3.6.72 \(\int \frac {4 a q x-3 b p x^2+a p x^4}{\sqrt {q+p x^3} (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4)} \, dx\)
Optimal. Leaf size=44 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {p x^3+q}}{\sqrt {c} \left (a x^2+b\right )}\right )}{\sqrt {c} \sqrt {d}} \]
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Rubi [F] time = 4.08, 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 {4 a q x-3 b p x^2+a p x^4}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
Int[(4*a*q*x - 3*b*p*x^2 + a*p*x^4)/(Sqrt[q + p*x^3]*(b^2*c + d*q + 2*a*b*c*x^2 + d*p*x^3 + a^2*c*x^4)),x]
[Out]
(2*Sqrt[2 + Sqrt[3]]*p^(2/3)*(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]])/(3^(1/4)*a*c*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]) - (p*(b^2*c + d*q)*Defer[Int][1/(Sqrt[q + p*x^3]*(b^2*c + d*q + 2*a*b*c*x^2 + d*p*x^3 +
a^2*c*x^4)), x])/(a*c) + 4*a*q*Defer[Int][x/(Sqrt[q + p*x^3]*(b^2*c + d*q + 2*a*b*c*x^2 + d*p*x^3 + a^2*c*x^4
)), x] - 5*b*p*Defer[Int][x^2/(Sqrt[q + p*x^3]*(b^2*c + d*q + 2*a*b*c*x^2 + d*p*x^3 + a^2*c*x^4)), x] - (d*p^2
*Defer[Int][x^3/(Sqrt[q + p*x^3]*(b^2*c + d*q + 2*a*b*c*x^2 + d*p*x^3 + a^2*c*x^4)), x])/(a*c)
Rubi steps
\begin {align*} \int \frac {4 a q x-3 b p x^2+a p x^4}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )} \, dx &=\int \frac {x \left (4 a q-3 b p x+a p x^3\right )}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )} \, dx\\ &=\int \left (\frac {p}{a c \sqrt {q+p x^3}}-\frac {p \left (b^2 c+d q\right )-4 a^2 c q x+5 a b c p x^2+d p^2 x^3}{a c \sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )}\right ) \, dx\\ &=-\frac {\int \frac {p \left (b^2 c+d q\right )-4 a^2 c q x+5 a b c p x^2+d p^2 x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )} \, dx}{a c}+\frac {p \int \frac {1}{\sqrt {q+p x^3}} \, dx}{a c}\\ &=\frac {2 \sqrt {2+\sqrt {3}} p^{2/3} \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} a c \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 {p \left (b^2 c+d q\right )}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )}-\frac {4 a^2 c q x}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )}+\frac {5 a b c p x^2}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )}+\frac {d p^2 x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )}\right ) \, dx}{a c}\\ &=\frac {2 \sqrt {2+\sqrt {3}} p^{2/3} \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} a c \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}}-(5 b p) \int \frac {x^2}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )} \, dx-\frac {\left (d p^2\right ) \int \frac {x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )} \, dx}{a c}+(4 a q) \int \frac {x}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )} \, dx-\frac {\left (p \left (b^2 c+d q\right )\right ) \int \frac {1}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x^2+d p x^3+a^2 c x^4\right )} \, dx}{a c}\\ \end {align*}
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Mathematica [C] time = 7.09, size = 8005, normalized size = 181.93 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(4*a*q*x - 3*b*p*x^2 + a*p*x^4)/(Sqrt[q + p*x^3]*(b^2*c + d*q + 2*a*b*c*x^2 + d*p*x^3 + a^2*c*x^4)),
x]
[Out]
Result too large to show
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IntegrateAlgebraic [A] time = 18.39, size = 44, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {q+p x^3}}{\sqrt {c} \left (b+a x^2\right )}\right )}{\sqrt {c} \sqrt {d}} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(4*a*q*x - 3*b*p*x^2 + a*p*x^4)/(Sqrt[q + p*x^3]*(b^2*c + d*q + 2*a*b*c*x^2 + d*p*x^3 + a^2
*c*x^4)),x]
[Out]
(-2*ArcTan[(Sqrt[d]*Sqrt[q + p*x^3])/(Sqrt[c]*(b + a*x^2))])/(Sqrt[c]*Sqrt[d])
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fricas [B] time = 3.91, size = 471, normalized size = 10.70 \begin {gather*} \left [-\frac {\sqrt {-c d} \log \left (\frac {a^{4} c^{2} x^{8} - 6 \, a^{2} c d p x^{7} - 12 \, a b c d p x^{5} + {\left (4 \, a^{3} b c^{2} + d^{2} p^{2}\right )} x^{6} + b^{4} c^{2} - 6 \, b^{2} c d q + 6 \, {\left (a^{2} b^{2} c^{2} - a^{2} c d q\right )} x^{4} + d^{2} q^{2} - 2 \, {\left (3 \, b^{2} c d p - d^{2} p q\right )} x^{3} + 4 \, {\left (a b^{3} c^{2} - 3 \, a b c d q\right )} x^{2} - 4 \, {\left (a^{3} c x^{6} + 3 \, a^{2} b c x^{4} - a d p x^{5} - b d p x^{3} + b^{3} c - b d q + {\left (3 \, a b^{2} c - a d q\right )} x^{2}\right )} \sqrt {p x^{3} + q} \sqrt {-c d}}{a^{4} c^{2} x^{8} + 2 \, a^{2} c d p x^{7} + 4 \, a b c d p x^{5} + {\left (4 \, a^{3} b c^{2} + d^{2} p^{2}\right )} x^{6} + b^{4} c^{2} + 2 \, b^{2} c d q + 2 \, {\left (3 \, a^{2} b^{2} c^{2} + a^{2} c d q\right )} x^{4} + d^{2} q^{2} + 2 \, {\left (b^{2} c d p + d^{2} p q\right )} x^{3} + 4 \, {\left (a b^{3} c^{2} + a b c d q\right )} x^{2}}\right )}{2 \, c d}, \frac {\sqrt {c d} \arctan \left (\frac {{\left (a^{2} c x^{4} + 2 \, a b c x^{2} - d p x^{3} + b^{2} c - d q\right )} \sqrt {p x^{3} + q} \sqrt {c d}}{2 \, {\left (a c d p x^{5} + b c d p x^{3} + a c d q x^{2} + 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^4-3*b*p*x^2+4*a*q*x)/(p*x^3+q)^(1/2)/(a^2*c*x^4+2*a*b*c*x^2+d*p*x^3+b^2*c+d*q),x, algorithm="
fricas")
[Out]
[-1/2*sqrt(-c*d)*log((a^4*c^2*x^8 - 6*a^2*c*d*p*x^7 - 12*a*b*c*d*p*x^5 + (4*a^3*b*c^2 + d^2*p^2)*x^6 + b^4*c^2
- 6*b^2*c*d*q + 6*(a^2*b^2*c^2 - a^2*c*d*q)*x^4 + d^2*q^2 - 2*(3*b^2*c*d*p - d^2*p*q)*x^3 + 4*(a*b^3*c^2 - 3*
a*b*c*d*q)*x^2 - 4*(a^3*c*x^6 + 3*a^2*b*c*x^4 - a*d*p*x^5 - b*d*p*x^3 + b^3*c - b*d*q + (3*a*b^2*c - a*d*q)*x^
2)*sqrt(p*x^3 + q)*sqrt(-c*d))/(a^4*c^2*x^8 + 2*a^2*c*d*p*x^7 + 4*a*b*c*d*p*x^5 + (4*a^3*b*c^2 + d^2*p^2)*x^6
+ b^4*c^2 + 2*b^2*c*d*q + 2*(3*a^2*b^2*c^2 + a^2*c*d*q)*x^4 + d^2*q^2 + 2*(b^2*c*d*p + d^2*p*q)*x^3 + 4*(a*b^3
*c^2 + a*b*c*d*q)*x^2))/(c*d), sqrt(c*d)*arctan(1/2*(a^2*c*x^4 + 2*a*b*c*x^2 - d*p*x^3 + b^2*c - d*q)*sqrt(p*x
^3 + q)*sqrt(c*d)/(a*c*d*p*x^5 + b*c*d*p*x^3 + a*c*d*q*x^2 + 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^4-3*b*p*x^2+4*a*q*x)/(p*x^3+q)^(1/2)/(a^2*c*x^4+2*a*b*c*x^2+d*p*x^3+b^2*c+d*q),x, algorithm="
giac")
[Out]
Timed out
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maple [C] time = 0.80, size = 3170, normalized size = 72.05
| | |
| method |
result |
size |
| | |
| default |
\(\text {Expression too large to display}\) |
\(3170\) |
| elliptic |
\(\text {Expression too large to display}\) |
\(3170\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a*p*x^4-3*b*p*x^2+4*a*q*x)/(p*x^3+q)^(1/2)/(a^2*c*x^4+2*a*b*c*x^2+d*p*x^3+b^2*c+d*q),x,method=_RETURNVERB
OSE)
[Out]
-2/3*I/a/c*3^(1/2)*(-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/a/c^2/p^2*2^(1/2)*sum((_a
lpha^3*d*p^2+5*_alpha^2*a*b*c*p-4*_alpha*a^2*c*q+b^2*c*p+d*p*q)/_alpha/(4*_alpha^2*a^2*c+3*_alpha*d*p+4*a*b*c)
/(a^6*q^4+2*a^3*b^3*p^2*q^2+b^6*p^4)*(-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^3+(-q*p^2)^(2/3)*b^4*d*p^4+2*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha^2*a^5*b*q^2*p^2*c+3*I*(-q*p^2)^(1/3)*p^4
*3^(1/2)*_alpha^2*a^2*b^2*d*q+4*I*(-q*p^2)^(1/3)*p^3*3^(1/2)*_alpha*a^3*b^3*q*c+2*p^2*(2*_alpha^3*a^5*b*c*p*q^
2-_alpha^3*a^2*b^4*c*p^3+_alpha^2*a^6*c*q^3-2*_alpha^2*a^3*b^3*c*p^2*q+2*_alpha^2*a^3*b*d*p^2*q^2-_alpha^2*b^4
*d*p^4+_alpha*a^4*b^2*c*p*q^2-2*_alpha*a*b^5*c*p^3+_alpha*a^4*d*p*q^3-2*_alpha*a*b^3*d*p^3*q-3*a^2*b^4*c*p^2*q
-3*a^2*b^2*d*p^2*q^2)+I*(-q*p^2)^(1/3)*3^(1/2)*a^4*b^2*q^2*p^2*c-3*(-q*p^2)^(1/3)*_alpha^3*a^4*b^2*c*p^3*q-3*(
-q*p^2)^(2/3)*_alpha^2*a^4*b^2*c*p^2*q-2*(-q*p^2)^(1/3)*_alpha^2*a^5*b*c*p^2*q^2-3*(-q*p^2)^(1/3)*_alpha^2*a^2
*b^2*d*p^4*q-4*(-q*p^2)^(1/3)*_alpha*a^3*b^3*c*p^3*q-3*(-q*p^2)^(2/3)*_alpha*a^2*b^2*d*p^3*q-2*(-q*p^2)^(1/3)*
_alpha*a^3*b*d*p^3*q^2-I*(-q*p^2)^(1/3)*p^5*3^(1/2)*_alpha*b^4*d+(-q*p^2)^(2/3)*_alpha^3*a^6*c*p*q^2+I*(-q*p^2
)^(1/3)*3^(1/2)*_alpha*a^6*q^3*p*c+I*(-q*p^2)^(2/3)*3^(1/2)*_alpha^3*a^6*q^2*p*c+I*(-q*p^2)^(2/3)*3^(1/2)*_alp
ha^2*a^4*d*q^2*p^2-2*I*(-q*p^2)^(2/3)*p^4*3^(1/2)*_alpha^2*a*b^3*d-3*I*(-q*p^2)^(2/3)*p^3*3^(1/2)*_alpha*a^2*b
^4*c-2*I*(-q*p^2)^(2/3)*p^3*3^(1/2)*_alpha^3*a^3*b^3*c-I*(-q*p^2)^(1/3)*p^4*3^(1/2)*_alpha^2*a^2*b^4*c+(-q*p^2
)^(1/3)*_alpha*b^4*d*p^5+2*(-q*p^2)^(1/3)*a*b^5*c*p^4-(-q*p^2)^(1/3)*a^4*d*p^2*q^3+I*(-q*p^2)^(1/3)*3^(1/2)*a^
4*d*q^3*p^2-2*I*(-q*p^2)^(1/3)*p^4*3^(1/2)*a*b^5*c-4*(-q*p^2)^(2/3)*a^3*b^3*c*p^2*q-(-q*p^2)^(1/3)*a^4*b^2*c*p
^2*q^2-2*(-q*p^2)^(2/3)*a^3*b*d*p^2*q^2+2*(-q*p^2)^(1/3)*a*b^3*d*p^4*q+I*(-q*p^2)^(2/3)*p^4*3^(1/2)*b^4*d-I*(-
q*p^2)^(2/3)*3^(1/2)*a^6*q^3*c-4*I*(-q*p^2)^(2/3)*3^(1/2)*a^3*b^3*q*p^2*c-2*I*(-q*p^2)^(2/3)*3^(1/2)*a^3*b*d*q
^2*p^2-2*I*(-q*p^2)^(1/3)*p^4*3^(1/2)*a*b^3*d*q-3*I*(-q*p^2)^(2/3)*p^3*3^(1/2)*_alpha*a^2*b^2*d*q+2*I*(-q*p^2)
^(1/3)*p^3*3^(1/2)*_alpha*a^3*b*d*q^2+3*I*(-q*p^2)^(1/3)*p^3*3^(1/2)*_alpha^3*a^4*b^2*q*c-3*I*(-q*p^2)^(2/3)*3
^(1/2)*_alpha^2*a^4*b^2*q*p^2*c-2*(-q*p^2)^(2/3)*_alpha^3*a^3*b^3*c*p^3+(-q*p^2)^(1/3)*_alpha^2*a^2*b^4*c*p^4+
(-q*p^2)^(2/3)*_alpha^2*a^4*d*p^2*q^2-2*(-q*p^2)^(2/3)*_alpha^2*a*b^3*d*p^4-3*(-q*p^2)^(2/3)*_alpha*a^2*b^4*c*
p^3-(-q*p^2)^(1/3)*_alpha*a^6*c*p*q^3)*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*(12*p^3*a^3*b^3*q^2*c-3*p^5*b^4*d*q+3*a^6*q^4*p*c+I*(-q*p^2)^
(2/3)*3^(1/2)*a^4*d*p*q^3-2*I*(-q*p^2)^(2/3)*3^(1/2)*a*b^5*c*p^3-4*I*3^(1/2)*a^3*b^3*c*p^3*q^2-2*I*3^(1/2)*a^3
*b*d*p^3*q^3+I*(-q*p^2)^(2/3)*3^(1/2)*a^4*b^2*c*p*q^2-6*p^3*(-q*p^2)^(2/3)*a*b^5*c+3*q^3*(-q*p^2)^(2/3)*a^4*d*
p+3*q^2*(-q*p^2)^(2/3)*a^4*b^2*p*c-6*p^3*(-q*p^2)^(2/3)*a*b^3*d*q+I*3^(1/2)*b^4*d*p^5*q-I*3^(1/2)*a^6*c*p*q^4+
4*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha*a*b^5*c*p^4-2*I*3^(1/2)*_alpha^2*a*b^3*d*p^5*q-3*I*3^(1/2)*_alpha*a^2*b^4*c*
p^4*q-2*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha*a^4*d*p^2*q^3-3*I*3^(1/2)*_alpha*a^2*b^2*d*p^4*q^2+6*p^3*a^3*b*d*q^3-3
*_alpha^3*a^6*q^3*p^2*c-3*p^3*_alpha^2*a^4*d*q^3+3*q^3*(-q*p^2)^(2/3)*_alpha*a^6*c-3*p^4*(-q*p^2)^(2/3)*_alpha
*b^4*d+2*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha^3*a^2*b^4*c*p^4-2*I*3^(1/2)*_alpha^3*a^3*b^3*c*p^4*q-I*(-q*p^2)^(2/3)
*3^(1/2)*_alpha^2*a^2*b^4*c*p^3-2*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha^2*a^6*c*p*q^3-3*I*3^(1/2)*_alpha^2*a^4*b^2*c
*p^3*q^2+9*(-q*p^2)^(2/3)*_alpha^3*a^4*b^2*p^2*q*c+6*q^2*(-q*p^2)^(2/3)*_alpha^2*a^5*b*p*c+9*p^3*(-q*p^2)^(2/3
)*_alpha^2*a^2*b^2*d*q+12*(-q*p^2)^(2/3)*_alpha*a^3*b^3*p^2*q*c+6*q^2*(-q*p^2)^(2/3)*_alpha*a^3*b*d*p^2+I*3^(1
/2)*_alpha^3*a^6*c*p^2*q^3+I*(-q*p^2)^(2/3)*3^(1/2)*_alpha*a^6*c*q^3+I*3^(1/2)*_alpha^2*a^4*d*p^3*q^3+2*I*(-q*
p^2)^(1/3)*3^(1/2)*_alpha^2*b^4*d*p^5-I*(-q*p^2)^(2/3)*3^(1/2)*_alpha*b^4*d*p^4+3*I*(-q*p^2)^(2/3)*3^(1/2)*_al
pha^3*a^4*b^2*c*p^2*q-4*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha^3*a^5*b*c*p^2*q^2+2*I*(-q*p^2)^(2/3)*3^(1/2)*_alpha^2*
a^5*b*c*p*q^2+4*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha^2*a^3*b^3*c*p^3*q+3*I*(-q*p^2)^(2/3)*3^(1/2)*_alpha^2*a^2*b^2*
d*p^3*q+4*I*(-q*p^2)^(2/3)*3^(1/2)*_alpha*a^3*b^3*c*p^2*q-4*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha^2*a^3*b*d*p^3*q^2-
2*I*(-q*p^2)^(1/3)*3^(1/2)*_alpha*a^4*b^2*c*p^2*q^2+2*I*(-q*p^2)^(2/3)*3^(1/2)*_alpha*a^3*b*d*p^2*q^2+4*I*(-q*
p^2)^(1/3)*3^(1/2)*_alpha*a*b^3*d*p^4*q+6*I*(-q*p^2)^(1/3)*3^(1/2)*a^2*b^4*c*p^3*q-2*I*(-q*p^2)^(2/3)*3^(1/2)*
a*b^3*d*p^3*q+6*I*(-q*p^2)^(1/3)*3^(1/2)*a^2*b^2*d*p^3*q^2+6*p^5*_alpha^2*a*b^3*d*q+9*p^4*_alpha*a^2*b^4*q*c+9
*p^4*_alpha*a^2*b^2*d*q^2-3*p^3*(-q*p^2)^(2/3)*_alpha^2*a^2*b^4*c+6*p^4*_alpha^3*a^3*b^3*q*c+9*p^3*_alpha^2*a^
4*b^2*q^2*c)/c/(a^6*q^4+2*a^3*b^3*p^2*q^2+b^6*p^4),(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^4*a^2*c+_Z^3*d*p+2*_Z^2*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^{4} - 3 \, b p x^{2} + 4 \, a q x}{{\left (a^{2} c x^{4} + 2 \, a b c x^{2} + d p x^{3} + 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^4-3*b*p*x^2+4*a*q*x)/(p*x^3+q)^(1/2)/(a^2*c*x^4+2*a*b*c*x^2+d*p*x^3+b^2*c+d*q),x, algorithm="
maxima")
[Out]
integrate((a*p*x^4 - 3*b*p*x^2 + 4*a*q*x)/((a^2*c*x^4 + 2*a*b*c*x^2 + d*p*x^3 + b^2*c + d*q)*sqrt(p*x^3 + q)),
x)
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mupad [B] time = 80.21, size = 1058, normalized size = 24.05 \begin {gather*} \frac {\ln \left (\frac {\left (-2\,a\,\sqrt {c}\,\sqrt {d}\,\sqrt {p\,x^3+q}+a^2\,c\,x^2\,1{}\mathrm {i}+a\,b\,c\,1{}\mathrm {i}-d\,p\,x\,1{}\mathrm {i}\right )\,\left (a^4\,b^2\,c^2\,q\,1{}\mathrm {i}+a^6\,c^2\,q\,x^4\,1{}\mathrm {i}-b^2\,d^2\,p^3\,x\,1{}\mathrm {i}+a^4\,c\,d\,q^2\,4{}\mathrm {i}+a\,b^3\,c\,d\,p^2\,1{}\mathrm {i}+a^3\,b^3\,c^2\,p\,x\,1{}\mathrm {i}+a^5\,b\,c^2\,p\,x^5\,1{}\mathrm {i}+a\,b\,d^2\,p^3\,x^3\,1{}\mathrm {i}+a^5\,b\,c^2\,q\,x^2\,2{}\mathrm {i}+a^4\,b^2\,c^2\,p\,x^3\,2{}\mathrm {i}+a^2\,d^2\,p^2\,q\,x^2\,1{}\mathrm {i}-a^2\,b^2\,c\,d\,p^2\,x^2\,1{}\mathrm {i}+a^4\,c\,d\,p\,q\,x^3\,2{}\mathrm {i}+2\,a\,b^2\,\sqrt {c}\,d^{3/2}\,p^2\,\sqrt {p\,x^3+q}+a^3\,b\,c\,d\,p^2\,x^4\,2{}\mathrm {i}+a^3\,b\,c\,d\,p\,q\,x\,2{}\mathrm {i}\right )\,\left (-a^3\,b^3\,c^2\,q\,2{}\mathrm {i}+a^3\,b^3\,c^2\,\left (p\,x^3+q\right )\,3{}\mathrm {i}+a^6\,c^2\,q\,x^6\,1{}\mathrm {i}-b^2\,d^2\,p^2\,\left (p\,x^3+q\right )\,1{}\mathrm {i}+b^4\,c\,d\,p^2\,1{}\mathrm {i}+a^3\,b\,c\,d\,q^2\,1{}\mathrm {i}+a^3\,b\,c\,d\,{\left (p\,x^3+q\right )}^2\,2{}\mathrm {i}-2\,a^3\,\sqrt {c}\,d^{3/2}\,q^2\,\sqrt {p\,x^3+q}-2\,b^3\,\sqrt {c}\,d^{3/2}\,p^2\,\sqrt {p\,x^3+q}+a^2\,b^4\,c^2\,p\,x\,1{}\mathrm {i}+a^5\,b\,c^2\,q\,x^4\,2{}\mathrm {i}+a^4\,c\,d\,q^2\,x^2\,1{}\mathrm {i}+a^5\,b\,c^2\,x^4\,\left (p\,x^3+q\right )\,1{}\mathrm {i}+a^4\,b^2\,c^2\,x^2\,\left (p\,x^3+q\right )\,3{}\mathrm {i}+a\,b\,d^2\,p^2\,x^2\,\left (p\,x^3+q\right )\,1{}\mathrm {i}+a^4\,c\,d\,q\,x^2\,\left (p\,x^3+q\right )\,2{}\mathrm {i}+a^2\,d^2\,p\,q\,x\,\left (p\,x^3+q\right )\,1{}\mathrm {i}+a\,b^3\,c\,d\,p^2\,x^2\,1{}\mathrm {i}+a^2\,b^2\,c\,d\,p\,x\,\left (p\,x^3+q\right )\,2{}\mathrm {i}\right )}{\left (c\,a^2\,x^4+2\,c\,a\,b\,x^2+c\,b^2+d\,p\,x^3+d\,q\right )\,\left (a^4\,c^2\,x^4+2\,a^3\,b\,c^2\,x^2+a^2\,b^2\,c^2+2\,a^2\,c\,d\,p\,x^3+4\,q\,a^2\,c\,d-2\,a\,b\,c\,d\,p\,x+d^2\,p^2\,x^2\right )\,\left (a^8\,c^2\,q^2\,x^4+2\,a^7\,b\,c^2\,p\,q\,x^5+2\,a^7\,b\,c^2\,q^2\,x^2+a^6\,b^2\,c^2\,p^2\,x^6+4\,a^6\,b^2\,c^2\,p\,q\,x^3+a^6\,b^2\,c^2\,q^2+2\,a^6\,c\,d\,p\,q^2\,x^3+4\,a^6\,c\,d\,q^3+2\,a^5\,b^3\,c^2\,p^2\,x^4+2\,a^5\,b^3\,c^2\,p\,q\,x+4\,a^5\,b\,c\,d\,p^2\,q\,x^4+6\,a^5\,b\,c\,d\,p\,q^2\,x+a^4\,b^4\,c^2\,p^2\,x^2+2\,a^4\,b^2\,c\,d\,p^3\,x^5+2\,a^4\,b^2\,c\,d\,p^2\,q\,x^2+a^4\,d^2\,p^2\,q^2\,x^2+2\,a^3\,b^3\,c\,d\,p^2\,q+2\,a^3\,b\,d^2\,p^3\,q\,x^3+2\,a^2\,b^4\,c\,d\,p^3\,x+a^2\,b^2\,d^2\,p^4\,x^4-2\,a^2\,b^2\,d^2\,p^3\,q\,x-2\,a\,b^3\,d^2\,p^4\,x^2+b^4\,d^2\,p^4\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^4 - 3*b*p*x^2 + 4*a*q*x)/((q + p*x^3)^(1/2)*(d*q + b^2*c + d*p*x^3 + a^2*c*x^4 + 2*a*b*c*x^2)),x)
[Out]
(log(((a^2*c*x^2*1i + a*b*c*1i - d*p*x*1i - 2*a*c^(1/2)*d^(1/2)*(q + p*x^3)^(1/2))*(a^4*b^2*c^2*q*1i + a^6*c^2
*q*x^4*1i - b^2*d^2*p^3*x*1i + a^4*c*d*q^2*4i + a*b^3*c*d*p^2*1i + a^3*b^3*c^2*p*x*1i + a^5*b*c^2*p*x^5*1i + a
*b*d^2*p^3*x^3*1i + a^5*b*c^2*q*x^2*2i + a^4*b^2*c^2*p*x^3*2i + a^2*d^2*p^2*q*x^2*1i - a^2*b^2*c*d*p^2*x^2*1i
+ a^4*c*d*p*q*x^3*2i + 2*a*b^2*c^(1/2)*d^(3/2)*p^2*(q + p*x^3)^(1/2) + a^3*b*c*d*p^2*x^4*2i + a^3*b*c*d*p*q*x*
2i)*(a^3*b^3*c^2*(q + p*x^3)*3i - a^3*b^3*c^2*q*2i + a^6*c^2*q*x^6*1i - b^2*d^2*p^2*(q + p*x^3)*1i + b^4*c*d*p
^2*1i + a^3*b*c*d*q^2*1i + a^3*b*c*d*(q + p*x^3)^2*2i - 2*a^3*c^(1/2)*d^(3/2)*q^2*(q + p*x^3)^(1/2) - 2*b^3*c^
(1/2)*d^(3/2)*p^2*(q + p*x^3)^(1/2) + a^2*b^4*c^2*p*x*1i + a^5*b*c^2*q*x^4*2i + a^4*c*d*q^2*x^2*1i + a^5*b*c^2
*x^4*(q + p*x^3)*1i + a^4*b^2*c^2*x^2*(q + p*x^3)*3i + a*b*d^2*p^2*x^2*(q + p*x^3)*1i + a^4*c*d*q*x^2*(q + p*x
^3)*2i + a^2*d^2*p*q*x*(q + p*x^3)*1i + a*b^3*c*d*p^2*x^2*1i + a^2*b^2*c*d*p*x*(q + p*x^3)*2i))/((d*q + b^2*c
+ d*p*x^3 + a^2*c*x^4 + 2*a*b*c*x^2)*(a^2*b^2*c^2 + a^4*c^2*x^4 + d^2*p^2*x^2 + 2*a^3*b*c^2*x^2 + 4*a^2*c*d*q
+ 2*a^2*c*d*p*x^3 - 2*a*b*c*d*p*x)*(b^4*d^2*p^4 + a^6*b^2*c^2*q^2 + a^8*c^2*q^2*x^4 + 4*a^6*c*d*q^3 + a^4*b^4*
c^2*p^2*x^2 + 2*a^5*b^3*c^2*p^2*x^4 + a^6*b^2*c^2*p^2*x^6 + a^2*b^2*d^2*p^4*x^4 + a^4*d^2*p^2*q^2*x^2 - 2*a*b^
3*d^2*p^4*x^2 + 2*a^7*b*c^2*q^2*x^2 + 2*a^4*b^2*c*d*p^3*x^5 + 4*a^6*b^2*c^2*p*q*x^3 - 2*a^2*b^2*d^2*p^3*q*x +
2*a^3*b*d^2*p^3*q*x^3 + 2*a^3*b^3*c*d*p^2*q + 2*a^2*b^4*c*d*p^3*x + 2*a^5*b^3*c^2*p*q*x + 2*a^7*b*c^2*p*q*x^5
+ 2*a^6*c*d*p*q^2*x^3 + 6*a^5*b*c*d*p*q^2*x + 4*a^5*b*c*d*p^2*q*x^4 + 2*a^4*b^2*c*d*p^2*q*x^2)))*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**4-3*b*p*x**2+4*a*q*x)/(p*x**3+q)**(1/2)/(a**2*c*x**4+2*a*b*c*x**2+d*p*x**3+b**2*c+d*q),x)
[Out]
Timed out
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