ((b*c - a*d)^3*(b*c*(1 - n) - a*d*(1 + 3*n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)
])/(a^2*b^4*n)

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])

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 413

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

Rubi steps

\begin {align*} \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx &=\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}+\frac {\int \frac {\left (c+d x^n\right )^2 \left (c (a d-b c (1-n))+d (a d (1+3 n)-b (c+2 c n)) x^n\right )}{a+b x^n} \, dx}{a b n}\\ &=\frac {d (a d (1+3 n)-b (c+2 c n)) x \left (c+d x^n\right )^2}{a b^2 n (1+2 n)}+\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}+\frac {\int \frac {\left (c+d x^n\right ) \left (c \left (2 a b c d (1+2 n)-a^2 d^2 (1+3 n)-b^2 c^2 \left (1+n-2 n^2\right )\right )-d \left (b^2 c^2 \left (1+3 n+2 n^2\right )-2 a b c d \left (1+4 n+5 n^2\right )+a^2 d^2 \left (1+5 n+6 n^2\right )\right ) x^n\right )}{a+b x^n} \, dx}{a b^2 n (1+2 n)}\\ &=-\frac {d \left (b^2 c^2 \left (1+3 n+2 n^2\right )-2 a b c d \left (1+4 n+5 n^2\right )+a^2 d^2 \left (1+5 n+6 n^2\right )\right ) x \left (c+d x^n\right )}{a b^3 n (1+n) (1+2 n)}+\frac {d (a d (1+3 n)-b (c+2 c n)) x \left (c+d x^n\right )^2}{a b^2 n (1+2 n)}+\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}+\frac {\int \frac {c \left (3 a b^2 c^2 d \left (1+3 n+2 n^2\right )+a^3 d^3 \left (1+5 n+6 n^2\right )-a^2 b c d^2 \left (3+12 n+13 n^2\right )-b^3 c^3 \left (1+2 n-n^2-2 n^3\right )\right )-d \left (b^3 c^3 \left (1+3 n+2 n^2\right )-a^3 d^3 \left (1+6 n+11 n^2+6 n^3\right )-a b^2 c^2 d \left (3+12 n+17 n^2+12 n^3\right )+a^2 b c d^2 \left (3+15 n+26 n^2+16 n^3\right )\right ) x^n}{a+b x^n} \, dx}{a b^3 n (1+n) (1+2 n)}\\ &=-\frac {d \left (b^3 c^3 \left (1+3 n+2 n^2\right )-a^3 d^3 \left (1+6 n+11 n^2+6 n^3\right )-a b^2 c^2 d \left (3+12 n+17 n^2+12 n^3\right )+a^2 b c d^2 \left (3+15 n+26 n^2+16 n^3\right )\right ) x}{a b^4 n (1+n) (1+2 n)}-\frac {d \left (b^2 c^2 \left (1+3 n+2 n^2\right )-2 a b c d \left (1+4 n+5 n^2\right )+a^2 d^2 \left (1+5 n+6 n^2\right )\right ) x \left (c+d x^n\right )}{a b^3 n (1+n) (1+2 n)}+\frac {d (a d (1+3 n)-b (c+2 c n)) x \left (c+d x^n\right )^2}{a b^2 n (1+2 n)}+\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}-\frac {\left ((b c-a d)^3 (b c (1-n)-a d (1+3 n))\right ) \int \frac {1}{a+b x^n} \, dx}{a b^4 n}\\ &=-\frac {d \left (b^3 c^3 \left (1+3 n+2 n^2\right )-a^3 d^3 \left (1+6 n+11 n^2+6 n^3\right )-a b^2 c^2 d \left (3+12 n+17 n^2+12 n^3\right )+a^2 b c d^2 \left (3+15 n+26 n^2+16 n^3\right )\right ) x}{a b^4 n (1+n) (1+2 n)}-\frac {d \left (b^2 c^2 \left (1+3 n+2 n^2\right )-2 a b c d \left (1+4 n+5 n^2\right )+a^2 d^2 \left (1+5 n+6 n^2\right )\right ) x \left (c+d x^n\right )}{a b^3 n (1+n) (1+2 n)}+\frac {d (a d (1+3 n)-b (c+2 c n)) x \left (c+d x^n\right )^2}{a b^2 n (1+2 n)}+\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}-\frac {(b c-a d)^3 (b c (1-n)-a d (1+3 n)) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2 b^4 n}\\ \end {align*}

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Mathematica [A]  time = 5.26, size = 217, normalized size = 0.64 \[ \frac {x \left (\frac {(b c-a d)^3 (a d (3 n+1)+b c (n-1)) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2 n}+\frac {(a d-b c)^3 (a d (3 n+1)+b c (n-1))}{a^2 n}+\frac {-a^4 d^4+4 a^3 b c d^3-6 a^2 b^2 c^2 d^2+4 a b^3 c^3 d+b^4 c^4 (n-1)}{a^2 n}+\frac {2 b d^3 x^n (2 b c-a d)}{n+1}+\frac {(b c-a d)^4}{a n \left (a+b x^n\right )}+\frac {b^2 d^4 x^{2 n}}{2 n+1}\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^n)^4/(a + b*x^n)^2,x]

[Out]

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

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fricas [F]  time = 1.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{4} x^{4 \, n} + 4 \, c d^{3} x^{3 \, n} + 6 \, c^{2} d^{2} x^{2 \, n} + 4 \, c^{3} d x^{n} + c^{4}}{b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{n} + c\right )}^{4}}{{\left (b x^{n} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^n)^4/(a+b*x^n)^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.62, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \,x^{n}+c \right )^{4}}{\left (b \,x^{n}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^n+c)^4/(b*x^n+a)^2,x)

[Out]

int((d*x^n+c)^4/(b*x^n+a)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -{\left (a^{4} d^{4} {\left (3 \, n + 1\right )} - 4 \, a^{3} b c d^{3} {\left (2 \, n + 1\right )} + 6 \, a^{2} b^{2} c^{2} d^{2} {\left (n + 1\right )} - b^{4} c^{4} {\left (n - 1\right )} - 4 \, a b^{3} c^{3} d\right )} \int \frac {1}{a b^{5} n x^{n} + a^{2} b^{4} n}\,{d x} + \frac {{\left (n^{2} + n\right )} a b^{3} d^{4} x x^{3 \, n} + {\left (4 \, {\left (2 \, n^{2} + n\right )} a b^{3} c d^{3} - {\left (3 \, n^{2} + n\right )} a^{2} b^{2} d^{4}\right )} x x^{2 \, n} + {\left (6 \, {\left (2 \, n^{3} + 3 \, n^{2} + n\right )} a b^{3} c^{2} d^{2} - 4 \, {\left (4 \, n^{3} + 4 \, n^{2} + n\right )} a^{2} b^{2} c d^{3} + {\left (6 \, n^{3} + 5 \, n^{2} + n\right )} a^{3} b d^{4}\right )} x x^{n} + {\left ({\left (2 \, n^{2} + 3 \, n + 1\right )} b^{4} c^{4} - 4 \, {\left (2 \, n^{2} + 3 \, n + 1\right )} a b^{3} c^{3} d + 6 \, {\left (2 \, n^{3} + 5 \, n^{2} + 4 \, n + 1\right )} a^{2} b^{2} c^{2} d^{2} - 4 \, {\left (4 \, n^{3} + 8 \, n^{2} + 5 \, n + 1\right )} a^{3} b c d^{3} + {\left (6 \, n^{3} + 11 \, n^{2} + 6 \, n + 1\right )} a^{4} d^{4}\right )} x}{{\left (2 \, n^{3} + 3 \, n^{2} + n\right )} a b^{5} x^{n} + {\left (2 \, n^{3} + 3 \, n^{2} + n\right )} a^{2} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^n)^4/(a+b*x^n)^2,x, algorithm="maxima")

[Out]

-(a^4*d^4*(3*n + 1) - 4*a^3*b*c*d^3*(2*n + 1) + 6*a^2*b^2*c^2*d^2*(n + 1) - b^4*c^4*(n - 1) - 4*a*b^3*c^3*d)*i
ntegrate(1/(a*b^5*n*x^n + a^2*b^4*n), x) + ((n^2 + n)*a*b^3*d^4*x*x^(3*n) + (4*(2*n^2 + n)*a*b^3*c*d^3 - (3*n^
2 + n)*a^2*b^2*d^4)*x*x^(2*n) + (6*(2*n^3 + 3*n^2 + n)*a*b^3*c^2*d^2 - 4*(4*n^3 + 4*n^2 + n)*a^2*b^2*c*d^3 + (
6*n^3 + 5*n^2 + n)*a^3*b*d^4)*x*x^n + ((2*n^2 + 3*n + 1)*b^4*c^4 - 4*(2*n^2 + 3*n + 1)*a*b^3*c^3*d + 6*(2*n^3
+ 5*n^2 + 4*n + 1)*a^2*b^2*c^2*d^2 - 4*(4*n^3 + 8*n^2 + 5*n + 1)*a^3*b*c*d^3 + (6*n^3 + 11*n^2 + 6*n + 1)*a^4*
d^4)*x)/((2*n^3 + 3*n^2 + n)*a*b^5*x^n + (2*n^3 + 3*n^2 + n)*a^2*b^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x^n\right )}^4}{{\left (a+b\,x^n\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^n)^4/(a + b*x^n)^2,x)

[Out]

int((c + d*x^n)^4/(a + b*x^n)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d x^{n}\right )^{4}}{\left (a + b x^{n}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x**n)**4/(a+b*x**n)**2,x)

[Out]

Integral((c + d*x**n)**4/(a + b*x**n)**2, x)

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