3.27.13 \(\int \frac {-3 a q+4 b p x^3+a p x^4}{\sqrt [3]{q+p x^4} (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4)} \, dx\)

Optimal. Leaf size=228 \[ \frac {\log \left (a^2 d^{2/3} x^2+\sqrt [3]{p x^4+q} \left (b \left (-\sqrt [3]{c}\right ) \sqrt [3]{d}-a \sqrt [3]{c} \sqrt [3]{d} x\right )+2 a b d^{2/3} x+b^2 d^{2/3}+c^{2/3} \left (p x^4+q\right )^{2/3}\right )}{2 c^{2/3} \sqrt [3]{d}}-\frac {\log \left (a \sqrt [3]{d} x+b \sqrt [3]{d}+\sqrt [3]{c} \sqrt [3]{p x^4+q}\right )}{c^{2/3} \sqrt [3]{d}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{c} \sqrt [3]{p x^4+q}}{-2 a \sqrt [3]{d} x-2 b \sqrt [3]{d}+\sqrt [3]{c} \sqrt [3]{p x^4+q}}\right )}{c^{2/3} \sqrt [3]{d}} \]

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Rubi [F]  time = 4.14, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-3 a q+4 b p x^3+a p x^4}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*a*q + 4*b*p*x^3 + a*p*x^4)/((q + p*x^4)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x + 3*a^2*b*d*x^2 + a^3*d*x^3 +
c*p*x^4)),x]

[Out]

(a*x*(1 + (p*x^4)/q)^(1/3)*Hypergeometric2F1[1/4, 1/3, 5/4, -((p*x^4)/q)])/(c*(q + p*x^4)^(1/3)) - (a*(b^3*d +
4*c*q)*Defer[Int][1/((q + p*x^4)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x + 3*a^2*b*d*x^2 + a^3*d*x^3 + c*p*x^4)), x]
)/c - (3*a^2*b^2*d*Defer[Int][x/((q + p*x^4)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x + 3*a^2*b*d*x^2 + a^3*d*x^3 + c*
p*x^4)), x])/c - (3*a^3*b*d*Defer[Int][x^2/((q + p*x^4)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x + 3*a^2*b*d*x^2 + a^3
*d*x^3 + c*p*x^4)), x])/c - ((a^4*d - 4*b*c*p)*Defer[Int][x^3/((q + p*x^4)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x +
3*a^2*b*d*x^2 + a^3*d*x^3 + c*p*x^4)), x])/c

Rubi steps

\begin {align*} \int \frac {-3 a q+4 b p x^3+a p x^4}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )} \, dx &=\int \left (\frac {a}{c \sqrt [3]{q+p x^4}}-\frac {a \left (b^3 d+4 c q\right )+3 a^2 b^2 d x+3 a^3 b d x^2+\left (a^4 d-4 b c p\right ) x^3}{c \sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )}\right ) \, dx\\ &=-\frac {\int \frac {a \left (b^3 d+4 c q\right )+3 a^2 b^2 d x+3 a^3 b d x^2+\left (a^4 d-4 b c p\right ) x^3}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )} \, dx}{c}+\frac {a \int \frac {1}{\sqrt [3]{q+p x^4}} \, dx}{c}\\ &=-\frac {\int \left (\frac {a \left (b^3 d+4 c q\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )}+\frac {3 a^2 b^2 d x}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )}+\frac {3 a^3 b d x^2}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )}+\frac {\left (a^4 d-4 b c p\right ) x^3}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )}\right ) \, dx}{c}+\frac {\left (a \sqrt [3]{1+\frac {p x^4}{q}}\right ) \int \frac {1}{\sqrt [3]{1+\frac {p x^4}{q}}} \, dx}{c \sqrt [3]{q+p x^4}}\\ &=\frac {a x \sqrt [3]{1+\frac {p x^4}{q}} \, _2F_1\left (\frac {1}{4},\frac {1}{3};\frac {5}{4};-\frac {p x^4}{q}\right )}{c \sqrt [3]{q+p x^4}}-\frac {\left (3 a^3 b d\right ) \int \frac {x^2}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )} \, dx}{c}-\frac {\left (3 a^2 b^2 d\right ) \int \frac {x}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )} \, dx}{c}-\frac {\left (a^4 d-4 b c p\right ) \int \frac {x^3}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )} \, dx}{c}-\frac {\left (a \left (b^3 d+4 c q\right )\right ) \int \frac {1}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )} \, dx}{c}\\ \end {align*}

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Mathematica [F]  time = 1.74, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-3 a q+4 b p x^3+a p x^4}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x+3 a^2 b d x^2+a^3 d x^3+c p x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(-3*a*q + 4*b*p*x^3 + a*p*x^4)/((q + p*x^4)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x + 3*a^2*b*d*x^2 + a^3*d
*x^3 + c*p*x^4)),x]

[Out]

Integrate[(-3*a*q + 4*b*p*x^3 + a*p*x^4)/((q + p*x^4)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x + 3*a^2*b*d*x^2 + a^3*d
*x^3 + c*p*x^4)), x]

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IntegrateAlgebraic [A]  time = 20.97, size = 228, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{c} \sqrt [3]{q+p x^4}}{-2 b \sqrt [3]{d}-2 a \sqrt [3]{d} x+\sqrt [3]{c} \sqrt [3]{q+p x^4}}\right )}{c^{2/3} \sqrt [3]{d}}-\frac {\log \left (b \sqrt [3]{d}+a \sqrt [3]{d} x+\sqrt [3]{c} \sqrt [3]{q+p x^4}\right )}{c^{2/3} \sqrt [3]{d}}+\frac {\log \left (b^2 d^{2/3}+2 a b d^{2/3} x+a^2 d^{2/3} x^2+\left (-b \sqrt [3]{c} \sqrt [3]{d}-a \sqrt [3]{c} \sqrt [3]{d} x\right ) \sqrt [3]{q+p x^4}+c^{2/3} \left (q+p x^4\right )^{2/3}\right )}{2 c^{2/3} \sqrt [3]{d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-3*a*q + 4*b*p*x^3 + a*p*x^4)/((q + p*x^4)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x + 3*a^2*b*d*x^
2 + a^3*d*x^3 + c*p*x^4)),x]

[Out]

-((Sqrt[3]*ArcTan[(Sqrt[3]*c^(1/3)*(q + p*x^4)^(1/3))/(-2*b*d^(1/3) - 2*a*d^(1/3)*x + c^(1/3)*(q + p*x^4)^(1/3
))])/(c^(2/3)*d^(1/3))) - Log[b*d^(1/3) + a*d^(1/3)*x + c^(1/3)*(q + p*x^4)^(1/3)]/(c^(2/3)*d^(1/3)) + Log[b^2
*d^(2/3) + 2*a*b*d^(2/3)*x + a^2*d^(2/3)*x^2 + (-(b*c^(1/3)*d^(1/3)) - a*c^(1/3)*d^(1/3)*x)*(q + p*x^4)^(1/3)
+ c^(2/3)*(q + p*x^4)^(2/3)]/(2*c^(2/3)*d^(1/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*p*x^4+4*b*p*x^3-3*a*q)/(p*x^4+q)^(1/3)/(a^3*d*x^3+3*a^2*b*d*x^2+c*p*x^4+3*a*b^2*d*x+b^3*d+c*q),x,
algorithm="fricas")

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*p*x^4+4*b*p*x^3-3*a*q)/(p*x^4+q)^(1/3)/(a^3*d*x^3+3*a^2*b*d*x^2+c*p*x^4+3*a*b^2*d*x+b^3*d+c*q),x,
algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {p a \,x^{4}+4 b p \,x^{3}-3 a q}{\left (p \,x^{4}+q \right )^{\frac {1}{3}} \left (a^{3} d \,x^{3}+3 a^{2} b d \,x^{2}+c p \,x^{4}+3 a \,b^{2} d x +b^{3} d +c q \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*p*x^4+4*b*p*x^3-3*a*q)/(p*x^4+q)^(1/3)/(a^3*d*x^3+3*a^2*b*d*x^2+c*p*x^4+3*a*b^2*d*x+b^3*d+c*q),x)

[Out]

int((a*p*x^4+4*b*p*x^3-3*a*q)/(p*x^4+q)^(1/3)/(a^3*d*x^3+3*a^2*b*d*x^2+c*p*x^4+3*a*b^2*d*x+b^3*d+c*q),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a p x^{4} + 4 \, b p x^{3} - 3 \, a q}{{\left (a^{3} d x^{3} + 3 \, a^{2} b d x^{2} + c p x^{4} + 3 \, a b^{2} d x + b^{3} d + c q\right )} {\left (p x^{4} + q\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*p*x^4+4*b*p*x^3-3*a*q)/(p*x^4+q)^(1/3)/(a^3*d*x^3+3*a^2*b*d*x^2+c*p*x^4+3*a*b^2*d*x+b^3*d+c*q),x,
algorithm="maxima")

[Out]

integrate((a*p*x^4 + 4*b*p*x^3 - 3*a*q)/((a^3*d*x^3 + 3*a^2*b*d*x^2 + c*p*x^4 + 3*a*b^2*d*x + b^3*d + c*q)*(p*
x^4 + q)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a\,p\,x^4+4\,b\,p\,x^3-3\,a\,q}{{\left (p\,x^4+q\right )}^{1/3}\,\left (d\,a^3\,x^3+3\,d\,a^2\,b\,x^2+3\,d\,a\,b^2\,x+d\,b^3+c\,p\,x^4+c\,q\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*p*x^4 - 3*a*q + 4*b*p*x^3)/((q + p*x^4)^(1/3)*(c*q + b^3*d + c*p*x^4 + a^3*d*x^3 + 3*a*b^2*d*x + 3*a^2*
b*d*x^2)),x)

[Out]

int((a*p*x^4 - 3*a*q + 4*b*p*x^3)/((q + p*x^4)^(1/3)*(c*q + b^3*d + c*p*x^4 + a^3*d*x^3 + 3*a*b^2*d*x + 3*a^2*
b*d*x^2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*p*x**4+4*b*p*x**3-3*a*q)/(p*x**4+q)**(1/3)/(a**3*d*x**3+3*a**2*b*d*x**2+c*p*x**4+3*a*b**2*d*x+b**
3*d+c*q),x)

[Out]

Timed out

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