3.28.8 \(\int \frac {(b+a x) (-3 a q+4 b p x^3+a p x^4)}{(q+p x^4)^{2/3} (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4)} \, dx\)

Optimal. Leaf size=247 \[ \frac {\log \left (a^2 \sqrt [3]{c} x+a b \sqrt [3]{c}+a \sqrt [3]{d} \sqrt [3]{p x^4+q}\right )}{c^{2/3} \sqrt [3]{d}}-\frac {\log \left (a^4 c^{2/3} x^2+2 a^3 b c^{2/3} x+a^2 b^2 c^{2/3}+a^2 d^{2/3} \left (p x^4+q\right )^{2/3}+\sqrt [3]{p x^4+q} \left (a^3 \left (-\sqrt [3]{c}\right ) \sqrt [3]{d} x-a^2 b \sqrt [3]{c} \sqrt [3]{d}\right )\right )}{2 c^{2/3} \sqrt [3]{d}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} a \sqrt [3]{c} x+\sqrt {3} b \sqrt [3]{c}}{a \sqrt [3]{c} x+b \sqrt [3]{c}-2 \sqrt [3]{d} \sqrt [3]{p x^4+q}}\right )}{c^{2/3} \sqrt [3]{d}} \]

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Rubi [F]  time = 6.22, 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 (-3 a q+4 b p x^3+a p x^4\right )}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-1/2*(3^(3/4)*Sqrt[2 - Sqrt[3]]*a^2*(q^(1/3) - (q + p*x^4)^(1/3))*Sqrt[(q^(2/3) + q^(1/3)*(q + p*x^4)^(1/3) +
(q + p*x^4)^(2/3))/((1 - Sqrt[3])*q^(1/3) - (q + p*x^4)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*q^(1/3) - (q
+ p*x^4)^(1/3))/((1 - Sqrt[3])*q^(1/3) - (q + p*x^4)^(1/3))], -7 + 4*Sqrt[3]])/(d*p*x^2*Sqrt[-((q^(1/3)*(q^(1
/3) - (q + p*x^4)^(1/3)))/((1 - Sqrt[3])*q^(1/3) - (q + p*x^4)^(1/3))^2)]) - (a*(a^4*c - 5*b*d*p)*x*(1 + (p*x^
4)/q)^(2/3)*Hypergeometric2F1[1/4, 2/3, 5/4, -((p*x^4)/q)])/(d^2*p*(q + p*x^4)^(2/3)) + (a*(a^4*c*(b^3*c + d*q
) - b*d*p*(5*b^3*c + 8*d*q))*Defer[Int][1/((q + p*x^4)^(2/3)*(b^3*c + d*q + 3*a*b^2*c*x + 3*a^2*b*c*x^2 + a^3*
c*x^3 + d*p*x^4)), x])/(d^2*p) + (a^2*(3*a^4*b^2*c^2 - 4*d*p*(4*b^3*c + d*q))*Defer[Int][x/((q + p*x^4)^(2/3)*
(b^3*c + d*q + 3*a*b^2*c*x + 3*a^2*b*c*x^2 + a^3*c*x^3 + d*p*x^4)), x])/(d^2*p) + (3*a^3*b*c*(a^4*c - 6*b*d*p)
*Defer[Int][x^2/((q + p*x^4)^(2/3)*(b^3*c + d*q + 3*a*b^2*c*x + 3*a^2*b*c*x^2 + a^3*c*x^3 + d*p*x^4)), x])/(d^
2*p) + ((a^8*c^2 - 8*a^4*b*c*d*p + 4*b^2*d^2*p^2)*Defer[Int][x^3/((q + p*x^4)^(2/3)*(b^3*c + d*q + 3*a*b^2*c*x
+ 3*a^2*b*c*x^2 + a^3*c*x^3 + d*p*x^4)), x])/(d^2*p)

Rubi steps

\begin {align*} \int \frac {(b+a x) \left (-3 a q+4 b p x^3+a p x^4\right )}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx &=\int \left (-\frac {a \left (a^4 c-5 b d p\right )}{d^2 p \left (q+p x^4\right )^{2/3}}+\frac {a^2 x}{d \left (q+p x^4\right )^{2/3}}+\frac {a \left (a^4 c \left (b^3 c+d q\right )-b d p \left (5 b^3 c+8 d q\right )\right )+a^2 \left (3 a^4 b^2 c^2-4 d p \left (4 b^3 c+d q\right )\right ) x+3 a^3 b c \left (a^4 c-6 b d p\right ) x^2+\left (a^8 c^2-8 a^4 b c d p+4 b^2 d^2 p^2\right ) x^3}{d^2 p \left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )}\right ) \, dx\\ &=\frac {a^2 \int \frac {x}{\left (q+p x^4\right )^{2/3}} \, dx}{d}+\frac {\int \frac {a \left (a^4 c \left (b^3 c+d q\right )-b d p \left (5 b^3 c+8 d q\right )\right )+a^2 \left (3 a^4 b^2 c^2-4 d p \left (4 b^3 c+d q\right )\right ) x+3 a^3 b c \left (a^4 c-6 b d p\right ) x^2+\left (a^8 c^2-8 a^4 b c d p+4 b^2 d^2 p^2\right ) x^3}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}-\frac {\left (a \left (a^4 c-5 b d p\right )\right ) \int \frac {1}{\left (q+p x^4\right )^{2/3}} \, dx}{d^2 p}\\ &=\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\left (q+p x^2\right )^{2/3}} \, dx,x,x^2\right )}{2 d}+\frac {\int \left (\frac {a \left (a^4 c \left (b^3 c+d q\right )-b d p \left (5 b^3 c+8 d q\right )\right )}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )}+\frac {a^2 \left (3 a^4 b^2 c^2-4 d p \left (4 b^3 c+d q\right )\right ) x}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )}+\frac {3 a^3 b c \left (a^4 c-6 b d p\right ) x^2}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )}+\frac {\left (a^8 c^2-8 a^4 b c d p+4 b^2 d^2 p^2\right ) x^3}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )}\right ) \, dx}{d^2 p}-\frac {\left (a \left (a^4 c-5 b d p\right ) \left (1+\frac {p x^4}{q}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {p x^4}{q}\right )^{2/3}} \, dx}{d^2 p \left (q+p x^4\right )^{2/3}}\\ &=-\frac {a \left (a^4 c-5 b d p\right ) x \left (1+\frac {p x^4}{q}\right )^{2/3} \, _2F_1\left (\frac {1}{4},\frac {2}{3};\frac {5}{4};-\frac {p x^4}{q}\right )}{d^2 p \left (q+p x^4\right )^{2/3}}+\frac {\left (3 a^3 b c \left (a^4 c-6 b d p\right )\right ) \int \frac {x^2}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a^8 c^2-8 a^4 b c d p+4 b^2 d^2 p^2\right ) \int \frac {x^3}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a^2 \left (3 a^4 b^2 c^2-4 d p \left (4 b^3 c+d q\right )\right )\right ) \int \frac {x}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a \left (a^4 c \left (b^3 c+d q\right )-b d p \left (5 b^3 c+8 d q\right )\right )\right ) \int \frac {1}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (3 a^2 \sqrt {p x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-q+x^3}} \, dx,x,\sqrt [3]{q+p x^4}\right )}{4 d p x^2}\\ &=-\frac {3^{3/4} \sqrt {2-\sqrt {3}} a^2 \left (\sqrt [3]{q}-\sqrt [3]{q+p x^4}\right ) \sqrt {\frac {q^{2/3}+\sqrt [3]{q} \sqrt [3]{q+p x^4}+\left (q+p x^4\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{q}-\sqrt [3]{q+p x^4}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{q}-\sqrt [3]{q+p x^4}}{\left (1-\sqrt {3}\right ) \sqrt [3]{q}-\sqrt [3]{q+p x^4}}\right )|-7+4 \sqrt {3}\right )}{2 d p x^2 \sqrt {-\frac {\sqrt [3]{q} \left (\sqrt [3]{q}-\sqrt [3]{q+p x^4}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{q}-\sqrt [3]{q+p x^4}\right )^2}}}-\frac {a \left (a^4 c-5 b d p\right ) x \left (1+\frac {p x^4}{q}\right )^{2/3} \, _2F_1\left (\frac {1}{4},\frac {2}{3};\frac {5}{4};-\frac {p x^4}{q}\right )}{d^2 p \left (q+p x^4\right )^{2/3}}+\frac {\left (3 a^3 b c \left (a^4 c-6 b d p\right )\right ) \int \frac {x^2}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a^8 c^2-8 a^4 b c d p+4 b^2 d^2 p^2\right ) \int \frac {x^3}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a^2 \left (3 a^4 b^2 c^2-4 d p \left (4 b^3 c+d q\right )\right )\right ) \int \frac {x}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a \left (a^4 c \left (b^3 c+d q\right )-b d p \left (5 b^3 c+8 d q\right )\right )\right ) \int \frac {1}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}\\ \end {align*}

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Mathematica [F]  time = 2.13, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(b+a x) \left (-3 a q+4 b p x^3+a p x^4\right )}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 23.11, size = 247, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \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^4}}\right )}{c^{2/3} \sqrt [3]{d}}+\frac {\log \left (a b \sqrt [3]{c}+a^2 \sqrt [3]{c} x+a \sqrt [3]{d} \sqrt [3]{q+p x^4}\right )}{c^{2/3} \sqrt [3]{d}}-\frac {\log \left (a^2 b^2 c^{2/3}+2 a^3 b c^{2/3} x+a^4 c^{2/3} x^2+\left (-a^2 b \sqrt [3]{c} \sqrt [3]{d}-a^3 \sqrt [3]{c} \sqrt [3]{d} x\right ) \sqrt [3]{q+p x^4}+a^2 d^{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[((b + a*x)*(-3*a*q + 4*b*p*x^3 + a*p*x^4))/((q + p*x^4)^(2/3)*(b^3*c + d*q + 3*a*b^2*c*x +
3*a^2*b*c*x^2 + a^3*c*x^3 + d*p*x^4)),x]

[Out]

(Sqrt[3]*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^4)^(1/
3))])/(c^(2/3)*d^(1/3)) + Log[a*b*c^(1/3) + a^2*c^(1/3)*x + a*d^(1/3)*(q + p*x^4)^(1/3)]/(c^(2/3)*d^(1/3)) - L
og[a^2*b^2*c^(2/3) + 2*a^3*b*c^(2/3)*x + a^4*c^(2/3)*x^2 + (-(a^2*b*c^(1/3)*d^(1/3)) - a^3*c^(1/3)*d^(1/3)*x)*
(q + p*x^4)^(1/3) + a^2*d^(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*x+b)*(a*p*x^4+4*b*p*x^3-3*a*q)/(p*x^4+q)^(2/3)/(a^3*c*x^3+3*a^2*b*c*x^2+d*p*x^4+3*a*b^2*c*x+b^3*c
+d*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*x+b)*(a*p*x^4+4*b*p*x^3-3*a*q)/(p*x^4+q)^(2/3)/(a^3*c*x^3+3*a^2*b*c*x^2+d*p*x^4+3*a*b^2*c*x+b^3*c
+d*q),x, algorithm="giac")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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