3.22.67 \(\int \frac {(-q+2 p x^3) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x (b x^2+a (q+p x^3)^2)} \, dx\) [2167]

Optimal. Leaf size=159 \[ \frac {\sqrt {b+2 a p q} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {b+2 a p q} x^2}{a q^2+b x^2+2 a p q x^3+a p^2 x^6+\left (a q+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}\right )}{a \sqrt {b}}-\frac {\log (x)}{a}+\frac {\log \left (q+p x^3+\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right )}{a} \]

[Out]

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

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Rubi [F]
time = 3.73, 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 \left (b x^2+a \left (q+p x^3\right )^2\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*(b*x^2 + a*(q + p*x^3)^2)),x]

[Out]

-(Defer[Int][Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6]/x, x]/(a*q)) + (b*Defer[Int][(x*Sqrt[q^2 - 2*p*q*x^2
+ 2*p*q*x^3 + p^2*x^6])/(a*q^2 + b*x^2 + 2*a*p*q*x^3 + a*p^2*x^6), x])/(a*q) + 4*p*Defer[Int][(x^2*Sqrt[q^2 -
2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6])/(a*q^2 + b*x^2 + 2*a*p*q*x^3 + a*p^2*x^6), x] + (p^2*Defer[Int][(x^5*Sqrt[q^
2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6])/(a*q^2 + b*x^2 + 2*a*p*q*x^3 + a*p^2*x^6), x])/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 \left (b x^2+a \left (q+p x^3\right )^2\right )} \, dx &=\int \left (-\frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q x}+\frac {x \left (b+4 a p q x+a p^2 x^4\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q \left (a q^2+b x^2+2 a p q x^3+a p^2 x^6\right )}\right ) \, dx\\ &=-\frac {\int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx}{a q}+\frac {\int \frac {x \left (b+4 a p q x+a p^2 x^4\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q^2+b x^2+2 a p q x^3+a p^2 x^6} \, dx}{a q}\\ &=-\frac {\int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx}{a q}+\frac {\int \left (\frac {b x \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q^2+b x^2+2 a p q x^3+a p^2 x^6}+\frac {4 a p q x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q^2+b x^2+2 a p q x^3+a p^2 x^6}+\frac {a p^2 x^5 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q^2+b x^2+2 a p q x^3+a p^2 x^6}\right ) \, dx}{a q}\\ &=(4 p) \int \frac {x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q^2+b x^2+2 a p q x^3+a p^2 x^6} \, dx-\frac {\int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx}{a q}+\frac {b \int \frac {x \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q^2+b x^2+2 a p q x^3+a p^2 x^6} \, dx}{a q}+\frac {p^2 \int \frac {x^5 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{a q^2+b x^2+2 a p q x^3+a p^2 x^6} \, dx}{q}\\ \end {align*}

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Mathematica [A]
time = 0.68, size = 128, normalized size = 0.81 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}}{q+p x^3}\right )+\frac {\sqrt {b+2 a p q} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {b+2 a p q} x^2}{b x^2+a \left (q+p x^3\right ) \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right )}\right )}{\sqrt {b}}}{a} \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*(b*x^2 + a*(q + p*x^3)^2)),x]

[Out]

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

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

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/(b*x^2+a*(p*x^3+q)^2),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/(b*x^2+a*(p*x^3+q)^2),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/(b*x^2+a*(p*x^3+q)^2),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)/(((p*x^3 + q)^2*a + b*x^2)*x), 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/(b*x^2+a*(p*x^3+q)^2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \left (a p^{2} x^{6} + 2 a p q x^{3} + a q^{2} + b x^{2}\right )}\, dx \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/(b*x**2+a*(p*x**3+q)**2),x)

[Out]

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

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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/(b*x^2+a*(p*x^3+q)^2),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)/(((p*x^3 + q)^2*a + b*x^2)*x), 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\,\left (a\,{\left (p\,x^3+q\right )}^2+b\,x^2\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*(a*(q + p*x^3)^2 + b*x^2)),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*(a*(q + p*x^3)^2 + b*x^2)), x)

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