3.553 \(\int \frac {x^5}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx\)

Optimal. Leaf size=73 \[ -\frac {2 d \sqrt {a+b x^3}}{3 b^2 c^2}-\frac {2 \left (a c^2-d^2\right ) \log \left (c \sqrt {a+b x^3}+d\right )}{3 b^2 c^3}+\frac {x^3}{3 b c} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2155, 697} \[ -\frac {2 \left (a c^2-d^2\right ) \log \left (c \sqrt {a+b x^3}+d\right )}{3 b^2 c^3}-\frac {2 d \sqrt {a+b x^3}}{3 b^2 c^2}+\frac {x^3}{3 b c} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a*c + b*c*x^3 + d*Sqrt[a + b*x^3]),x]

[Out]

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

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2155

Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[1/n, Subst[Int
[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a + b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c
- a*d, 0] && IntegerQ[(m + 1)/n]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 63, normalized size = 0.86 \[ \frac {\left (2 d^2-2 a c^2\right ) \log \left (c \sqrt {a+b x^3}+d\right )+c \left (b c x^3-2 d \sqrt {a+b x^3}\right )}{3 b^2 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a*c + b*c*x^3 + d*Sqrt[a + b*x^3]),x]

[Out]

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

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fricas [A]  time = 0.41, size = 118, normalized size = 1.62 \[ \frac {b c^{2} x^{3} - 2 \, \sqrt {b x^{3} + a} c d - {\left (a c^{2} - d^{2}\right )} \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) - {\left (a c^{2} - d^{2}\right )} \log \left (\sqrt {b x^{3} + a} c + d\right ) + {\left (a c^{2} - d^{2}\right )} \log \left (\sqrt {b x^{3} + a} c - d\right )}{3 \, b^{2} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x, algorithm="fricas")

[Out]

1/3*(b*c^2*x^3 - 2*sqrt(b*x^3 + a)*c*d - (a*c^2 - d^2)*log(b*c^2*x^3 + a*c^2 - d^2) - (a*c^2 - d^2)*log(sqrt(b
*x^3 + a)*c + d) + (a*c^2 - d^2)*log(sqrt(b*x^3 + a)*c - d))/(b^2*c^3)

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giac [A]  time = 0.33, size = 72, normalized size = 0.99 \[ -\frac {\frac {2 \, {\left (a c^{2} - d^{2}\right )} \log \left ({\left | \sqrt {b x^{3} + a} c + d \right |}\right )}{b c^{3}} - \frac {{\left (b x^{3} + a\right )} b c - 2 \, \sqrt {b x^{3} + a} b d}{b^{2} c^{2}}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x, algorithm="giac")

[Out]

-1/3*(2*(a*c^2 - d^2)*log(abs(sqrt(b*x^3 + a)*c + d))/(b*c^3) - ((b*x^3 + a)*b*c - 2*sqrt(b*x^3 + a)*b*d)/(b^2
*c^2))/b

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maple [C]  time = 0.03, size = 943, normalized size = 12.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x)

[Out]

-2/3*d*(b*x^3+a)^(1/2)/b^2/c^2+1/3*I/b^4/d*2^(1/2)*sum((-a*b^2)^(1/3)*(1/2*I*(2*x+((-a*b^2)^(1/3)-I*3^(1/2)*(-
a*b^2)^(1/3))/b)/(-a*b^2)^(1/3)*b)^(1/2)*((x-(-a*b^2)^(1/3)/b)/(-3*(-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3))*b)
^(1/2)*(-1/2*I*(2*x+((-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3))/b)/(-a*b^2)^(1/3)*b)^(1/2)/(b*x^3+a)^(1/2)*(2*_a
lpha^2*b^2+I*(-a*b^2)^(1/3)*3^(1/2)*_alpha*b-(-a*b^2)^(1/3)*_alpha*b-I*(-a*b^2)^(2/3)*3^(1/2)-(-a*b^2)^(2/3))*
EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1
/2),-1/2*(2*I*(-a*b^2)^(1/3)*3^(1/2)*_alpha^2*b+I*3^(1/2)*a*b-3*a*b-I*(-a*b^2)^(2/3)*3^(1/2)*_alpha-3*(-a*b^2)
^(2/3)*_alpha)/b*c^2/d^2,(I*3^(1/2)*(-a*b^2)^(1/3)/(-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/b)^(
1/2)),_alpha=RootOf(_Z^3*b*c^2+a*c^2-d^2))*a-1/3*I*d/b^4/c^2*2^(1/2)*sum((-a*b^2)^(1/3)*(1/2*I*(2*x+((-a*b^2)^
(1/3)-I*3^(1/2)*(-a*b^2)^(1/3))/b)/(-a*b^2)^(1/3)*b)^(1/2)*((x-(-a*b^2)^(1/3)/b)/(-3*(-a*b^2)^(1/3)+I*3^(1/2)*
(-a*b^2)^(1/3))*b)^(1/2)*(-1/2*I*(2*x+((-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3))/b)/(-a*b^2)^(1/3)*b)^(1/2)/(b*
x^3+a)^(1/2)*(2*_alpha^2*b^2+I*(-a*b^2)^(1/3)*3^(1/2)*_alpha*b-(-a*b^2)^(1/3)*_alpha*b-I*(-a*b^2)^(2/3)*3^(1/2
)-(-a*b^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)/(-
a*b^2)^(1/3)*b)^(1/2),-1/2*(2*I*(-a*b^2)^(1/3)*3^(1/2)*_alpha^2*b+I*3^(1/2)*a*b-3*a*b-I*(-a*b^2)^(2/3)*3^(1/2)
*_alpha-3*(-a*b^2)^(2/3)*_alpha)/b*c^2/d^2,(I*3^(1/2)*(-a*b^2)^(1/3)/(-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*
b^2)^(1/3)/b)/b)^(1/2)),_alpha=RootOf(_Z^3*b*c^2+a*c^2-d^2))-1/3*a/c/b^2*ln(b*c^2*x^3+a*c^2-d^2)+1/3*x^3/b/c+1
/3/b^2/c^3*d^2*ln(b*c^2*x^3+a*c^2-d^2)

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maxima [A]  time = 0.58, size = 62, normalized size = 0.85 \[ \frac {\frac {{\left (b x^{3} + a\right )} c - 2 \, \sqrt {b x^{3} + a} d}{c^{2}} - \frac {2 \, {\left (a c^{2} - d^{2}\right )} \log \left (\sqrt {b x^{3} + a} c + d\right )}{c^{3}}}{3 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x, algorithm="maxima")

[Out]

1/3*(((b*x^3 + a)*c - 2*sqrt(b*x^3 + a)*d)/c^2 - 2*(a*c^2 - d^2)*log(sqrt(b*x^3 + a)*c + d)/c^3)/b^2

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mupad [B]  time = 3.56, size = 119, normalized size = 1.63 \[ \frac {x^3}{3\,b\,c}-\frac {2\,d\,\sqrt {b\,x^3+a}}{3\,b^2\,c^2}+\frac {\ln \left (\frac {d-c\,\sqrt {b\,x^3+a}}{d+c\,\sqrt {b\,x^3+a}}\right )\,\left (a\,c^2-d^2\right )}{3\,b^2\,c^3}-\frac {\ln \left (b\,c^2\,x^3+a\,c^2-d^2\right )\,\left (a\,c^2-d^2\right )}{3\,b^2\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a*c + d*(a + b*x^3)^(1/2) + b*c*x^3),x)

[Out]

x^3/(3*b*c) - (2*d*(a + b*x^3)^(1/2))/(3*b^2*c^2) + (log((d - c*(a + b*x^3)^(1/2))/(d + c*(a + b*x^3)^(1/2)))*
(a*c^2 - d^2))/(3*b^2*c^3) - (log(a*c^2 - d^2 + b*c^2*x^3)*(a*c^2 - d^2))/(3*b^2*c^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(a*c+b*c*x**3+d*(b*x**3+a)**(1/2)),x)

[Out]

Timed out

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