Integrand size = 28, antiderivative size = 70 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{15/2}} \, dx=-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{7 a (e x)^{17/2}}+\frac {2 (2 b c-7 a d) \left (a x^2+b x^3\right )^{5/2}}{35 a^2 (e x)^{15/2}} \] Output:
-2/7*c*e*(b*x^3+a*x^2)^(5/2)/a/(e*x)^(17/2)+2/35*(-7*a*d+2*b*c)*(b*x^3+a*x ^2)^(5/2)/a^2/(e*x)^(15/2)
Time = 0.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.61 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{15/2}} \, dx=-\frac {2 e \left (x^2 (a+b x)\right )^{5/2} (5 a c-2 b c x+7 a d x)}{35 a^2 (e x)^{17/2}} \] Input:
Integrate[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/(e*x)^(15/2),x]
Output:
(-2*e*(x^2*(a + b*x))^(5/2)*(5*a*c - 2*b*c*x + 7*a*d*x))/(35*a^2*(e*x)^(17 /2))
Time = 0.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1944, 1920}
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.
\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{3/2} (c+d x)}{(e x)^{15/2}} \, dx\) |
\(\Big \downarrow \) 1944 |
\(\displaystyle -\frac {(2 b c-7 a d) \int \frac {\left (b x^3+a x^2\right )^{3/2}}{(e x)^{13/2}}dx}{7 a e}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{7 a (e x)^{17/2}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {2 \left (a x^2+b x^3\right )^{5/2} (2 b c-7 a d)}{35 a^2 (e x)^{15/2}}-\frac {2 c e \left (a x^2+b x^3\right )^{5/2}}{7 a (e x)^{17/2}}\) |
Input:
Int[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/(e*x)^(15/2),x]
Output:
(-2*c*e*(a*x^2 + b*x^3)^(5/2))/(7*a*(e*x)^(17/2)) + (2*(2*b*c - 7*a*d)*(a* x^2 + b*x^3)^(5/2))/(35*a^2*(e*x)^(15/2))
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[ n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 0.47 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(-\frac {2 x \left (b x +a \right ) \left (7 a d x -2 c b x +5 a c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{35 a^{2} \left (e x \right )^{\frac {15}{2}}}\) | \(45\) |
orering | \(-\frac {2 x \left (b x +a \right ) \left (7 a d x -2 c b x +5 a c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{35 a^{2} \left (e x \right )^{\frac {15}{2}}}\) | \(45\) |
default | \(-\frac {2 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (7 a b d \,x^{2}-2 b^{2} c \,x^{2}+7 a^{2} d x +3 a b c x +5 a^{2} c \right )}{35 x^{6} a^{2} e^{7} \sqrt {e x}}\) | \(67\) |
risch | \(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (7 a \,b^{2} d \,x^{3}-2 b^{3} c \,x^{3}+14 a^{2} b d \,x^{2}+a \,b^{2} c \,x^{2}+7 a^{3} d x +8 a^{2} b c x +5 c \,a^{3}\right )}{35 e^{7} x^{4} \sqrt {e x}\, a^{2}}\) | \(88\) |
Input:
int((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(15/2),x,method=_RETURNVERBOSE)
Output:
-2/35*x*(b*x+a)*(7*a*d*x-2*b*c*x+5*a*c)*(b*x^3+a*x^2)^(3/2)/a^2/(e*x)^(15/ 2)
Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.27 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{15/2}} \, dx=-\frac {2 \, {\left (5 \, a^{3} c - {\left (2 \, b^{3} c - 7 \, a b^{2} d\right )} x^{3} + {\left (a b^{2} c + 14 \, a^{2} b d\right )} x^{2} + {\left (8 \, a^{2} b c + 7 \, a^{3} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{35 \, a^{2} e^{8} x^{5}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(15/2),x, algorithm="fricas")
Output:
-2/35*(5*a^3*c - (2*b^3*c - 7*a*b^2*d)*x^3 + (a*b^2*c + 14*a^2*b*d)*x^2 + (8*a^2*b*c + 7*a^3*d)*x)*sqrt(b*x^3 + a*x^2)*sqrt(e*x)/(a^2*e^8*x^5)
Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{15/2}} \, dx=\text {Timed out} \] Input:
integrate((d*x+c)*(b*x**3+a*x**2)**(3/2)/(e*x)**(15/2),x)
Output:
Timed out
\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{15/2}} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} {\left (d x + c\right )}}{\left (e x\right )^{\frac {15}{2}}} \,d x } \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(15/2),x, algorithm="maxima")
Output:
integrate((b*x^3 + a*x^2)^(3/2)*(d*x + c)/(e*x)^(15/2), x)
Time = 0.25 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.50 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{15/2}} \, dx=\frac {2 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{5} {\left (\frac {{\left (2 \, a b^{3} c e^{3} \mathrm {sgn}\left (x\right ) - 7 \, a^{2} b^{2} d e^{3} \mathrm {sgn}\left (x\right )\right )} {\left (b x + a\right )}}{a^{3}} - \frac {7 \, {\left (a^{2} b^{3} c e^{3} \mathrm {sgn}\left (x\right ) - a^{3} b^{2} d e^{3} \mathrm {sgn}\left (x\right )\right )}}{a^{3}}\right )}}{35 \, {\left ({\left (b x + a\right )} b e - a b e\right )}^{\frac {7}{2}} e^{7} {\left | b \right |}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(15/2),x, algorithm="giac")
Output:
2/35*(b*x + a)^(5/2)*b^5*((2*a*b^3*c*e^3*sgn(x) - 7*a^2*b^2*d*e^3*sgn(x))* (b*x + a)/a^3 - 7*(a^2*b^3*c*e^3*sgn(x) - a^3*b^2*d*e^3*sgn(x))/a^3)/(((b* x + a)*b*e - a*b*e)^(7/2)*e^7*abs(b))
Time = 9.49 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.41 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{15/2}} \, dx=-\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {2\,a\,c}{7\,e^7}+\frac {x\,\left (14\,d\,a^3+16\,b\,c\,a^2\right )}{35\,a^2\,e^7}-\frac {x^3\,\left (4\,b^3\,c-14\,a\,b^2\,d\right )}{35\,a^2\,e^7}+\frac {2\,b\,x^2\,\left (14\,a\,d+b\,c\right )}{35\,a\,e^7}\right )}{x^4\,\sqrt {e\,x}} \] Input:
int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/(e*x)^(15/2),x)
Output:
-((a*x^2 + b*x^3)^(1/2)*((2*a*c)/(7*e^7) + (x*(14*a^3*d + 16*a^2*b*c))/(35 *a^2*e^7) - (x^3*(4*b^3*c - 14*a*b^2*d))/(35*a^2*e^7) + (2*b*x^2*(14*a*d + b*c))/(35*a*e^7)))/(x^4*(e*x)^(1/2))
Time = 0.21 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.19 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{(e x)^{15/2}} \, dx=\frac {2 \sqrt {e}\, \left (-5 \sqrt {x}\, \sqrt {b x +a}\, a^{3} c -7 \sqrt {x}\, \sqrt {b x +a}\, a^{3} d x -8 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b c x -14 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b d \,x^{2}-\sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} c \,x^{2}-7 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} d \,x^{3}+2 \sqrt {x}\, \sqrt {b x +a}\, b^{3} c \,x^{3}-3 \sqrt {b}\, a \,b^{2} d \,x^{4}-2 \sqrt {b}\, b^{3} c \,x^{4}\right )}{35 a^{2} e^{8} x^{4}} \] Input:
int((d*x+c)*(b*x^3+a*x^2)^(3/2)/(e*x)^(15/2),x)
Output:
(2*sqrt(e)*( - 5*sqrt(x)*sqrt(a + b*x)*a**3*c - 7*sqrt(x)*sqrt(a + b*x)*a* *3*d*x - 8*sqrt(x)*sqrt(a + b*x)*a**2*b*c*x - 14*sqrt(x)*sqrt(a + b*x)*a** 2*b*d*x**2 - sqrt(x)*sqrt(a + b*x)*a*b**2*c*x**2 - 7*sqrt(x)*sqrt(a + b*x) *a*b**2*d*x**3 + 2*sqrt(x)*sqrt(a + b*x)*b**3*c*x**3 - 3*sqrt(b)*a*b**2*d* x**4 - 2*sqrt(b)*b**3*c*x**4))/(35*a**2*e**8*x**4)