Integrand size = 61, antiderivative size = 42 \[ \int \frac {-2 a q+3 b p x^2+a p x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt {d} \sqrt {q+p x^3}}{\sqrt {c} (b+a x)}\right )}{\sqrt {c} \sqrt {d}} \] Output:
2*arctan(d^(1/2)*(p*x^3+q)^(1/2)/c^(1/2)/(a*x+b))/c^(1/2)/d^(1/2)
Time = 5.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {-2 a q+3 b p x^2+a p x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt {d} \sqrt {q+p x^3}}{\sqrt {c} (b+a x)}\right )}{\sqrt {c} \sqrt {d}} \] Input:
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]
Output:
(2*ArcTan[(Sqrt[d]*Sqrt[q + p*x^3])/(Sqrt[c]*(b + a*x))])/(Sqrt[c]*Sqrt[d] )
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.
| \(\displaystyle \int \frac {a p x^3-2 a q+3 b p x^2}{\sqrt {p x^3+q} \left (a^2 c x^2+2 a b c x+b^2 c+d p x^3+d q\right )} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {a}{d \sqrt {p x^3+q}}-\frac {x^2 \left (a^3 c-3 b d p\right )+2 a^2 b c x+a \left (b^2 c+3 d q\right )}{d \sqrt {p x^3+q} \left (a^2 c x^2+2 a b c x+b^2 c+d p x^3+d q\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^2 b c \int \frac {x}{\sqrt {p x^3+q} \left (d p x^3+a^2 c x^2+2 a b c x+b^2 c+d q\right )}dx}{d}-\frac {a \left (b^2 c+3 d q\right ) \int \frac {1}{\sqrt {p x^3+q} \left (d p x^3+a^2 c x^2+2 a b c x+b^2 c+d q\right )}dx}{d}-\frac {\left (a^3 c-3 b d p\right ) \int \frac {x^2}{\sqrt {p x^3+q} \left (d p x^3+a^2 c x^2+2 a b c x+b^2 c+d q\right )}dx}{d}+\frac {2 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{p} x+\sqrt [3]{q}\right ) \sqrt {\frac {p^{2/3} x^2-\sqrt [3]{p} \sqrt [3]{q} x+q^{2/3}}{\left (\sqrt [3]{p} x+\left (1+\sqrt {3}\right ) \sqrt [3]{q}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{p} x+\left (1-\sqrt {3}\right ) \sqrt [3]{q}}{\sqrt [3]{p} x+\left (1+\sqrt {3}\right ) \sqrt [3]{q}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d \sqrt [3]{p} \sqrt {\frac {\sqrt [3]{q} \left (\sqrt [3]{p} x+\sqrt [3]{q}\right )}{\left (\sqrt [3]{p} x+\left (1+\sqrt {3}\right ) \sqrt [3]{q}\right )^2}} \sqrt {p x^3+q}}\) |
Input:
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]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.61 (sec) , antiderivative size = 2262, normalized size of antiderivative = 53.86
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(2262\) |
| default | \(\text {Expression too large to display}\) | \(2264\) |
Input:
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)
Output:
-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((_alpha^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) *(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)^(2/3)*a ^6*c*q+(-q*p^2)^(2/3)*b^4*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)^(1/3)*3^(1/2)*_alpha^2*a^4*d*q*p^2+2* p^2*(-2*_alpha^2*a^3*b*d*p*q-_alpha^2*b^4*d*p^2-2*_alpha*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*...
Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (32) = 64\).
Time = 0.80 (sec) , antiderivative size = 459, normalized size of antiderivative = 10.93 \[ \int \frac {-2 a q+3 b p x^2+a p x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx=\left [-\frac {\sqrt {-c d} \log \left (-\frac {6 \, a^{2} c d p x^{5} - d^{2} p^{2} x^{6} - b^{4} c^{2} + 6 \, b^{2} c d q - {\left (a^{4} c^{2} - 12 \, a b c d p\right )} x^{4} - d^{2} q^{2} - 2 \, {\left (2 \, a^{3} b c^{2} - 3 \, b^{2} c d p + d^{2} p q\right )} x^{3} - 6 \, {\left (a^{2} b^{2} c^{2} - a^{2} c d q\right )} x^{2} + 4 \, {\left (a d p x^{4} - 3 \, a^{2} b c x^{2} - b^{3} c - {\left (a^{3} c - b d p\right )} x^{3} + b d q - {\left (3 \, a b^{2} c - a d q\right )} x\right )} \sqrt {p x^{3} + q} \sqrt {-c d} - 4 \, {\left (a b^{3} c^{2} - 3 \, a b c d q\right )} x}{2 \, a^{2} c d p x^{5} + d^{2} p^{2} x^{6} + b^{4} c^{2} + 2 \, b^{2} c d q + {\left (a^{4} c^{2} + 4 \, a b c d p\right )} x^{4} + d^{2} q^{2} + 2 \, {\left (2 \, a^{3} b c^{2} + b^{2} c d p + d^{2} p q\right )} x^{3} + 2 \, {\left (3 \, a^{2} b^{2} c^{2} + a^{2} c d q\right )} x^{2} + 4 \, {\left (a b^{3} c^{2} + a b c d q\right )} x}\right )}{2 \, c d}, \frac {\sqrt {c d} \arctan \left (-\frac {{\left (a^{2} c x^{2} - d p x^{3} + 2 \, a b c x + b^{2} c - d q\right )} \sqrt {p x^{3} + q} \sqrt {c d}}{2 \, {\left (a c d p x^{4} + b c d p x^{3} + a c d q x + b c d q\right )}}\right )}{c d}\right ] \] Input:
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="fricas")
Output:
[-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)]
Timed out. \[ \int \frac {-2 a q+3 b p x^2+a p x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx=\text {Timed out} \] Input:
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)
Output:
Timed out
\[ \int \frac {-2 a q+3 b p x^2+a p x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx=\int { \frac {a p x^{3} + 3 \, b p x^{2} - 2 \, a q}{{\left (a^{2} c x^{2} + d p x^{3} + 2 \, a b c x + b^{2} c + d q\right )} \sqrt {p x^{3} + q}} \,d x } \] Input:
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="maxima")
Output:
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)
Timed out. \[ \int \frac {-2 a q+3 b p x^2+a p x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx=\text {Timed out} \] Input:
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")
Output:
Timed out
Time = 19.38 (sec) , antiderivative size = 456, normalized size of antiderivative = 10.86 \[ \int \frac {-2 a q+3 b p x^2+a p x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx=\frac {\ln \left (\frac {\left (a^3\,b\,c\,1{}\mathrm {i}-b^2\,d\,p\,1{}\mathrm {i}+a^4\,c\,x\,1{}\mathrm {i}+2\,a^3\,\sqrt {c}\,\sqrt {d}\,\sqrt {p\,x^3+q}-a^2\,d\,p\,x^2\,1{}\mathrm {i}+a\,b\,d\,p\,x\,1{}\mathrm {i}\right )\,\left (a^3\,b^3\,c^2\,1{}\mathrm {i}+a^6\,c^2\,x^3\,1{}\mathrm {i}+a^4\,b^2\,c^2\,x\,3{}\mathrm {i}+a^5\,b\,c^2\,x^2\,3{}\mathrm {i}-b^4\,c\,d\,p\,1{}\mathrm {i}+b^2\,d^2\,p\,\left (p\,x^3+q\right )\,1{}\mathrm {i}+a^2\,d^2\,p\,x^2\,\left (p\,x^3+q\right )\,1{}\mathrm {i}+a^3\,b\,c\,d\,q\,1{}\mathrm {i}+a^3\,b\,c\,d\,\left (p\,x^3+q\right )\,2{}\mathrm {i}+a^4\,c\,d\,q\,x\,1{}\mathrm {i}+a^4\,c\,d\,x\,\left (p\,x^3+q\right )\,2{}\mathrm {i}+2\,a^3\,\sqrt {c}\,d^{3/2}\,q\,\sqrt {p\,x^3+q}-2\,b^3\,\sqrt {c}\,d^{3/2}\,p\,\sqrt {p\,x^3+q}-a\,b^3\,c\,d\,p\,x\,1{}\mathrm {i}-a\,b\,d^2\,p\,x\,\left (p\,x^3+q\right )\,1{}\mathrm {i}\right )}{\left (c\,a^2\,x^2+2\,c\,a\,b\,x+c\,b^2+d\,p\,x^3+d\,q\right )\,\left (a^8\,c^2\,x^2+2\,a^7\,b\,c^2\,x+a^6\,b^2\,c^2+2\,a^6\,c\,d\,p\,x^3+4\,q\,a^6\,c\,d+a^4\,d^2\,p^2\,x^4-2\,a^3\,b^3\,c\,d\,p-2\,a^3\,b\,d^2\,p^2\,x^3+3\,a^2\,b^2\,d^2\,p^2\,x^2-2\,a\,b^3\,d^2\,p^2\,x+b^4\,d^2\,p^2\right )}\right )\,1{}\mathrm {i}}{\sqrt {c}\,\sqrt {d}} \] Input:
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)
Output:
(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))
\[ \int \frac {-2 a q+3 b p x^2+a p x^3}{\sqrt {q+p x^3} \left (b^2 c+d q+2 a b c x+a^2 c x^2+d p x^3\right )} \, dx=-2 \left (\int \frac {\sqrt {p \,x^{3}+q}}{a^{2} c p \,x^{5}+d \,p^{2} x^{6}+2 a b c p \,x^{4}+b^{2} c p \,x^{3}+a^{2} c q \,x^{2}+2 d p q \,x^{3}+2 a b c q x +b^{2} c q +d \,q^{2}}d x \right ) a q +\left (\int \frac {\sqrt {p \,x^{3}+q}\, x^{3}}{a^{2} c p \,x^{5}+d \,p^{2} x^{6}+2 a b c p \,x^{4}+b^{2} c p \,x^{3}+a^{2} c q \,x^{2}+2 d p q \,x^{3}+2 a b c q x +b^{2} c q +d \,q^{2}}d x \right ) a p +3 \left (\int \frac {\sqrt {p \,x^{3}+q}\, x^{2}}{a^{2} c p \,x^{5}+d \,p^{2} x^{6}+2 a b c p \,x^{4}+b^{2} c p \,x^{3}+a^{2} c q \,x^{2}+2 d p q \,x^{3}+2 a b c q x +b^{2} c q +d \,q^{2}}d x \right ) b p \] Input:
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)
Output:
- 2*int(sqrt(p*x**3 + q)/(a**2*c*p*x**5 + a**2*c*q*x**2 + 2*a*b*c*p*x**4 + 2*a*b*c*q*x + b**2*c*p*x**3 + b**2*c*q + d*p**2*x**6 + 2*d*p*q*x**3 + d* q**2),x)*a*q + int((sqrt(p*x**3 + q)*x**3)/(a**2*c*p*x**5 + a**2*c*q*x**2 + 2*a*b*c*p*x**4 + 2*a*b*c*q*x + b**2*c*p*x**3 + b**2*c*q + d*p**2*x**6 + 2*d*p*q*x**3 + d*q**2),x)*a*p + 3*int((sqrt(p*x**3 + q)*x**2)/(a**2*c*p*x* *5 + a**2*c*q*x**2 + 2*a*b*c*p*x**4 + 2*a*b*c*q*x + b**2*c*p*x**3 + b**2*c *q + d*p**2*x**6 + 2*d*p*q*x**3 + d*q**2),x)*b*p