3.28.81 \(\int \frac {-a q x+b p x^3}{\sqrt {q+p x^4} (b^2 c+d q+2 a b c x^2+(a^2 c+d p) x^4)} \, dx\)
Optimal. Leaf size=267 \[ -\frac {1}{4} \text {RootSum}\left [\text {$\#$1}^4 a^2 c+\text {$\#$1}^4 d p+4 \text {$\#$1}^3 a b c \sqrt {p}-2 \text {$\#$1}^2 a^2 c q+4 \text {$\#$1}^2 b^2 c p+2 \text {$\#$1}^2 d p q-4 \text {$\#$1} a b c \sqrt {p} q+a^2 c q^2+d p q^2\& ,\frac {\text {$\#$1}^2 (-b) p \log \left (-\text {$\#$1}+\sqrt {p x^4+q}+\sqrt {p} x^2\right )+2 \text {$\#$1} a \sqrt {p} q \log \left (-\text {$\#$1}+\sqrt {p x^4+q}+\sqrt {p} x^2\right )+b p q \log \left (-\text {$\#$1}+\sqrt {p x^4+q}+\sqrt {p} x^2\right )}{\text {$\#$1}^3 a^2 c+\text {$\#$1}^3 d p+3 \text {$\#$1}^2 a b c \sqrt {p}-\text {$\#$1} a^2 c q+2 \text {$\#$1} b^2 c p+\text {$\#$1} d p q-a b c \sqrt {p} q}\& \right ] \]
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Rubi [A] time = 1.15, antiderivative size = 44, normalized size of antiderivative = 0.16,
number of steps used = 4, number of rules used = 4, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used =
{1593, 6715, 1031, 205} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {c} \left (a x^2+b\right )}{\sqrt {d} \sqrt {p x^4+q}}\right )}{2 \sqrt {c} \sqrt {d}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(-(a*q*x) + b*p*x^3)/(Sqrt[q + p*x^4]*(b^2*c + d*q + 2*a*b*c*x^2 + (a^2*c + d*p)*x^4)),x]
[Out]
-1/2*ArcTan[(Sqrt[c]*(b + a*x^2))/(Sqrt[d]*Sqrt[q + p*x^4])]/(Sqrt[c]*Sqrt[d])
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 1031
Int[((g_) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*
g*(g*b - 2*a*h), Subst[Int[1/Simp[g*(g*b - 2*a*h)*(b^2 - 4*a*c) - b*d*x^2, x], x], x, Simp[g*b - 2*a*h - (b*h
- 2*g*c)*x, x]/Sqrt[d + f*x^2]], x] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[b*h^2*d -
2*g*h*(c*d - a*f) - g^2*b*f, 0]
Rule 1593
Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]
Rule 6715
Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]
Rubi steps
\begin {align*} \int \frac {-a q x+b p x^3}{\sqrt {q+p x^4} \left (b^2 c+d q+2 a b c x^2+\left (a^2 c+d p\right ) x^4\right )} \, dx &=\int \frac {x \left (-a q+b p x^2\right )}{\sqrt {q+p x^4} \left (b^2 c+d q+2 a b c x^2+\left (a^2 c+d p\right ) x^4\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {-a q+b p x}{\sqrt {q+p x^2} \left (b^2 c+d q+2 a b c x+\left (a^2 c+d p\right ) x^2\right )} \, dx,x,x^2\right )\\ &=-\left (\left (2 a b q \left (b^2 c p+a^2 c q+d p q\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-8 a b d q \left (b^2 c p+a^2 c q+d p q\right )^2-2 a b c q x^2} \, dx,x,-\frac {2 \left (b^2 c p+a^2 c q+d p q\right ) \left (b+a x^2\right )}{\sqrt {q+p x^4}}\right )\right )\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {c} \left (b+a x^2\right )}{\sqrt {d} \sqrt {q+p x^4}}\right )}{2 \sqrt {c} \sqrt {d}}\\ \end {align*}
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Mathematica [B] time = 1.46, size = 573, normalized size = 2.15 \begin {gather*} \frac {\sqrt {c} \left (\frac {\left (a^3 c q-b \sqrt {d} p \sqrt {b^2 (-c) p-q \left (a^2 c+d p\right )}+a p \left (b^2 c+d q\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {d} p x^2 \sqrt {b^2 (-c) p-q \left (a^2 c+d p\right )}+a^2 c q-a b c p x^2+d p q}{\sqrt {c} \sqrt {p x^4+q} \sqrt {a^4 c q+a^2 p \left (b^2 c+d q\right )-2 a b \sqrt {d} p \sqrt {b^2 (-c) p-q \left (a^2 c+d p\right )}-b^2 d p^2}}\right )}{2 c \sqrt {a^4 c q+a^2 p \left (b^2 c+d q\right )-2 a b \sqrt {d} p \sqrt {b^2 (-c) p-q \left (a^2 c+d p\right )}-b^2 d p^2}}-\frac {\left (a^3 c q+b \sqrt {d} p \sqrt {b^2 (-c) p-q \left (a^2 c+d p\right )}+a p \left (b^2 c+d q\right )\right ) \tanh ^{-1}\left (\frac {-\sqrt {d} p x^2 \sqrt {b^2 (-c) p-q \left (a^2 c+d p\right )}+a^2 c q-a b c p x^2+d p q}{\sqrt {c} \sqrt {p x^4+q} \sqrt {a^4 c q+a^2 p \left (b^2 c+d q\right )+2 a b \sqrt {d} p \sqrt {b^2 (-c) p-q \left (a^2 c+d p\right )}-b^2 d p^2}}\right )}{2 c \sqrt {a^4 c q+a^2 p \left (b^2 c+d q\right )+2 a b \sqrt {d} p \sqrt {b^2 (-c) p-q \left (a^2 c+d p\right )}-b^2 d p^2}}\right )}{2 \sqrt {d} \sqrt {b^2 (-c) p-q \left (a^2 c+d p\right )}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(-(a*q*x) + b*p*x^3)/(Sqrt[q + p*x^4]*(b^2*c + d*q + 2*a*b*c*x^2 + (a^2*c + d*p)*x^4)),x]
[Out]
(Sqrt[c]*(-1/2*((a^3*c*q + a*p*(b^2*c + d*q) + b*Sqrt[d]*p*Sqrt[-(b^2*c*p) - (a^2*c + d*p)*q])*ArcTanh[(a^2*c*
q + d*p*q - a*b*c*p*x^2 - Sqrt[d]*p*Sqrt[-(b^2*c*p) - (a^2*c + d*p)*q]*x^2)/(Sqrt[c]*Sqrt[-(b^2*d*p^2) + a^4*c
*q + a^2*p*(b^2*c + d*q) + 2*a*b*Sqrt[d]*p*Sqrt[-(b^2*c*p) - (a^2*c + d*p)*q]]*Sqrt[q + p*x^4])])/(c*Sqrt[-(b^
2*d*p^2) + a^4*c*q + a^2*p*(b^2*c + d*q) + 2*a*b*Sqrt[d]*p*Sqrt[-(b^2*c*p) - (a^2*c + d*p)*q]]) + ((a^3*c*q +
a*p*(b^2*c + d*q) - b*Sqrt[d]*p*Sqrt[-(b^2*c*p) - (a^2*c + d*p)*q])*ArcTanh[(a^2*c*q + d*p*q - a*b*c*p*x^2 + S
qrt[d]*p*Sqrt[-(b^2*c*p) - (a^2*c + d*p)*q]*x^2)/(Sqrt[c]*Sqrt[-(b^2*d*p^2) + a^4*c*q + a^2*p*(b^2*c + d*q) -
2*a*b*Sqrt[d]*p*Sqrt[-(b^2*c*p) - (a^2*c + d*p)*q]]*Sqrt[q + p*x^4])])/(2*c*Sqrt[-(b^2*d*p^2) + a^4*c*q + a^2*
p*(b^2*c + d*q) - 2*a*b*Sqrt[d]*p*Sqrt[-(b^2*c*p) - (a^2*c + d*p)*q]])))/(2*Sqrt[d]*Sqrt[-(b^2*c*p) - (a^2*c +
d*p)*q])
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IntegrateAlgebraic [A] time = 0.57, size = 269, normalized size = 1.01 \begin {gather*} -\frac {1}{4} \text {RootSum}\left [a^2 c q^2+d p q^2+4 a b c \sqrt {p} q \text {$\#$1}+4 b^2 c p \text {$\#$1}^2-2 a^2 c q \text {$\#$1}^2+2 d p q \text {$\#$1}^2-4 a b c \sqrt {p} \text {$\#$1}^3+a^2 c \text {$\#$1}^4+d p \text {$\#$1}^4\&,\frac {b p q \log \left (-\sqrt {p} x^2+\sqrt {q+p x^4}-\text {$\#$1}\right )-2 a \sqrt {p} q \log \left (-\sqrt {p} x^2+\sqrt {q+p x^4}-\text {$\#$1}\right ) \text {$\#$1}-b p \log \left (-\sqrt {p} x^2+\sqrt {q+p x^4}-\text {$\#$1}\right ) \text {$\#$1}^2}{a b c \sqrt {p} q+2 b^2 c p \text {$\#$1}-a^2 c q \text {$\#$1}+d p q \text {$\#$1}-3 a b c \sqrt {p} \text {$\#$1}^2+a^2 c \text {$\#$1}^3+d p \text {$\#$1}^3}\&\right ] \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(-(a*q*x) + b*p*x^3)/(Sqrt[q + p*x^4]*(b^2*c + d*q + 2*a*b*c*x^2 + (a^2*c + d*p)*x^4)),x]
[Out]
-1/4*RootSum[a^2*c*q^2 + d*p*q^2 + 4*a*b*c*Sqrt[p]*q*#1 + 4*b^2*c*p*#1^2 - 2*a^2*c*q*#1^2 + 2*d*p*q*#1^2 - 4*a
*b*c*Sqrt[p]*#1^3 + a^2*c*#1^4 + d*p*#1^4 & , (b*p*q*Log[-(Sqrt[p]*x^2) + Sqrt[q + p*x^4] - #1] - 2*a*Sqrt[p]*
q*Log[-(Sqrt[p]*x^2) + Sqrt[q + p*x^4] - #1]*#1 - b*p*Log[-(Sqrt[p]*x^2) + Sqrt[q + p*x^4] - #1]*#1^2)/(a*b*c*
Sqrt[p]*q + 2*b^2*c*p*#1 - a^2*c*q*#1 + d*p*q*#1 - 3*a*b*c*Sqrt[p]*#1^2 + a^2*c*#1^3 + d*p*#1^3) & ]
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fricas [B] time = 1.03, size = 452, normalized size = 1.69 \begin {gather*} \left [-\frac {\sqrt {-c d} \log \left (\frac {{\left (a^{4} c^{2} - 6 \, a^{2} c d p + d^{2} p^{2}\right )} x^{8} + 4 \, {\left (a^{3} b c^{2} - 3 \, a b c d p\right )} x^{6} + b^{4} c^{2} - 6 \, b^{2} c d q + 2 \, {\left (3 \, a^{2} b^{2} c^{2} - 3 \, b^{2} c d p - {\left (3 \, a^{2} c d - d^{2} p\right )} q\right )} x^{4} + d^{2} q^{2} + 4 \, {\left (a b^{3} c^{2} - 3 \, a b c d q\right )} x^{2} + 4 \, {\left ({\left (a^{3} c - a d p\right )} x^{6} + {\left (3 \, a^{2} b c - b d p\right )} x^{4} + b^{3} c - b d q + {\left (3 \, a b^{2} c - a d q\right )} x^{2}\right )} \sqrt {p x^{4} + q} \sqrt {-c d}}{{\left (a^{4} c^{2} + 2 \, a^{2} c d p + d^{2} p^{2}\right )} x^{8} + 4 \, {\left (a^{3} b c^{2} + a b c d p\right )} x^{6} + b^{4} c^{2} + 2 \, b^{2} c d q + 2 \, {\left (3 \, a^{2} b^{2} c^{2} + b^{2} c d p + {\left (a^{2} c d + d^{2} p\right )} q\right )} x^{4} + d^{2} q^{2} + 4 \, {\left (a b^{3} c^{2} + a b c d q\right )} x^{2}}\right )}{8 \, c d}, \frac {\sqrt {c d} \arctan \left (-\frac {{\left (2 \, a b c x^{2} + {\left (a^{2} c - d p\right )} x^{4} + b^{2} c - d q\right )} \sqrt {p x^{4} + q} \sqrt {c d}}{2 \, {\left (a c d p x^{6} + b c d p x^{4} + a c d q x^{2} + b c d q\right )}}\right )}{4 \, c d}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*p*x^3-a*q*x)/(p*x^4+q)^(1/2)/(b^2*c+d*q+2*a*b*c*x^2+(a^2*c+d*p)*x^4),x, algorithm="fricas")
[Out]
[-1/8*sqrt(-c*d)*log(((a^4*c^2 - 6*a^2*c*d*p + d^2*p^2)*x^8 + 4*(a^3*b*c^2 - 3*a*b*c*d*p)*x^6 + b^4*c^2 - 6*b^
2*c*d*q + 2*(3*a^2*b^2*c^2 - 3*b^2*c*d*p - (3*a^2*c*d - d^2*p)*q)*x^4 + d^2*q^2 + 4*(a*b^3*c^2 - 3*a*b*c*d*q)*
x^2 + 4*((a^3*c - a*d*p)*x^6 + (3*a^2*b*c - b*d*p)*x^4 + b^3*c - b*d*q + (3*a*b^2*c - a*d*q)*x^2)*sqrt(p*x^4 +
q)*sqrt(-c*d))/((a^4*c^2 + 2*a^2*c*d*p + d^2*p^2)*x^8 + 4*(a^3*b*c^2 + a*b*c*d*p)*x^6 + b^4*c^2 + 2*b^2*c*d*q
+ 2*(3*a^2*b^2*c^2 + b^2*c*d*p + (a^2*c*d + d^2*p)*q)*x^4 + d^2*q^2 + 4*(a*b^3*c^2 + a*b*c*d*q)*x^2))/(c*d),
1/4*sqrt(c*d)*arctan(-1/2*(2*a*b*c*x^2 + (a^2*c - d*p)*x^4 + b^2*c - d*q)*sqrt(p*x^4 + q)*sqrt(c*d)/(a*c*d*p*x
^6 + b*c*d*p*x^4 + 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((b*p*x^3-a*q*x)/(p*x^4+q)^(1/2)/(b^2*c+d*q+2*a*b*c*x^2+(a^2*c+d*p)*x^4),x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 0.32, size = 4832, normalized size = 18.10
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| method |
result |
size |
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| default |
\(\text {Expression too large to display}\) |
\(4832\) |
| elliptic |
\(\text {Expression too large to display}\) |
\(4832\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*p*x^3-a*q*x)/(p*x^4+q)^(1/2)/(b^2*c+d*q+2*a*b*c*x^2+(a^2*c+d*p)*x^4),x,method=_RETURNVERBOSE)
[Out]
-1/4/(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)/(-a^2*c-d*p)/(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*
q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*ln((2*c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^
2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2-2*p*(a*b*c-(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x
^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*p))+2*(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(
-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*(p*(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/
2))/(-a^2*c-d*p))^2-2*p*(a*b*c-(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c
*p+d*p*q))^(1/2))/(-a^2*c-d*p))+c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2
)*a*b*p)/(a^2*c+d*p)^2)^(1/2))/(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*p)))*a^3*c*q-1/4/(-d
*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)/(-a^2*c-d*p)/(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^2*c*
p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*ln((2*c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^
2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2-2*p*(a*b*c-(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(-a*b
*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*p))+2*(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*
c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*(p*(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^
2*c-d*p))^2-2*p*(a*b*c-(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q
))^(1/2))/(-a^2*c-d*p))+c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)
/(a^2*c+d*p)^2)^(1/2))/(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*p)))*a*b^2*c*p-1/4/(-d*(a^2*
c*q+b^2*c*p+d*p*q))^(1/2)/(-a^2*c-d*p)/(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^2*c*p+d*p*
q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*ln((2*c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^2*c*p+
d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2-2*p*(a*b*c-(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(-a*b*c+(-d
*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*p))+2*(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^
2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*(p*(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*
p))^2-2*p*(a*b*c-(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/
2))/(-a^2*c-d*p))+c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*
c+d*p)^2)^(1/2))/(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*p)))*a*d*p*q+1/4/(-a^2*c-d*p)/(c*(
a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*ln((2
*c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2-2*p*(a*b
*c-(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*
p))+2*(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(
1/2)*(p*(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*p))^2-2*p*(a*b*c-(-d*(a^2*c*q+b^2*c*p+d*p*q
))^(1/2))/(a^2*c+d*p)*(x^2+(-a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*p))+c*(a^4*c*q+a^2*b^2*c*p+a^
2*d*p*q-b^2*d*p^2-2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2))/(x^2+(-a*b*c+(-d*(a^2*c*q+
b^2*c*p+d*p*q))^(1/2))/(-a^2*c-d*p)))*b*p-1/4/(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)/(a^2*c+d*p)/(c*(a^4*c*q+a^2*b
^2*c*p+a^2*d*p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*ln((2*c*(a^4*c*q+a
^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2-2*p*(a*b*c+(-d*(a^2*c
*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p))+2*(c*(a^4*c
*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*(p*(x^2+(a
*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p))^2-2*p*(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+
d*p)*(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p))+c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2+2
*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2))/(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2
))/(a^2*c+d*p)))*a^3*c*q-1/4/(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)/(a^2*c+d*p)/(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-
b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*ln((2*c*(a^4*c*q+a^2*b^2*c*p+a^2*d*
p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2-2*p*(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q)
)^(1/2))/(a^2*c+d*p)*(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p))+2*(c*(a^4*c*q+a^2*b^2*c*p+a^
2*d*p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*(p*(x^2+(a*b*c+(-d*(a^2*c*q
+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p))^2-2*p*(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(a*b*c+
(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p))+c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2
*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2))/(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)))*
a*b^2*c*p-1/4/(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)/(a^2*c+d*p)/(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2+2*(-d
*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*ln((2*c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2+2
*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2-2*p*(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c
+d*p)*(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p))+2*(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p
^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2)*(p*(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q)
)^(1/2))/(a^2*c+d*p))^2-2*p*(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(a*b*c+(-d*(a^2*c*q+b^
2*c*p+d*p*q))^(1/2))/(a^2*c+d*p))+c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1
/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2))/(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)))*a*d*p*q-1/4/(a^
2*c+d*p)/(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2
)^(1/2)*ln((2*c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*
p)^2-2*p*(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)
)/(a^2*c+d*p))+2*(c*(a^4*c*q+a^2*b^2*c*p+a^2*d*p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*
c+d*p)^2)^(1/2)*(p*(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p))^2-2*p*(a*b*c+(-d*(a^2*c*q+b^2*
c*p+d*p*q))^(1/2))/(a^2*c+d*p)*(x^2+(a*b*c+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p))+c*(a^4*c*q+a^2*b^2
*c*p+a^2*d*p*q-b^2*d*p^2+2*(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*a*b*p)/(a^2*c+d*p)^2)^(1/2))/(x^2+(a*b*c+(-d*(a^
2*c*q+b^2*c*p+d*p*q))^(1/2))/(a^2*c+d*p)))*b*p
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {b p x^{3} - a q x}{{\left (2 \, a b c x^{2} + {\left (a^{2} c + d p\right )} x^{4} + b^{2} c + d q\right )} \sqrt {p x^{4} + q}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*p*x^3-a*q*x)/(p*x^4+q)^(1/2)/(b^2*c+d*q+2*a*b*c*x^2+(a^2*c+d*p)*x^4),x, algorithm="maxima")
[Out]
integrate((b*p*x^3 - a*q*x)/((2*a*b*c*x^2 + (a^2*c + d*p)*x^4 + b^2*c + d*q)*sqrt(p*x^4 + q)), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {b\,p\,x^3-a\,q\,x}{\sqrt {p\,x^4+q}\,\left (c\,b^2+2\,a\,c\,b\,x^2+\left (c\,a^2+d\,p\right )\,x^4+d\,q\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*p*x^3 - a*q*x)/((q + p*x^4)^(1/2)*(d*q + x^4*(d*p + a^2*c) + b^2*c + 2*a*b*c*x^2)),x)
[Out]
int((b*p*x^3 - a*q*x)/((q + p*x^4)^(1/2)*(d*q + x^4*(d*p + a^2*c) + b^2*c + 2*a*b*c*x^2)), x)
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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((b*p*x**3-a*q*x)/(p*x**4+q)**(1/2)/(b**2*c+d*q+2*a*b*c*x**2+(a**2*c+d*p)*x**4),x)
[Out]
Timed out
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