Integrand size = 56, antiderivative size = 267 \[ \int \frac {-a q x+b p x^3}{\sqrt {q+p x^4} \left (b^2 c+d q+2 a b c x^2+\left (a^2 c+d p\right ) x^4\right )} \, dx=-\frac {1}{4} \text {RootSum}\left [a^2 c q^2+d p q^2-4 a b c \sqrt {p} q \text {$\#$1}+4 b^2 c p \text {$\#$1}^2-2 a^2 c q \text {$\#$1}^2+2 d p q \text {$\#$1}^2+4 a b c \sqrt {p} \text {$\#$1}^3+a^2 c \text {$\#$1}^4+d p \text {$\#$1}^4\&,\frac {b p q \log \left (\sqrt {p} x^2+\sqrt {q+p x^4}-\text {$\#$1}\right )+2 a \sqrt {p} q \log \left (\sqrt {p} x^2+\sqrt {q+p x^4}-\text {$\#$1}\right ) \text {$\#$1}-b p \log \left (\sqrt {p} x^2+\sqrt {q+p x^4}-\text {$\#$1}\right ) \text {$\#$1}^2}{-a b c \sqrt {p} q+2 b^2 c p \text {$\#$1}-a^2 c q \text {$\#$1}+d p q \text {$\#$1}+3 a b c \sqrt {p} \text {$\#$1}^2+a^2 c \text {$\#$1}^3+d p \text {$\#$1}^3}\&\right ] \]
Time = 0.50 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.14 \[ \int \frac {-a q x+b p x^3}{\sqrt {q+p x^4} \left (b^2 c+d q+2 a b c x^2+\left (a^2 c+d p\right ) x^4\right )} \, dx=-\frac {1}{4} \text {RootSum}\left [b^2 c p^2+d p^2 q+4 a b c p \sqrt {q} \text {$\#$1}-2 b^2 c p \text {$\#$1}^2+4 a^2 c q \text {$\#$1}^2+2 d p q \text {$\#$1}^2-4 a b c \sqrt {q} \text {$\#$1}^3+b^2 c \text {$\#$1}^4+d q \text {$\#$1}^4\&,\frac {-a p q \log \left (x^2\right )+a p q \log \left (-\sqrt {q}+\sqrt {q+p x^4}-x^2 \text {$\#$1}\right )+2 b p \sqrt {q} \log \left (x^2\right ) \text {$\#$1}-2 b p \sqrt {q} \log \left (-\sqrt {q}+\sqrt {q+p x^4}-x^2 \text {$\#$1}\right ) \text {$\#$1}+a q \log \left (x^2\right ) \text {$\#$1}^2-a q \log \left (-\sqrt {q}+\sqrt {q+p x^4}-x^2 \text {$\#$1}\right ) \text {$\#$1}^2}{a b c p \sqrt {q}-b^2 c p \text {$\#$1}+2 a^2 c q \text {$\#$1}+d p q \text {$\#$1}-3 a b c \sqrt {q} \text {$\#$1}^2+b^2 c \text {$\#$1}^3+d q \text {$\#$1}^3}\&\right ] \]
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]
-1/4*RootSum[b^2*c*p^2 + d*p^2*q + 4*a*b*c*p*Sqrt[q]*#1 - 2*b^2*c*p*#1^2 + 4*a^2*c*q*#1^2 + 2*d*p*q*#1^2 - 4*a*b*c*Sqrt[q]*#1^3 + b^2*c*#1^4 + d*q*# 1^4 & , (-(a*p*q*Log[x^2]) + a*p*q*Log[-Sqrt[q] + Sqrt[q + p*x^4] - x^2*#1 ] + 2*b*p*Sqrt[q]*Log[x^2]*#1 - 2*b*p*Sqrt[q]*Log[-Sqrt[q] + Sqrt[q + p*x^ 4] - x^2*#1]*#1 + a*q*Log[x^2]*#1^2 - a*q*Log[-Sqrt[q] + Sqrt[q + p*x^4] - x^2*#1]*#1^2)/(a*b*c*p*Sqrt[q] - b^2*c*p*#1 + 2*a^2*c*q*#1 + d*p*q*#1 - 3 *a*b*c*Sqrt[q]*#1^2 + b^2*c*#1^3 + d*q*#1^3) & ]
Time = 1.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {2027, 7266, 25, 1364, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {b p x^3-a q x}{\sqrt {p x^4+q} \left (x^4 \left (a^2 c+d p\right )+2 a b c x^2+b^2 c+d q\right )} \, dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle \int \frac {x \left (b p x^2-a q\right )}{\sqrt {p x^4+q} \left (x^4 \left (a^2 c+d p\right )+2 a b c x^2+b^2 c+d q\right )}dx\) |
\(\Big \downarrow \) 7266 |
\(\displaystyle \frac {1}{2} \int -\frac {a q-b p x^2}{\sqrt {p x^4+q} \left (\left (c a^2+d p\right ) x^4+2 a b c x^2+b^2 c+d q\right )}dx^2\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {a q-b p x^2}{\sqrt {p x^4+q} \left (\left (c a^2+d p\right ) x^4+2 a b c x^2+b^2 c+d q\right )}dx^2\) |
\(\Big \downarrow \) 1364 |
\(\displaystyle 2 a b q \left (a^2 c q+b^2 c p+d p q\right ) \int \frac {1}{-2 a b c q x^4-8 a b d q \left (c q a^2+b^2 c p+d p q\right )^2}d\frac {2 \left (c q a^2+b^2 c p+d p q\right ) \left (a x^2+b\right )}{\sqrt {p x^4+q}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {\arctan \left (\frac {\sqrt {c} \left (a x^2+b\right )}{\sqrt {d} \sqrt {p x^4+q}}\right )}{2 \sqrt {c} \sqrt {d}}\) |
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]
3.28.81.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((g_) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f _.)*(x_)^2]), x_Symbol] :> Simp[-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]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Time = 1.33 (sec) , antiderivative size = 1303, normalized size of antiderivative = 4.88
method | result | size |
default | \(\text {Expression too large to display}\) | \(1303\) |
elliptic | \(\text {Expression too large to display}\) | \(1303\) |
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)
-1/4*(a^3*c*q+a*b^2*c*p+a*d*p*q+(-d*(a^2*c*q+b^2*c*p+d*p*q))^(1/2)*b*p)/(- 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)*((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*p-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)))+1/4*(-a^3*c*q-a*b^2*c*p-a*d*p*q+(-d*(a^ 2*c*q+b^2*c*p+d*p*q))^(1/2)*b*p)/(-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...
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.59 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.69 \[ \int \frac {-a q x+b p x^3}{\sqrt {q+p x^4} \left (b^2 c+d q+2 a b c x^2+\left (a^2 c+d p\right ) x^4\right )} \, dx=\left [-\frac {\sqrt {-c d} \log \left (\frac {{\left (a^{4} c^{2} - 6 \, a^{2} c d p + d^{2} p^{2}\right )} x^{8} + 4 \, {\left (a^{3} b c^{2} - 3 \, a b c d p\right )} x^{6} + b^{4} c^{2} - 6 \, b^{2} c d q + 2 \, {\left (3 \, a^{2} b^{2} c^{2} - 3 \, b^{2} c d p - {\left (3 \, a^{2} c d - d^{2} p\right )} q\right )} x^{4} + d^{2} q^{2} + 4 \, {\left (a b^{3} c^{2} - 3 \, a b c d q\right )} x^{2} + 4 \, {\left ({\left (a^{3} c - a d p\right )} x^{6} + {\left (3 \, a^{2} b c - b d p\right )} x^{4} + b^{3} c - b d q + {\left (3 \, a b^{2} c - a d q\right )} x^{2}\right )} \sqrt {p x^{4} + q} \sqrt {-c d}}{{\left (a^{4} c^{2} + 2 \, a^{2} c d p + d^{2} p^{2}\right )} x^{8} + 4 \, {\left (a^{3} b c^{2} + a b c d p\right )} x^{6} + b^{4} c^{2} + 2 \, b^{2} c d q + 2 \, {\left (3 \, a^{2} b^{2} c^{2} + b^{2} c d p + {\left (a^{2} c d + d^{2} p\right )} q\right )} x^{4} + d^{2} q^{2} + 4 \, {\left (a b^{3} c^{2} + a b c d q\right )} x^{2}}\right )}{8 \, c d}, \frac {\sqrt {c d} \arctan \left (-\frac {{\left (2 \, a b c x^{2} + {\left (a^{2} c - d p\right )} x^{4} + b^{2} c - d q\right )} \sqrt {p x^{4} + q} \sqrt {c d}}{2 \, {\left (a c d p x^{6} + b c d p x^{4} + a c d q x^{2} + b c d q\right )}}\right )}{4 \, c d}\right ] \]
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")
[-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)]
Not integrable
Time = 16.51 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.21 \[ \int \frac {-a q x+b p x^3}{\sqrt {q+p x^4} \left (b^2 c+d q+2 a b c x^2+\left (a^2 c+d p\right ) x^4\right )} \, dx=\int \frac {x \left (- a q + b p x^{2}\right )}{\sqrt {p x^{4} + q} \left (a^{2} c x^{4} + 2 a b c x^{2} + b^{2} c + d p x^{4} + d q\right )}\, dx \]
Integral(x*(-a*q + b*p*x**2)/(sqrt(p*x**4 + q)*(a**2*c*x**4 + 2*a*b*c*x**2 + b**2*c + d*p*x**4 + d*q)), x)
Not integrable
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.21 \[ \int \frac {-a q x+b p x^3}{\sqrt {q+p x^4} \left (b^2 c+d q+2 a b c x^2+\left (a^2 c+d p\right ) x^4\right )} \, dx=\int { \frac {b p x^{3} - a q x}{{\left (2 \, a b c x^{2} + {\left (a^{2} c + d p\right )} x^{4} + b^{2} c + d q\right )} \sqrt {p x^{4} + q}} \,d x } \]
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")
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)
Timed out. \[ \int \frac {-a q x+b p x^3}{\sqrt {q+p x^4} \left (b^2 c+d q+2 a b c x^2+\left (a^2 c+d p\right ) x^4\right )} \, dx=\text {Timed out} \]
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")
Not integrable
Time = 8.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.21 \[ \int \frac {-a q x+b p x^3}{\sqrt {q+p x^4} \left (b^2 c+d q+2 a b c x^2+\left (a^2 c+d p\right ) x^4\right )} \, dx=\int \frac {b\,p\,x^3-a\,q\,x}{\sqrt {p\,x^4+q}\,\left (c\,b^2+2\,a\,c\,b\,x^2+\left (c\,a^2+d\,p\right )\,x^4+d\,q\right )} \,d x \]