3.23.96 \(\int \frac {(-q+2 p x^3) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^2 (a q+b x+a p x^3)} \, dx\) [2296]

Optimal. Leaf size=175 \[ \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a x}+\frac {2 \sqrt {-b^2+2 a^2 p q} \text {ArcTan}\left (\frac {\sqrt {-b^2+2 a^2 p q} x}{a q+b x+a p x^3+a \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}\right )}{a^2}+\frac {b \log (x)}{a^2}-\frac {b \log \left (q+p x^3+\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right )}{a^2} \]

[Out]

(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)/a/x+2*(2*a^2*p*q-b^2)^(1/2)*arctan((2*a^2*p*q-b^2)^(1/2)*x/(a*q+b*x+a*
p*x^3+a*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)))/a^2+b*ln(x)/a^2-b*ln(q+p*x^3+(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^
2)^(1/2))/a^2

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Rubi [F]
time = 1.93, 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 {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^2 \left (a q+b x+a p x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-q + 2*p*x^3)*Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6])/(x^2*(a*q + b*x + a*p*x^3)),x]

[Out]

-(Defer[Int][Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6]/x^2, x]/a) + (b*Defer[Int][Sqrt[q^2 - 2*p*q*x^2 + 2*p
*q*x^3 + p^2*x^6]/x, x])/(a^2*q) - (b^2*Defer[Int][Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6]/(a*q + b*x + a*
p*x^3), x])/(a^2*q) + 3*p*Defer[Int][(x*Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6])/(a*q + b*x + a*p*x^3), x]
- (b*p*Defer[Int][(x^2*Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6])/(a*q + b*x + a*p*x^3), x])/(a*q)

Rubi steps

\begin {align*} \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^2 \left (a q+b x+a p x^3\right )} \, dx &=\int \left (-\frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a x^2}+\frac {b \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a^2 q x}+\frac {\left (-b^2+3 a^2 p q x-a b p x^2\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a^2 q \left (a q+b x+a p x^3\right )}\right ) \, dx\\ &=-\frac {\int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^2} \, dx}{a}+\frac {\int \frac {\left (-b^2+3 a^2 p q x-a b p x^2\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q+b x+a p x^3} \, dx}{a^2 q}+\frac {b \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx}{a^2 q}\\ &=-\frac {\int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^2} \, dx}{a}+\frac {\int \left (-\frac {b^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q+b x+a p x^3}+\frac {3 a^2 p q x \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q+b x+a p x^3}-\frac {a b p x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q+b x+a p x^3}\right ) \, dx}{a^2 q}+\frac {b \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx}{a^2 q}\\ &=-\frac {\int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^2} \, dx}{a}+(3 p) \int \frac {x \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q+b x+a p x^3} \, dx+\frac {b \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx}{a^2 q}-\frac {b^2 \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q+b x+a p x^3} \, dx}{a^2 q}-\frac {(b p) \int \frac {x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q+b x+a p x^3} \, dx}{a q}\\ \end {align*}

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Mathematica [A]
time = 0.76, size = 156, normalized size = 0.89 \begin {gather*} -\frac {-a \sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}-2 \sqrt {-b^2+2 a^2 p q} x \text {ArcTan}\left (\frac {\sqrt {-b^2+2 a^2 p q} x}{b x+a \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right )}\right )+b x \tanh ^{-1}\left (\frac {\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}}{q+p x^3}\right )}{a^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-q + 2*p*x^3)*Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6])/(x^2*(a*q + b*x + a*p*x^3)),x]

[Out]

-((-(a*Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6]) - 2*Sqrt[-b^2 + 2*a^2*p*q]*x*ArcTan[(Sqrt[-b^2 + 2*a^2*p*q]*x
)/(b*x + a*(q + p*x^3 + Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6]))] + b*x*ArcTanh[Sqrt[q^2 + 2*p*q*(-1 + x)*x^
2 + p^2*x^6]/(q + p*x^3)])/(a^2*x))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (2 p \,x^{3}-q \right ) \sqrt {p^{2} x^{6}+2 p q \,x^{3}-2 p q \,x^{2}+q^{2}}}{x^{2} \left (a p \,x^{3}+a q +b x \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*p*x^3-q)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)/x^2/(a*p*x^3+a*q+b*x),x)

[Out]

int((2*p*x^3-q)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)/x^2/(a*p*x^3+a*q+b*x),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*p*x^3-q)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)/x^2/(a*p*x^3+a*q+b*x),x, algorithm="maxima")

[Out]

integrate(sqrt(p^2*x^6 + 2*p*q*x^3 - 2*p*q*x^2 + q^2)*(2*p*x^3 - q)/((a*p*x^3 + a*q + b*x)*x^2), x)

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Fricas [F(-1)] Timed out
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((2*p*x^3-q)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)/x^2/(a*p*x^3+a*q+b*x),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
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((2*p*x**3-q)*(p**2*x**6+2*p*q*x**3-2*p*q*x**2+q**2)**(1/2)/x**2/(a*p*x**3+a*q+b*x),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*p*x^3-q)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)/x^2/(a*p*x^3+a*q+b*x),x, algorithm="giac")

[Out]

integrate(sqrt(p^2*x^6 + 2*p*q*x^3 - 2*p*q*x^2 + q^2)*(2*p*x^3 - q)/((a*p*x^3 + a*q + b*x)*x^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {\left (q-2\,p\,x^3\right )\,\sqrt {p^2\,x^6+2\,p\,q\,x^3-2\,p\,q\,x^2+q^2}}{x^2\,\left (a\,p\,x^3+b\,x+a\,q\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((q - 2*p*x^3)*(p^2*x^6 + q^2 - 2*p*q*x^2 + 2*p*q*x^3)^(1/2))/(x^2*(a*q + b*x + a*p*x^3)),x)

[Out]

int(-((q - 2*p*x^3)*(p^2*x^6 + q^2 - 2*p*q*x^2 + 2*p*q*x^3)^(1/2))/(x^2*(a*q + b*x + a*p*x^3)), x)

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