3.11.46 \(\int \frac {x^{-1+3 n} (a+b x^n)^2}{c+d x^n} \, dx\) [1046]

Optimal. Leaf size=118 \[ -\frac {c (b c-a d)^2 x^n}{d^4 n}+\frac {(b c-a d)^2 x^{2 n}}{2 d^3 n}-\frac {b (b c-2 a d) x^{3 n}}{3 d^2 n}+\frac {b^2 x^{4 n}}{4 d n}+\frac {c^2 (b c-a d)^2 \log \left (c+d x^n\right )}{d^5 n} \]

[Out]

-c*(-a*d+b*c)^2*x^n/d^4/n+1/2*(-a*d+b*c)^2*x^(2*n)/d^3/n-1/3*b*(-2*a*d+b*c)*x^(3*n)/d^2/n+1/4*b^2*x^(4*n)/d/n+
c^2*(-a*d+b*c)^2*ln(c+d*x^n)/d^5/n

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Rubi [A]
time = 0.07, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {457, 90} \begin {gather*} \frac {c^2 (b c-a d)^2 \log \left (c+d x^n\right )}{d^5 n}-\frac {c x^n (b c-a d)^2}{d^4 n}+\frac {x^{2 n} (b c-a d)^2}{2 d^3 n}-\frac {b x^{3 n} (b c-2 a d)}{3 d^2 n}+\frac {b^2 x^{4 n}}{4 d n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(b*c - a*d)^2*x^n)/(d^4*n)) + ((b*c - a*d)^2*x^(2*n))/(2*d^3*n) - (b*(b*c - 2*a*d)*x^(3*n))/(3*d^2*n) + (
b^2*x^(4*n))/(4*d*n) + (c^2*(b*c - a*d)^2*Log[c + d*x^n])/(d^5*n)

Rule 90

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 457

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

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 125, normalized size = 1.06 \begin {gather*} \frac {d x^n \left (6 a^2 d^2 \left (-2 c+d x^n\right )+4 a b d \left (6 c^2-3 c d x^n+2 d^2 x^{2 n}\right )+b^2 \left (-12 c^3+6 c^2 d x^n-4 c d^2 x^{2 n}+3 d^3 x^{3 n}\right )\right )+12 c^2 (b c-a d)^2 \log \left (c+d x^n\right )}{12 d^5 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x^n*(6*a^2*d^2*(-2*c + d*x^n) + 4*a*b*d*(6*c^2 - 3*c*d*x^n + 2*d^2*x^(2*n)) + b^2*(-12*c^3 + 6*c^2*d*x^n -
4*c*d^2*x^(2*n) + 3*d^3*x^(3*n))) + 12*c^2*(b*c - a*d)^2*Log[c + d*x^n])/(12*d^5*n)

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Maple [A]
time = 0.37, size = 157, normalized size = 1.33

method	result	size
norman	\(\frac {b^{2} {\mathrm e}^{4 n \ln \left (x \right )}}{4 d n}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) {\mathrm e}^{2 n \ln \left (x \right )}}{2 d^{3} n}+\frac {b \left (2 a d -b c \right ) {\mathrm e}^{3 n \ln \left (x \right )}}{3 d^{2} n}-\frac {c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) {\mathrm e}^{n \ln \left (x \right )}}{d^{4} n}+\frac {c^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (c +d \,{\mathrm e}^{n \ln \left (x \right )}\right )}{d^{5} n}\)	\(157\)
risch	\(\frac {b^{2} x^{4 n}}{4 d n}+\frac {2 b \,x^{3 n} a}{3 d n}-\frac {b^{2} x^{3 n} c}{3 d^{2} n}+\frac {x^{2 n} a^{2}}{2 d n}-\frac {x^{2 n} a b c}{d^{2} n}+\frac {x^{2 n} b^{2} c^{2}}{2 d^{3} n}-\frac {c \,x^{n} a^{2}}{d^{2} n}+\frac {2 c^{2} x^{n} a b}{d^{3} n}-\frac {c^{3} x^{n} b^{2}}{d^{4} n}+\frac {c^{2} \ln \left (x^{n}+\frac {c}{d}\right ) a^{2}}{d^{3} n}-\frac {2 c^{3} \ln \left (x^{n}+\frac {c}{d}\right ) a b}{d^{4} n}+\frac {c^{4} \ln \left (x^{n}+\frac {c}{d}\right ) b^{2}}{d^{5} n}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+3*n)*(a+b*x^n)^2/(c+d*x^n),x,method=_RETURNVERBOSE)

[Out]

1/4*b^2/d/n*exp(n*ln(x))^4+1/2/d^3*(a^2*d^2-2*a*b*c*d+b^2*c^2)/n*exp(n*ln(x))^2+1/3*b*(2*a*d-b*c)/d^2/n*exp(n*
ln(x))^3-c*(a^2*d^2-2*a*b*c*d+b^2*c^2)/d^4/n*exp(n*ln(x))+c^2/d^5*(a^2*d^2-2*a*b*c*d+b^2*c^2)/n*ln(c+d*exp(n*l
n(x)))

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Maxima [A]
time = 0.33, size = 192, normalized size = 1.63 \begin {gather*} \frac {1}{12} \, b^{2} {\left (\frac {12 \, c^{4} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{5} n} + \frac {3 \, d^{3} x^{4 \, n} - 4 \, c d^{2} x^{3 \, n} + 6 \, c^{2} d x^{2 \, n} - 12 \, c^{3} x^{n}}{d^{4} n}\right )} - \frac {1}{3} \, a b {\left (\frac {6 \, c^{3} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{4} n} - \frac {2 \, d^{2} x^{3 \, n} - 3 \, c d x^{2 \, n} + 6 \, c^{2} x^{n}}{d^{3} n}\right )} + \frac {1}{2} \, a^{2} {\left (\frac {2 \, c^{2} \log \left (\frac {d x^{n} + c}{d}\right )}{d^{3} n} + \frac {d x^{2 \, n} - 2 \, c x^{n}}{d^{2} n}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)*(a+b*x^n)^2/(c+d*x^n),x, algorithm="maxima")

[Out]

1/12*b^2*(12*c^4*log((d*x^n + c)/d)/(d^5*n) + (3*d^3*x^(4*n) - 4*c*d^2*x^(3*n) + 6*c^2*d*x^(2*n) - 12*c^3*x^n)
/(d^4*n)) - 1/3*a*b*(6*c^3*log((d*x^n + c)/d)/(d^4*n) - (2*d^2*x^(3*n) - 3*c*d*x^(2*n) + 6*c^2*x^n)/(d^3*n)) +
1/2*a^2*(2*c^2*log((d*x^n + c)/d)/(d^3*n) + (d*x^(2*n) - 2*c*x^n)/(d^2*n))

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Fricas [A]
time = 2.61, size = 146, normalized size = 1.24 \begin {gather*} \frac {3 \, b^{2} d^{4} x^{4 \, n} - 4 \, {\left (b^{2} c d^{3} - 2 \, a b d^{4}\right )} x^{3 \, n} + 6 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2 \, n} - 12 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{n} + 12 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} \log \left (d x^{n} + c\right )}{12 \, d^{5} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)*(a+b*x^n)^2/(c+d*x^n),x, algorithm="fricas")

[Out]

1/12*(3*b^2*d^4*x^(4*n) - 4*(b^2*c*d^3 - 2*a*b*d^4)*x^(3*n) + 6*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^(2*n)
- 12*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*x^n + 12*(b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*log(d*x^n + c))/(d
^5*n)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (99) = 198\).
time = 27.03, size = 258, normalized size = 2.19 \begin {gather*} \begin {cases} \frac {\left (a + b\right )^{2} \log {\left (x \right )}}{c} & \text {for}\: d = 0 \wedge n = 0 \\\frac {\left (a + b\right )^{2} \log {\left (x \right )}}{c + d} & \text {for}\: n = 0 \\\frac {\frac {a^{2} x^{3 n}}{3 n} + \frac {a b x^{4 n}}{2 n} + \frac {b^{2} x^{5 n}}{5 n}}{c} & \text {for}\: d = 0 \\\frac {a^{2} c^{2} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{3} n} - \frac {a^{2} c x^{n}}{d^{2} n} + \frac {a^{2} x^{2 n}}{2 d n} - \frac {2 a b c^{3} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{4} n} + \frac {2 a b c^{2} x^{n}}{d^{3} n} - \frac {a b c x^{2 n}}{d^{2} n} + \frac {2 a b x^{3 n}}{3 d n} + \frac {b^{2} c^{4} \log {\left (\frac {c}{d} + x^{n} \right )}}{d^{5} n} - \frac {b^{2} c^{3} x^{n}}{d^{4} n} + \frac {b^{2} c^{2} x^{2 n}}{2 d^{3} n} - \frac {b^{2} c x^{3 n}}{3 d^{2} n} + \frac {b^{2} x^{4 n}}{4 d n} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((a + b)**2*log(x)/c, Eq(d, 0) & Eq(n, 0)), ((a + b)**2*log(x)/(c + d), Eq(n, 0)), ((a**2*x**(3*n)/(
3*n) + a*b*x**(4*n)/(2*n) + b**2*x**(5*n)/(5*n))/c, Eq(d, 0)), (a**2*c**2*log(c/d + x**n)/(d**3*n) - a**2*c*x*
*n/(d**2*n) + a**2*x**(2*n)/(2*d*n) - 2*a*b*c**3*log(c/d + x**n)/(d**4*n) + 2*a*b*c**2*x**n/(d**3*n) - a*b*c*x
**(2*n)/(d**2*n) + 2*a*b*x**(3*n)/(3*d*n) + b**2*c**4*log(c/d + x**n)/(d**5*n) - b**2*c**3*x**n/(d**4*n) + b**
2*c**2*x**(2*n)/(2*d**3*n) - b**2*c*x**(3*n)/(3*d**2*n) + b**2*x**(4*n)/(4*d*n), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)*(a+b*x^n)^2/(c+d*x^n),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{3\,n-1}\,{\left (a+b\,x^n\right )}^2}{c+d\,x^n} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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