\(\int \frac {(-2 q+p x^3) \sqrt {q+p x^3}}{x^2 (a q+b x^2+a p x^3)} \, dx\) [724]

Optimal result

Integrand size = 41, antiderivative size = 56 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\frac {2 \sqrt {q+p x^3}}{a x}+\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a} \sqrt {q+p x^3}}\right )}{a^{3/2}} \]

[Out]

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

Rubi [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx \]

[In]

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

[Out]

(2*Sqrt[q + p*x^3])/(a*x) - (6*p^(1/3)*Sqrt[q + p*x^3])/(a*((1 + Sqrt[3])*q^(1/3) + p^(1/3)*x)) + (3*3^(1/4)*S
qrt[2 - Sqrt[3]]*p^(1/3)*q^(1/3)*(q^(1/3) + p^(1/3)*x)*Sqrt[(q^(2/3) - p^(1/3)*q^(1/3)*x + p^(2/3)*x^2)/((1 +
Sqrt[3])*q^(1/3) + p^(1/3)*x)^2]*EllipticE[ArcSin[((1 - Sqrt[3])*q^(1/3) + p^(1/3)*x)/((1 + Sqrt[3])*q^(1/3) +
p^(1/3)*x)], -7 - 4*Sqrt[3]])/(a*Sqrt[(q^(1/3)*(q^(1/3) + p^(1/3)*x))/((1 + Sqrt[3])*q^(1/3) + p^(1/3)*x)^2]*
Sqrt[q + p*x^3]) - (2*Sqrt[2]*3^(3/4)*p^(1/3)*q^(1/3)*(q^(1/3) + p^(1/3)*x)*Sqrt[(q^(2/3) - p^(1/3)*q^(1/3)*x
+ p^(2/3)*x^2)/((1 + Sqrt[3])*q^(1/3) + p^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*q^(1/3) + p^(1/3)*x)/((1
+ Sqrt[3])*q^(1/3) + p^(1/3)*x)], -7 - 4*Sqrt[3]])/(a*Sqrt[(q^(1/3)*(q^(1/3) + p^(1/3)*x))/((1 + Sqrt[3])*q^(
1/3) + p^(1/3)*x)^2]*Sqrt[q + p*x^3]) + (2*b*Defer[Int][Sqrt[q + p*x^3]/(a*q + b*x^2 + a*p*x^3), x])/a + 3*p*D
efer[Int][(x*Sqrt[q + p*x^3])/(a*q + b*x^2 + a*p*x^3), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 \sqrt {q+p x^3}}{a x^2}+\frac {(2 b+3 a p x) \sqrt {q+p x^3}}{a \left (a q+b x^2+a p x^3\right )}\right ) \, dx \\ & = \frac {\int \frac {(2 b+3 a p x) \sqrt {q+p x^3}}{a q+b x^2+a p x^3} \, dx}{a}-\frac {2 \int \frac {\sqrt {q+p x^3}}{x^2} \, dx}{a} \\ & = \frac {2 \sqrt {q+p x^3}}{a x}+\frac {\int \left (\frac {2 b \sqrt {q+p x^3}}{a q+b x^2+a p x^3}+\frac {3 a p x \sqrt {q+p x^3}}{a q+b x^2+a p x^3}\right ) \, dx}{a}-\frac {(3 p) \int \frac {x}{\sqrt {q+p x^3}} \, dx}{a} \\ & = \frac {2 \sqrt {q+p x^3}}{a x}+\frac {(2 b) \int \frac {\sqrt {q+p x^3}}{a q+b x^2+a p x^3} \, dx}{a}-\frac {\left (3 p^{2/3}\right ) \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x}{\sqrt {q+p x^3}} \, dx}{a}+(3 p) \int \frac {x \sqrt {q+p x^3}}{a q+b x^2+a p x^3} \, dx+\frac {\left (3 \left (1-\sqrt {3}\right ) p^{2/3} \sqrt [3]{q}\right ) \int \frac {1}{\sqrt {q+p x^3}} \, dx}{a} \\ & = \frac {2 \sqrt {q+p x^3}}{a x}-\frac {6 \sqrt [3]{p} \sqrt {q+p x^3}}{a \left (\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x\right )}+\frac {3 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{p} \sqrt [3]{q} \left (\sqrt [3]{q}+\sqrt [3]{p} x\right ) \sqrt {\frac {q^{2/3}-\sqrt [3]{p} \sqrt [3]{q} x+p^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x}\right )|-7-4 \sqrt {3}\right )}{a \sqrt {\frac {\sqrt [3]{q} \left (\sqrt [3]{q}+\sqrt [3]{p} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x\right )^2}} \sqrt {q+p x^3}}-\frac {2 \sqrt {2} 3^{3/4} \sqrt [3]{p} \sqrt [3]{q} \left (\sqrt [3]{q}+\sqrt [3]{p} x\right ) \sqrt {\frac {q^{2/3}-\sqrt [3]{p} \sqrt [3]{q} x+p^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x}\right ),-7-4 \sqrt {3}\right )}{a \sqrt {\frac {\sqrt [3]{q} \left (\sqrt [3]{q}+\sqrt [3]{p} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{q}+\sqrt [3]{p} x\right )^2}} \sqrt {q+p x^3}}+\frac {(2 b) \int \frac {\sqrt {q+p x^3}}{a q+b x^2+a p x^3} \, dx}{a}+(3 p) \int \frac {x \sqrt {q+p x^3}}{a q+b x^2+a p x^3} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\frac {2 \sqrt {q+p x^3}}{a x}+\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a} \sqrt {q+p x^3}}\right )}{a^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 6.64 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {2 \sqrt {p \,x^{3}+q}}{a x}-\frac {2 b \arctan \left (\frac {a \sqrt {p \,x^{3}+q}}{x \sqrt {a b}}\right )}{a \sqrt {a b}}\)	\(50\)
default	\(-\frac {2 \left (b \arctan \left (\frac {a \sqrt {p \,x^{3}+q}}{x \sqrt {a b}}\right ) x -\sqrt {p \,x^{3}+q}\, \sqrt {a b}\right )}{a x \sqrt {a b}}\)	\(54\)
pseudoelliptic	\(-\frac {2 \left (b \arctan \left (\frac {a \sqrt {p \,x^{3}+q}}{x \sqrt {a b}}\right ) x -\sqrt {p \,x^{3}+q}\, \sqrt {a b}\right )}{a x \sqrt {a b}}\)	\(54\)
elliptic	\(\text {Expression too large to display}\)	\(877\)

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\int { \frac {\sqrt {p x^{3} + q} {\left (p x^{3} - 2 \, q\right )}}{{\left (a p x^{3} + b x^{2} + a q\right )} x^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\int { \frac {\sqrt {p x^{3} + q} {\left (p x^{3} - 2 \, q\right )}}{{\left (a p x^{3} + b x^{2} + a q\right )} x^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.40 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.82 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{x^2 \left (a q+b x^2+a p x^3\right )} \, dx=\frac {2\,\sqrt {p\,x^3+q}}{a\,x}+\frac {\sqrt {b}\,\ln \left (\frac {a^5\,b\,p^4\,x^2-a^6\,p^4\,\left (p\,x^3+q\right )+a^{11/2}\,\sqrt {b}\,p^4\,x\,\sqrt {p\,x^3+q}\,2{}\mathrm {i}}{4\,b^2\,q\,x^2+4\,a\,b\,q\,\left (p\,x^3+q\right )}\right )\,1{}\mathrm {i}}{a^{3/2}} \]

[In]

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

[Out]

(2*(q + p*x^3)^(1/2))/(a*x) + (b^(1/2)*log((a^5*b*p^4*x^2 - a^6*p^4*(q + p*x^3) + a^(11/2)*b^(1/2)*p^4*x*(q +
p*x^3)^(1/2)*2i)/(4*b^2*q*x^2 + 4*a*b*q*(q + p*x^3)))*1i)/a^(3/2)