3.28.22 \(\int \frac {(b+a x) (-a q+b p x^2)}{(q+p x^3)^{2/3} (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+(a^3 c+d p) x^3)} \, dx\)

Optimal. Leaf size=250 \[ -\frac {\log \left (a^2 b^2 c^{2/3} x^2+2 a b^3 c^{2/3} x+\sqrt [3]{p x^3+q} \left (b^3 \left (-\sqrt [3]{c}\right ) \sqrt [3]{d}-a b^2 \sqrt [3]{c} \sqrt [3]{d} x\right )+b^4 c^{2/3}+b^2 d^{2/3} \left (p x^3+q\right )^{2/3}\right )}{6 c^{2/3} \sqrt [3]{d}}+\frac {\log \left (a b \sqrt [3]{c} x+b^2 \sqrt [3]{c}+b \sqrt [3]{d} \sqrt [3]{p x^3+q}\right )}{3 c^{2/3} \sqrt [3]{d}}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} a \sqrt [3]{c} x+\sqrt {3} b \sqrt [3]{c}}{a \sqrt [3]{c} x+b \sqrt [3]{c}-2 \sqrt [3]{d} \sqrt [3]{p x^3+q}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{d}} \]

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Rubi [F]  time = 3.57, 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 {(b+a x) \left (-a q+b p x^2\right )}{\left (q+p x^3\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+\left (a^3 c+d p\right ) x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 1.54, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(b+a x) \left (-a q+b p x^2\right )}{\left (q+p x^3\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+\left (a^3 c+d p\right ) x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 17.80, size = 250, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {3} b \sqrt [3]{c}+\sqrt {3} a \sqrt [3]{c} x}{b \sqrt [3]{c}+a \sqrt [3]{c} x-2 \sqrt [3]{d} \sqrt [3]{q+p x^3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{d}}+\frac {\log \left (b^2 \sqrt [3]{c}+a b \sqrt [3]{c} x+b \sqrt [3]{d} \sqrt [3]{q+p x^3}\right )}{3 c^{2/3} \sqrt [3]{d}}-\frac {\log \left (b^4 c^{2/3}+2 a b^3 c^{2/3} x+a^2 b^2 c^{2/3} x^2+\left (-b^3 \sqrt [3]{c} \sqrt [3]{d}-a b^2 \sqrt [3]{c} \sqrt [3]{d} x\right ) \sqrt [3]{q+p x^3}+b^2 d^{2/3} \left (q+p x^3\right )^{2/3}\right )}{6 c^{2/3} \sqrt [3]{d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[3]*b*c^(1/3) + Sqrt[3]*a*c^(1/3)*x)/(b*c^(1/3) + a*c^(1/3)*x - 2*d^(1/3)*(q + p*x^3)^(1/3))]/(Sqr
t[3]*c^(2/3)*d^(1/3)) + Log[b^2*c^(1/3) + a*b*c^(1/3)*x + b*d^(1/3)*(q + p*x^3)^(1/3)]/(3*c^(2/3)*d^(1/3)) - L
og[b^4*c^(2/3) + 2*a*b^3*c^(2/3)*x + a^2*b^2*c^(2/3)*x^2 + (-(b^3*c^(1/3)*d^(1/3)) - a*b^2*c^(1/3)*d^(1/3)*x)*
(q + p*x^3)^(1/3) + b^2*d^(2/3)*(q + p*x^3)^(2/3)]/(6*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*x+b)*(b*p*x^2-a*q)/(p*x^3+q)^(2/3)/(b^3*c+d*q+3*a*b^2*c*x+3*a^2*b*c*x^2+(a^3*c+d*p)*x^3),x, algor
ithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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