3.1269 \(\int \frac {x^{14}}{a+b x^5} \, dx\)

Optimal. Leaf size=40 \[ \frac {a^2 \log \left (a+b x^5\right )}{5 b^3}-\frac {a x^5}{5 b^2}+\frac {x^{10}}{10 b} \]

[Out]

-1/5*a*x^5/b^2+1/10*x^10/b+1/5*a^2*ln(b*x^5+a)/b^3

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Rubi [A]  time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {266, 43} \[ \frac {a^2 \log \left (a+b x^5\right )}{5 b^3}-\frac {a x^5}{5 b^2}+\frac {x^{10}}{10 b} \]

Antiderivative was successfully verified.

[In]

Int[x^14/(a + b*x^5),x]

[Out]

-(a*x^5)/(5*b^2) + x^10/(10*b) + (a^2*Log[a + b*x^5])/(5*b^3)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

Rubi steps

\begin {align*} \int \frac {x^{14}}{a+b x^5} \, dx &=\frac {1}{5} \operatorname {Subst}\left (\int \frac {x^2}{a+b x} \, dx,x,x^5\right )\\ &=\frac {1}{5} \operatorname {Subst}\left (\int \left (-\frac {a}{b^2}+\frac {x}{b}+\frac {a^2}{b^2 (a+b x)}\right ) \, dx,x,x^5\right )\\ &=-\frac {a x^5}{5 b^2}+\frac {x^{10}}{10 b}+\frac {a^2 \log \left (a+b x^5\right )}{5 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 40, normalized size = 1.00 \[ \frac {a^2 \log \left (a+b x^5\right )}{5 b^3}-\frac {a x^5}{5 b^2}+\frac {x^{10}}{10 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^14/(a + b*x^5),x]

[Out]

-1/5*(a*x^5)/b^2 + x^10/(10*b) + (a^2*Log[a + b*x^5])/(5*b^3)

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fricas [A]  time = 0.55, size = 33, normalized size = 0.82 \[ \frac {b^{2} x^{10} - 2 \, a b x^{5} + 2 \, a^{2} \log \left (b x^{5} + a\right )}{10 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^5+a),x, algorithm="fricas")

[Out]

1/10*(b^2*x^10 - 2*a*b*x^5 + 2*a^2*log(b*x^5 + a))/b^3

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giac [A]  time = 0.17, size = 35, normalized size = 0.88 \[ \frac {a^{2} \log \left ({\left | b x^{5} + a \right |}\right )}{5 \, b^{3}} + \frac {b x^{10} - 2 \, a x^{5}}{10 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^5+a),x, algorithm="giac")

[Out]

1/5*a^2*log(abs(b*x^5 + a))/b^3 + 1/10*(b*x^10 - 2*a*x^5)/b^2

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maple [A]  time = 0.00, size = 35, normalized size = 0.88 \[ \frac {x^{10}}{10 b}-\frac {a \,x^{5}}{5 b^{2}}+\frac {a^{2} \ln \left (b \,x^{5}+a \right )}{5 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(b*x^5+a),x)

[Out]

-1/5*a*x^5/b^2+1/10*x^10/b+1/5*a^2*ln(b*x^5+a)/b^3

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maxima [A]  time = 1.06, size = 34, normalized size = 0.85 \[ \frac {a^{2} \log \left (b x^{5} + a\right )}{5 \, b^{3}} + \frac {b x^{10} - 2 \, a x^{5}}{10 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(b*x^5+a),x, algorithm="maxima")

[Out]

1/5*a^2*log(b*x^5 + a)/b^3 + 1/10*(b*x^10 - 2*a*x^5)/b^2

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mupad [B]  time = 1.06, size = 33, normalized size = 0.82 \[ \frac {2\,a^2\,\ln \left (b\,x^5+a\right )+b^2\,x^{10}-2\,a\,b\,x^5}{10\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(a + b*x^5),x)

[Out]

(2*a^2*log(a + b*x^5) + b^2*x^10 - 2*a*b*x^5)/(10*b^3)

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sympy [A]  time = 0.60, size = 32, normalized size = 0.80 \[ \frac {a^{2} \log {\left (a + b x^{5} \right )}}{5 b^{3}} - \frac {a x^{5}}{5 b^{2}} + \frac {x^{10}}{10 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**14/(b*x**5+a),x)

[Out]

a**2*log(a + b*x**5)/(5*b**3) - a*x**5/(5*b**2) + x**10/(10*b)

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