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

Optimal. Leaf size=42 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {p x^3+q}}{\sqrt {c} (a x+b)}\right )}{\sqrt {c} \sqrt {d}} \]

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Rubi [F]  time = 2.63, 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 {-2 a q+3 b p x^2+a p x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {-2 a q+3 b p x^2+a p x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx &=\int \left (\frac {a}{d \sqrt {q+p x^3}}-\frac {a \left (b^2 c+3 d q\right )+2 a^2 b c x+\left (a^3 c-3 b d p\right ) x^2}{d \sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )}\right ) \, dx\\ &=-\frac {\int \frac {a \left (b^2 c+3 d q\right )+2 a^2 b c x+\left (a^3 c-3 b d p\right ) x^2}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx}{d}+\frac {a \int \frac {1}{\sqrt {q+p x^3}} \, dx}{d}\\ &=\frac {2 \sqrt {2+\sqrt {3}} a \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}} F\left (\sin ^{-1}\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 )}{\sqrt [4]{3} d \sqrt [3]{p} \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 {\int \left (\frac {a \left (b^2 c+3 d q\right )}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )}+\frac {2 a^2 b c x}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )}+\frac {\left (a^3 c-3 b d p\right ) x^2}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )}\right ) \, dx}{d}\\ &=\frac {2 \sqrt {2+\sqrt {3}} a \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}} F\left (\sin ^{-1}\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 )}{\sqrt [4]{3} d \sqrt [3]{p} \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 {\left (2 a^2 b c\right ) \int \frac {x}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx}{d}-\frac {\left (a^3 c-3 b d p\right ) \int \frac {x^2}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx}{d}-\frac {\left (a \left (b^2 c+3 d q\right )\right ) \int \frac {1}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx}{d}\\ \end {align*}

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Mathematica [C]  time = 6.79, size = 5105, normalized size = 121.55 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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IntegrateAlgebraic [A]  time = 6.70, size = 42, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {q+p x^3}}{\sqrt {c} (b+a x)}\right )}{\sqrt {c} \sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.68, size = 459, normalized size = 10.93 \begin {gather*} \left [-\frac {\sqrt {-c d} \log \left (-\frac {6 \, a^{2} c d p x^{5} - d^{2} p^{2} x^{6} - b^{4} c^{2} + 6 \, b^{2} c d q - {\left (a^{4} c^{2} - 12 \, a b c d p\right )} x^{4} - d^{2} q^{2} - 2 \, {\left (2 \, a^{3} b c^{2} - 3 \, b^{2} c d p + d^{2} p q\right )} x^{3} - 6 \, {\left (a^{2} b^{2} c^{2} - a^{2} c d q\right )} x^{2} + 4 \, {\left (a d p x^{4} - 3 \, a^{2} b c x^{2} - b^{3} c - {\left (a^{3} c - b d p\right )} x^{3} + b d q - {\left (3 \, a b^{2} c - a d q\right )} x\right )} \sqrt {p x^{3} + q} \sqrt {-c d} - 4 \, {\left (a b^{3} c^{2} - 3 \, a b c d q\right )} x}{2 \, a^{2} c d p x^{5} + d^{2} p^{2} x^{6} + b^{4} c^{2} + 2 \, b^{2} c d q + {\left (a^{4} c^{2} + 4 \, a b c d p\right )} x^{4} + d^{2} q^{2} + 2 \, {\left (2 \, a^{3} b c^{2} + b^{2} c d p + d^{2} p q\right )} x^{3} + 2 \, {\left (3 \, a^{2} b^{2} c^{2} + a^{2} c d q\right )} x^{2} + 4 \, {\left (a b^{3} c^{2} + a b c d q\right )} x}\right )}{2 \, c d}, \frac {\sqrt {c d} \arctan \left (-\frac {{\left (a^{2} c x^{2} - d p x^{3} + 2 \, a b c x + b^{2} c - d q\right )} \sqrt {p x^{3} + q} \sqrt {c d}}{2 \, {\left (a c d p x^{4} + b c d p x^{3} + a c d q x + b c d q\right )}}\right )}{c d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-c*d)*log(-(6*a^2*c*d*p*x^5 - d^2*p^2*x^6 - b^4*c^2 + 6*b^2*c*d*q - (a^4*c^2 - 12*a*b*c*d*p)*x^4 -
d^2*q^2 - 2*(2*a^3*b*c^2 - 3*b^2*c*d*p + d^2*p*q)*x^3 - 6*(a^2*b^2*c^2 - a^2*c*d*q)*x^2 + 4*(a*d*p*x^4 - 3*a^2
*b*c*x^2 - b^3*c - (a^3*c - b*d*p)*x^3 + b*d*q - (3*a*b^2*c - a*d*q)*x)*sqrt(p*x^3 + q)*sqrt(-c*d) - 4*(a*b^3*
c^2 - 3*a*b*c*d*q)*x)/(2*a^2*c*d*p*x^5 + d^2*p^2*x^6 + b^4*c^2 + 2*b^2*c*d*q + (a^4*c^2 + 4*a*b*c*d*p)*x^4 + d
^2*q^2 + 2*(2*a^3*b*c^2 + b^2*c*d*p + d^2*p*q)*x^3 + 2*(3*a^2*b^2*c^2 + a^2*c*d*q)*x^2 + 4*(a*b^3*c^2 + a*b*c*
d*q)*x))/(c*d), sqrt(c*d)*arctan(-1/2*(a^2*c*x^2 - d*p*x^3 + 2*a*b*c*x + b^2*c - d*q)*sqrt(p*x^3 + q)*sqrt(c*d
)/(a*c*d*p*x^4 + b*c*d*p*x^3 + a*c*d*q*x + b*c*d*q))/(c*d)]

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

[Out]

Timed out

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maple [C]  time = 0.77, size = 2262, normalized size = 53.86

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(2262\)
default	\(\text {Expression too large to display}\)	\(2264\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*I*a/d*3^(1/2)/p*(-q*p^2)^(1/3)*(I*(x+1/2/p*(-q*p^2)^(1/3)-1/2*I*3^(1/2)/p*(-q*p^2)^(1/3))*3^(1/2)*p/(-q*p
^2)^(1/3))^(1/2)*((x-1/p*(-q*p^2)^(1/3))/(-3/2/p*(-q*p^2)^(1/3)+1/2*I*3^(1/2)/p*(-q*p^2)^(1/3)))^(1/2)*(-I*(x+
1/2/p*(-q*p^2)^(1/3)+1/2*I*3^(1/2)/p*(-q*p^2)^(1/3))*3^(1/2)*p/(-q*p^2)^(1/3))^(1/2)/(p*x^3+q)^(1/2)*EllipticF
(1/3*3^(1/2)*(I*(x+1/2/p*(-q*p^2)^(1/3)-1/2*I*3^(1/2)/p*(-q*p^2)^(1/3))*3^(1/2)*p/(-q*p^2)^(1/3))^(1/2),(I*3^(
1/2)/p*(-q*p^2)^(1/3)/(-3/2/p*(-q*p^2)^(1/3)+1/2*I*3^(1/2)/p*(-q*p^2)^(1/3)))^(1/2))-I/d/p^2/c*2^(1/2)*sum((_a
lpha^2*a^3*c-3*_alpha^2*b*d*p+2*_alpha*a^2*b*c+a*b^2*c+3*a*d*q)/(3*_alpha^2*d*p+2*_alpha*a^2*c+2*a*b*c)/(a^6*q
^2-2*a^3*b^3*p*q+b^6*p^2)*(-q*p^2)^(1/3)*(1/2*I*p*(2*x+1/p*(-I*3^(1/2)*(-q*p^2)^(1/3)+(-q*p^2)^(1/3)))/(-q*p^2
)^(1/3))^(1/2)*(p*(x-1/p*(-q*p^2)^(1/3))/(-3*(-q*p^2)^(1/3)+I*3^(1/2)*(-q*p^2)^(1/3)))^(1/2)*(-1/2*I*p*(2*x+1/
p*(I*3^(1/2)*(-q*p^2)^(1/3)+(-q*p^2)^(1/3)))/(-q*p^2)^(1/3))^(1/2)/(p*x^3+q)^(1/2)*(-(-q*p^2)^(2/3)*a^6*c*q+(-
q*p^2)^(2/3)*b^4*d*p^2+4*(-q*p^2)^(2/3)*a^3*b^3*c*p-3*(-q*p^2)^(1/3)*a^2*b^4*c*p^2+4*I*(-q*p^2)^(2/3)*3^(1/2)*
a^3*b^3*p*c+3*I*(-q*p^2)^(1/3)*3^(1/2)*a^2*b^4*p^2*c+2*p^2*(-2*_alpha^2*a^3*b*d*p*q-_alpha^2*b^4*d*p^2-2*_alph
a*a^5*b*c*q-_alpha*a^2*b^4*c*p+3*_alpha*a^2*b^2*d*p*q-a^4*b^2*c*q-2*a*b^5*c*p-a^4*d*q^2-2*a*b^3*d*p*q)+2*I*(-q
*p^2)^(2/3)*3^(1/2)*a^3*b*d*q*p+3*I*(-q*p^2)^(1/3)*3^(1/2)*a^2*b^2*d*q*p^2+2*(-q*p^2)^(1/3)*_alpha*a^3*b*d*p^2
*q-I*(-q*p^2)^(1/3)*p^3*3^(1/2)*_alpha*b^4*d+2*(-q*p^2)^(2/3)*a^3*b*d*p*q-3*(-q*p^2)^(1/3)*a^2*b^2*d*p^2*q+I*(
-q*p^2)^(2/3)*3^(1/2)*b^4*d*p^2-I*(-q*p^2)^(2/3)*3^(1/2)*a^6*q*c+(-q*p^2)^(1/3)*_alpha*b^4*d*p^3+3*(-q*p^2)^(2
/3)*_alpha^2*a^2*b^2*d*p^2+3*(-q*p^2)^(2/3)*_alpha*a^4*b^2*c*p-(-q*p^2)^(1/3)*_alpha^2*a^4*d*p^2*q-2*(-q*p^2)^
(1/3)*_alpha^2*a*b^3*d*p^3-(-q*p^2)^(1/3)*_alpha*a^6*c*p*q-2*(-q*p^2)^(1/3)*_alpha*a^3*b^3*c*p^2-(-q*p^2)^(2/3
)*_alpha*a^4*d*p*q-2*(-q*p^2)^(2/3)*_alpha*a*b^3*d*p^2+3*I*(-q*p^2)^(2/3)*3^(1/2)*_alpha^2*a^2*b^2*d*p^2+3*I*(
-q*p^2)^(2/3)*3^(1/2)*_alpha*a^4*b^2*p*c+2*I*(-q*p^2)^(1/3)*p^3*3^(1/2)*_alpha^2*a*b^3*d+2*I*(-q*p^2)^(1/3)*3^
(1/2)*_alpha*a^3*b^3*p^2*c-I*(-q*p^2)^(2/3)*3^(1/2)*_alpha*a^4*d*q*p-2*I*(-q*p^2)^(2/3)*3^(1/2)*_alpha*a*b^3*d
*p^2+I*(-q*p^2)^(1/3)*3^(1/2)*_alpha^2*a^4*d*q*p^2+I*(-q*p^2)^(1/3)*3^(1/2)*_alpha*a^6*q*p*c-2*I*(-q*p^2)^(1/3
)*3^(1/2)*_alpha*a^3*b*d*q*p^2)*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/p*(-q*p^2)^(1/3)-1/2*I*3^(1/2)/p*(-q*p^2)^(1/
3))*3^(1/2)*p/(-q*p^2)^(1/3))^(1/2),-1/2/p*(-2*I*3^(1/2)*a^3*b*d*p^2*q^2-3*I*(-q*p^2)^(2/3)*3^(1/2)*a^2*b^4*c*
p-4*I*(-q*p^2)^(1/3)*3^(1/2)*a*b^5*c*p^2-4*I*3^(1/2)*a^3*b^3*c*p^2*q-2*I*(-q*p^2)^(1/3)*3^(1/2)*a^4*d*p*q^2+3*
p^3*b^4*d*q-3*a^6*q^2*p*c-4*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha*a^5*b*c*p*q+2*I*(-q*p^2)^(2/3)*3^(1/2)*_alpha*a^3*
b*d*p*q+6*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha*a^2*b^2*d*p^2*q+12*a^3*b^3*p^2*q*c+I*(-q*p^2)^(2/3)*3^(1/2)*_alpha*b
^4*d*p^2+I*3^(1/2)*_alpha*a^4*d*p^2*q^2-I*(-q*p^2)^(2/3)*3^(1/2)*_alpha*a^6*c*q-2*I*(-q*p^2)^(1/3)*3^(1/2)*_al
pha^2*b^4*d*p^3+6*(-q*p^2)^(2/3)*_alpha*a^3*b*d*p*q+6*a^3*b*d*q^2*p^2+9*p^3*_alpha^2*a^2*b^2*d*q+9*_alpha*a^4*
b^2*p^2*q*c-6*p^3*_alpha*a*b^3*d*q-3*(-q*p^2)^(2/3)*a^4*d*_alpha^2*p*q-6*(-q*p^2)^(2/3)*a*d*_alpha^2*b^3*p^2-6
*(-q*p^2)^(2/3)*_alpha*a^3*b^3*p*c-9*(-q*p^2)^(2/3)*a^2*b^4*p*c-3*_alpha*a^4*d*q^2*p^2-3*(-q*p^2)^(2/3)*_alpha
*a^6*q*c+3*(-q*p^2)^(2/3)*_alpha*b^4*d*p^2-9*(-q*p^2)^(2/3)*a^2*b^2*d*p*q+I*3^(1/2)*a^6*c*p*q^2-I*3^(1/2)*b^4*
d*p^3*q-2*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha*a^2*b^4*c*p^2-3*I*3^(1/2)*_alpha^2*a^2*b^2*d*p^3*q-3*I*3^(1/2)*_alph
a*a^4*b^2*c*p^2*q+2*I*3^(1/2)*_alpha*a*b^3*d*p^3*q-I*(-q*p^2)^(2/3)*3^(1/2)*_alpha^2*a^4*d*p*q-2*I*(-q*p^2)^(1
/3)*3^(1/2)*a^4*b^2*c*p*q-3*I*(-q*p^2)^(2/3)*3^(1/2)*a^2*b^2*d*p*q-4*I*(-q*p^2)^(1/3)*3^(1/2)*a*b^3*d*p^2*q-2*
I*(-q*p^2)^(2/3)*3^(1/2)*_alpha^2*a*b^3*d*p^2-2*I*(-q*p^2)^(2/3)*3^(1/2)*_alpha*a^3*b^3*c*p-4*I*(-q*p^2)^(1/3)
*3^(1/2)*_alpha^2*a^3*b*d*p^2*q)/(a^6*q^2-2*a^3*b^3*p*q+b^6*p^2)/c,(I*3^(1/2)/p*(-q*p^2)^(1/3)/(-3/2/p*(-q*p^2
)^(1/3)+1/2*I*3^(1/2)/p*(-q*p^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d*p+_Z^2*a^2*c+2*_Z*a*b*c+b^2*c+d*q))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 10.90, size = 456, normalized size = 10.86 \begin {gather*} \frac {\ln \left (\frac {\left (a^3\,b\,c\,1{}\mathrm {i}-b^2\,d\,p\,1{}\mathrm {i}+a^4\,c\,x\,1{}\mathrm {i}+2\,a^3\,\sqrt {c}\,\sqrt {d}\,\sqrt {p\,x^3+q}-a^2\,d\,p\,x^2\,1{}\mathrm {i}+a\,b\,d\,p\,x\,1{}\mathrm {i}\right )\,\left (a^3\,b^3\,c^2\,1{}\mathrm {i}+a^6\,c^2\,x^3\,1{}\mathrm {i}+a^4\,b^2\,c^2\,x\,3{}\mathrm {i}+a^5\,b\,c^2\,x^2\,3{}\mathrm {i}-b^4\,c\,d\,p\,1{}\mathrm {i}+b^2\,d^2\,p\,\left (p\,x^3+q\right )\,1{}\mathrm {i}+a^2\,d^2\,p\,x^2\,\left (p\,x^3+q\right )\,1{}\mathrm {i}+a^3\,b\,c\,d\,q\,1{}\mathrm {i}+a^3\,b\,c\,d\,\left (p\,x^3+q\right )\,2{}\mathrm {i}+a^4\,c\,d\,q\,x\,1{}\mathrm {i}+a^4\,c\,d\,x\,\left (p\,x^3+q\right )\,2{}\mathrm {i}+2\,a^3\,\sqrt {c}\,d^{3/2}\,q\,\sqrt {p\,x^3+q}-2\,b^3\,\sqrt {c}\,d^{3/2}\,p\,\sqrt {p\,x^3+q}-a\,b^3\,c\,d\,p\,x\,1{}\mathrm {i}-a\,b\,d^2\,p\,x\,\left (p\,x^3+q\right )\,1{}\mathrm {i}\right )}{\left (c\,a^2\,x^2+2\,c\,a\,b\,x+c\,b^2+d\,p\,x^3+d\,q\right )\,\left (a^8\,c^2\,x^2+2\,a^7\,b\,c^2\,x+a^6\,b^2\,c^2+2\,a^6\,c\,d\,p\,x^3+4\,q\,a^6\,c\,d+a^4\,d^2\,p^2\,x^4-2\,a^3\,b^3\,c\,d\,p-2\,a^3\,b\,d^2\,p^2\,x^3+3\,a^2\,b^2\,d^2\,p^2\,x^2-2\,a\,b^3\,d^2\,p^2\,x+b^4\,d^2\,p^2\right )}\right )\,1{}\mathrm {i}}{\sqrt {c}\,\sqrt {d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(((a^3*b*c*1i - b^2*d*p*1i + a^4*c*x*1i + 2*a^3*c^(1/2)*d^(1/2)*(q + p*x^3)^(1/2) - a^2*d*p*x^2*1i + a*b*d
*p*x*1i)*(a^3*b^3*c^2*1i + a^6*c^2*x^3*1i + a^4*b^2*c^2*x*3i + a^5*b*c^2*x^2*3i - b^4*c*d*p*1i + b^2*d^2*p*(q
+ p*x^3)*1i + a^2*d^2*p*x^2*(q + p*x^3)*1i + a^3*b*c*d*q*1i + a^3*b*c*d*(q + p*x^3)*2i + a^4*c*d*q*x*1i + a^4*
c*d*x*(q + p*x^3)*2i + 2*a^3*c^(1/2)*d^(3/2)*q*(q + p*x^3)^(1/2) - 2*b^3*c^(1/2)*d^(3/2)*p*(q + p*x^3)^(1/2) -
a*b^3*c*d*p*x*1i - a*b*d^2*p*x*(q + p*x^3)*1i))/((d*q + b^2*c + d*p*x^3 + a^2*c*x^2 + 2*a*b*c*x)*(a^6*b^2*c^2
+ b^4*d^2*p^2 + a^8*c^2*x^2 + 4*a^6*c*d*q + a^4*d^2*p^2*x^4 + 2*a^7*b*c^2*x - 2*a^3*b^3*c*d*p + 2*a^6*c*d*p*x
^3 + 3*a^2*b^2*d^2*p^2*x^2 - 2*a*b^3*d^2*p^2*x - 2*a^3*b*d^2*p^2*x^3)))*1i)/(c^(1/2)*d^(1/2))

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

[Out]

Timed out

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