Integrand size = 30, antiderivative size = 107 \[ \int (c x)^{3/2} \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 a A (c x)^{5/2}}{5 c}+\frac {2 a B (c x)^{7/2}}{7 c^2}+\frac {2 (A b+a C) (c x)^{9/2}}{9 c^3}+\frac {2 (b B+a D) (c x)^{11/2}}{11 c^4}+\frac {2 b C (c x)^{13/2}}{13 c^5}+\frac {2 b D (c x)^{15/2}}{15 c^6} \] Output:
2/5*a*A*(c*x)^(5/2)/c+2/7*a*B*(c*x)^(7/2)/c^2+2/9*(A*b+C*a)*(c*x)^(9/2)/c^ 3+2/11*(B*b+D*a)*(c*x)^(11/2)/c^4+2/13*b*C*(c*x)^(13/2)/c^5+2/15*b*D*(c*x) ^(15/2)/c^6
Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.58 \[ \int (c x)^{3/2} \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 x (c x)^{3/2} \left (9009 a A+7 b x^2 \left (715 A+585 B x+495 C x^2+429 D x^3\right )+65 a x (99 B+7 x (11 C+9 D x))\right )}{45045} \] Input:
Integrate[(c*x)^(3/2)*(a + b*x^2)*(A + B*x + C*x^2 + D*x^3),x]
Output:
(2*x*(c*x)^(3/2)*(9009*a*A + 7*b*x^2*(715*A + 585*B*x + 495*C*x^2 + 429*D* x^3) + 65*a*x*(99*B + 7*x*(11*C + 9*D*x))))/45045
Time = 0.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2333, 2009}
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.
\(\displaystyle \int (c x)^{3/2} \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle \int \left (\frac {(c x)^{7/2} (a C+A b)}{c^2}+a A (c x)^{3/2}+\frac {(c x)^{9/2} (a D+b B)}{c^3}+\frac {a B (c x)^{5/2}}{c}+\frac {b D (c x)^{13/2}}{c^5}+\frac {b C (c x)^{11/2}}{c^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (c x)^{9/2} (a C+A b)}{9 c^3}+\frac {2 a A (c x)^{5/2}}{5 c}+\frac {2 (c x)^{11/2} (a D+b B)}{11 c^4}+\frac {2 a B (c x)^{7/2}}{7 c^2}+\frac {2 b D (c x)^{15/2}}{15 c^6}+\frac {2 b C (c x)^{13/2}}{13 c^5}\) |
Input:
Int[(c*x)^(3/2)*(a + b*x^2)*(A + B*x + C*x^2 + D*x^3),x]
Output:
(2*a*A*(c*x)^(5/2))/(5*c) + (2*a*B*(c*x)^(7/2))/(7*c^2) + (2*(A*b + a*C)*( c*x)^(9/2))/(9*c^3) + (2*(b*B + a*D)*(c*x)^(11/2))/(11*c^4) + (2*b*C*(c*x) ^(13/2))/(13*c^5) + (2*b*D*(c*x)^(15/2))/(15*c^6)
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 0.39 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.55
method | result | size |
pseudoelliptic | \(\frac {2 \sqrt {c x}\, \left (\frac {D b \,x^{5}}{3}+\frac {5 C b \,x^{4}}{13}+\frac {5 \left (B b +D a \right ) x^{3}}{11}+\frac {5 \left (A b +C a \right ) x^{2}}{9}+\frac {5 B a x}{7}+A a \right ) x^{2} c}{5}\) | \(59\) |
gosper | \(\frac {2 x \left (3003 D b \,x^{5}+3465 C b \,x^{4}+4095 b B \,x^{3}+4095 D a \,x^{3}+5005 A b \,x^{2}+5005 C a \,x^{2}+6435 B a x +9009 A a \right ) \left (c x \right )^{\frac {3}{2}}}{45045}\) | \(61\) |
orering | \(\frac {2 x \left (3003 D b \,x^{5}+3465 C b \,x^{4}+4095 b B \,x^{3}+4095 D a \,x^{3}+5005 A b \,x^{2}+5005 C a \,x^{2}+6435 B a x +9009 A a \right ) \left (c x \right )^{\frac {3}{2}}}{45045}\) | \(61\) |
trager | \(\frac {2 c \,x^{2} \left (3003 D b \,x^{5}+3465 C b \,x^{4}+4095 b B \,x^{3}+4095 D a \,x^{3}+5005 A b \,x^{2}+5005 C a \,x^{2}+6435 B a x +9009 A a \right ) \sqrt {c x}}{45045}\) | \(64\) |
derivativedivides | \(\frac {\frac {2 D b \left (c x \right )^{\frac {15}{2}}}{15}+\frac {2 b C c \left (c x \right )^{\frac {13}{2}}}{13}+\frac {2 \left (B b \,c^{2}+a \,c^{2} D\right ) \left (c x \right )^{\frac {11}{2}}}{11}+\frac {2 \left (A b \,c^{3}+C a \,c^{3}\right ) \left (c x \right )^{\frac {9}{2}}}{9}+\frac {2 a \,c^{4} B \left (c x \right )^{\frac {7}{2}}}{7}+\frac {2 a \,c^{5} A \left (c x \right )^{\frac {5}{2}}}{5}}{c^{6}}\) | \(90\) |
default | \(\frac {\frac {2 D b \left (c x \right )^{\frac {15}{2}}}{15}+\frac {2 b C c \left (c x \right )^{\frac {13}{2}}}{13}+\frac {2 \left (B b \,c^{2}+a \,c^{2} D\right ) \left (c x \right )^{\frac {11}{2}}}{11}+\frac {2 \left (A b \,c^{3}+C a \,c^{3}\right ) \left (c x \right )^{\frac {9}{2}}}{9}+\frac {2 a \,c^{4} B \left (c x \right )^{\frac {7}{2}}}{7}+\frac {2 a \,c^{5} A \left (c x \right )^{\frac {5}{2}}}{5}}{c^{6}}\) | \(90\) |
Input:
int((c*x)^(3/2)*(b*x^2+a)*(D*x^3+C*x^2+B*x+A),x,method=_RETURNVERBOSE)
Output:
2/5*(c*x)^(1/2)*(1/3*D*b*x^5+5/13*C*b*x^4+5/11*(B*b+D*a)*x^3+5/9*(A*b+C*a) *x^2+5/7*B*a*x+A*a)*x^2*c
Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.62 \[ \int (c x)^{3/2} \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2}{45045} \, {\left (3003 \, D b c x^{7} + 3465 \, C b c x^{6} + 4095 \, {\left (D a + B b\right )} c x^{5} + 6435 \, B a c x^{3} + 5005 \, {\left (C a + A b\right )} c x^{4} + 9009 \, A a c x^{2}\right )} \sqrt {c x} \] Input:
integrate((c*x)^(3/2)*(b*x^2+a)*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
Output:
2/45045*(3003*D*b*c*x^7 + 3465*C*b*c*x^6 + 4095*(D*a + B*b)*c*x^5 + 6435*B *a*c*x^3 + 5005*(C*a + A*b)*c*x^4 + 9009*A*a*c*x^2)*sqrt(c*x)
Time = 0.30 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.24 \[ \int (c x)^{3/2} \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 A a x \left (c x\right )^{\frac {3}{2}}}{5} + \frac {2 A b x^{3} \left (c x\right )^{\frac {3}{2}}}{9} + \frac {2 B a x^{2} \left (c x\right )^{\frac {3}{2}}}{7} + \frac {2 B b x^{4} \left (c x\right )^{\frac {3}{2}}}{11} + \frac {2 C a x^{3} \left (c x\right )^{\frac {3}{2}}}{9} + \frac {2 C b x^{5} \left (c x\right )^{\frac {3}{2}}}{13} + \frac {2 D a x^{4} \left (c x\right )^{\frac {3}{2}}}{11} + \frac {2 D b x^{6} \left (c x\right )^{\frac {3}{2}}}{15} \] Input:
integrate((c*x)**(3/2)*(b*x**2+a)*(D*x**3+C*x**2+B*x+A),x)
Output:
2*A*a*x*(c*x)**(3/2)/5 + 2*A*b*x**3*(c*x)**(3/2)/9 + 2*B*a*x**2*(c*x)**(3/ 2)/7 + 2*B*b*x**4*(c*x)**(3/2)/11 + 2*C*a*x**3*(c*x)**(3/2)/9 + 2*C*b*x**5 *(c*x)**(3/2)/13 + 2*D*a*x**4*(c*x)**(3/2)/11 + 2*D*b*x**6*(c*x)**(3/2)/15
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78 \[ \int (c x)^{3/2} \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (3003 \, \left (c x\right )^{\frac {15}{2}} D b + 3465 \, \left (c x\right )^{\frac {13}{2}} C b c + 6435 \, \left (c x\right )^{\frac {7}{2}} B a c^{4} + 9009 \, \left (c x\right )^{\frac {5}{2}} A a c^{5} + 4095 \, {\left (D a + B b\right )} \left (c x\right )^{\frac {11}{2}} c^{2} + 5005 \, {\left (C a + A b\right )} \left (c x\right )^{\frac {9}{2}} c^{3}\right )}}{45045 \, c^{6}} \] Input:
integrate((c*x)^(3/2)*(b*x^2+a)*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
Output:
2/45045*(3003*(c*x)^(15/2)*D*b + 3465*(c*x)^(13/2)*C*b*c + 6435*(c*x)^(7/2 )*B*a*c^4 + 9009*(c*x)^(5/2)*A*a*c^5 + 4095*(D*a + B*b)*(c*x)^(11/2)*c^2 + 5005*(C*a + A*b)*(c*x)^(9/2)*c^3)/c^6
Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93 \[ \int (c x)^{3/2} \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2}{45045} \, {\left (3003 \, \sqrt {c x} D b x^{7} + 3465 \, \sqrt {c x} C b x^{6} + 4095 \, \sqrt {c x} D a x^{5} + 4095 \, \sqrt {c x} B b x^{5} + 5005 \, \sqrt {c x} C a x^{4} + 5005 \, \sqrt {c x} A b x^{4} + 6435 \, \sqrt {c x} B a x^{3} + 9009 \, \sqrt {c x} A a x^{2}\right )} c \] Input:
integrate((c*x)^(3/2)*(b*x^2+a)*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
Output:
2/45045*(3003*sqrt(c*x)*D*b*x^7 + 3465*sqrt(c*x)*C*b*x^6 + 4095*sqrt(c*x)* D*a*x^5 + 4095*sqrt(c*x)*B*b*x^5 + 5005*sqrt(c*x)*C*a*x^4 + 5005*sqrt(c*x) *A*b*x^4 + 6435*sqrt(c*x)*B*a*x^3 + 9009*sqrt(c*x)*A*a*x^2)*c
Timed out. \[ \int (c x)^{3/2} \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\int {\left (c\,x\right )}^{3/2}\,\left (b\,x^2+a\right )\,\left (A+B\,x+C\,x^2+x^3\,D\right ) \,d x \] Input:
int((c*x)^(3/2)*(a + b*x^2)*(A + B*x + C*x^2 + x^3*D),x)
Output:
int((c*x)^(3/2)*(a + b*x^2)*(A + B*x + C*x^2 + x^3*D), x)
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.60 \[ \int (c x)^{3/2} \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \sqrt {x}\, \sqrt {c}\, c \,x^{2} \left (3003 b d \,x^{5}+3465 b c \,x^{4}+4095 a d \,x^{3}+4095 b^{2} x^{3}+5005 a b \,x^{2}+5005 a c \,x^{2}+6435 a b x +9009 a^{2}\right )}{45045} \] Input:
int((c*x)^(3/2)*(b*x^2+a)*(D*x^3+C*x^2+B*x+A),x)
Output:
(2*sqrt(x)*sqrt(c)*c*x**2*(9009*a**2 + 5005*a*b*x**2 + 6435*a*b*x + 5005*a *c*x**2 + 4095*a*d*x**3 + 4095*b**2*x**3 + 3465*b*c*x**4 + 3003*b*d*x**5)) /45045