Integrand size = 28, antiderivative size = 109 \[ \int \frac {(e x)^{7/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2 c e (e x)^{5/2}}{a \left (a x^2+b x^3\right )^{3/2}}-\frac {2 (4 b c-a d) (e x)^{7/2}}{3 a^2 \left (a x^2+b x^3\right )^{3/2}}-\frac {4 (4 b c-a d) e^2 (e x)^{3/2}}{3 a^3 \sqrt {a x^2+b x^3}} \] Output:
-2*c*e*(e*x)^(5/2)/a/(b*x^3+a*x^2)^(3/2)-2/3*(-a*d+4*b*c)*(e*x)^(7/2)/a^2/ (b*x^3+a*x^2)^(3/2)-4/3*(-a*d+4*b*c)*e^2*(e*x)^(3/2)/a^3/(b*x^3+a*x^2)^(1/ 2)
Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.56 \[ \int \frac {(e x)^{7/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 e (e x)^{5/2} \left (-8 b^2 c x^2-3 a^2 (c-d x)+2 a b x (-6 c+d x)\right )}{3 a^3 \left (x^2 (a+b x)\right )^{3/2}} \] Input:
Integrate[((e*x)^(7/2)*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x]
Output:
(2*e*(e*x)^(5/2)*(-8*b^2*c*x^2 - 3*a^2*(c - d*x) + 2*a*b*x*(-6*c + d*x)))/ (3*a^3*(x^2*(a + b*x))^(3/2))
Time = 0.49 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.13, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1943, 1921, 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.
\(\displaystyle \int \frac {(e x)^{7/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1943 |
\(\displaystyle \frac {e^2 (4 b c-a d) \int \frac {(e x)^{3/2}}{\left (b x^3+a x^2\right )^{3/2}}dx}{3 a b}+\frac {2 e (e x)^{5/2} (b c-a d)}{3 a b \left (a x^2+b x^3\right )^{3/2}}\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {e^2 (4 b c-a d) \left (\frac {2 e^2 \int \frac {1}{\sqrt {e x} \sqrt {b x^3+a x^2}}dx}{a}+\frac {2 e \sqrt {e x}}{a \sqrt {a x^2+b x^3}}\right )}{3 a b}+\frac {2 e (e x)^{5/2} (b c-a d)}{3 a b \left (a x^2+b x^3\right )^{3/2}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {e^2 (4 b c-a d) \left (\frac {2 e \sqrt {e x}}{a \sqrt {a x^2+b x^3}}-\frac {4 e^3 \sqrt {a x^2+b x^3}}{a^2 (e x)^{3/2}}\right )}{3 a b}+\frac {2 e (e x)^{5/2} (b c-a d)}{3 a b \left (a x^2+b x^3\right )^{3/2}}\) |
Input:
Int[((e*x)^(7/2)*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x]
Output:
(2*(b*c - a*d)*e*(e*x)^(5/2))/(3*a*b*(a*x^2 + b*x^3)^(3/2)) + ((4*b*c - a* d)*e^2*((2*e*Sqrt[e*x])/(a*Sqrt[a*x^2 + b*x^3]) - (4*e^3*Sqrt[a*x^2 + b*x^ 3])/(a^2*(e*x)^(3/2))))/(3*a*b)
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[((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] + Simp[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1))) In t[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n} , x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/( n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1))) Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 ] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
Time = 0.42 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.61
method | result | size |
gosper | \(-\frac {2 x \left (b x +a \right ) \left (-2 a b d \,x^{2}+8 b^{2} c \,x^{2}-3 a^{2} d x +12 a b c x +3 a^{2} c \right ) \left (e x \right )^{\frac {7}{2}}}{3 a^{3} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\) | \(67\) |
orering | \(-\frac {2 x \left (b x +a \right ) \left (-2 a b d \,x^{2}+8 b^{2} c \,x^{2}-3 a^{2} d x +12 a b c x +3 a^{2} c \right ) \left (e x \right )^{\frac {7}{2}}}{3 a^{3} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\) | \(67\) |
default | \(-\frac {2 x^{4} \left (b x +a \right ) \left (-2 a b d \,x^{2}+8 b^{2} c \,x^{2}-3 a^{2} d x +12 a b c x +3 a^{2} c \right ) \sqrt {e x}\, e^{3}}{3 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} a^{3}}\) | \(72\) |
risch | \(-\frac {2 c \left (b x +a \right ) e^{4} x}{a^{3} \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}+\frac {2 \left (2 a b d x -5 b^{2} c x +3 a^{2} d -6 a b c \right ) x^{2} e^{4}}{3 \left (b x +a \right ) a^{3} \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}\) | \(92\) |
Input:
int((e*x)^(7/2)*(d*x+c)/(b*x^3+a*x^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*x*(b*x+a)*(-2*a*b*d*x^2+8*b^2*c*x^2-3*a^2*d*x+12*a*b*c*x+3*a^2*c)*(e* x)^(7/2)/a^3/(b*x^3+a*x^2)^(5/2)
Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \frac {(e x)^{7/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, a^{2} c e^{3} + 2 \, {\left (4 \, b^{2} c - a b d\right )} e^{3} x^{2} + 3 \, {\left (4 \, a b c - a^{2} d\right )} e^{3} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{3 \, {\left (a^{3} b^{2} x^{4} + 2 \, a^{4} b x^{3} + a^{5} x^{2}\right )}} \] Input:
integrate((e*x)^(7/2)*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="fricas")
Output:
-2/3*(3*a^2*c*e^3 + 2*(4*b^2*c - a*b*d)*e^3*x^2 + 3*(4*a*b*c - a^2*d)*e^3* x)*sqrt(b*x^3 + a*x^2)*sqrt(e*x)/(a^3*b^2*x^4 + 2*a^4*b*x^3 + a^5*x^2)
\[ \int \frac {(e x)^{7/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int \frac {\left (e x\right )^{\frac {7}{2}} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((e*x)**(7/2)*(d*x+c)/(b*x**3+a*x**2)**(5/2),x)
Output:
Integral((e*x)**(7/2)*(c + d*x)/(x**2*(a + b*x))**(5/2), x)
\[ \int \frac {(e x)^{7/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((e*x)^(7/2)*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="maxima")
Output:
integrate((d*x + c)*(e*x)^(7/2)/(b*x^3 + a*x^2)^(5/2), x)
Time = 0.39 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.57 \[ \int \frac {(e x)^{7/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {4 \, \sqrt {b e} c e^{6}}{{\left (a e^{2} - {\left (\sqrt {b e} \sqrt {e x} - \sqrt {b e^{2} x + a e^{2}}\right )}^{2}\right )} a^{2} {\left | e \right |} \mathrm {sgn}\left (x\right )} - \frac {2 \, \sqrt {e x} {\left (\frac {{\left (5 \, a^{2} b^{3} c e^{7} {\left | e \right |} \mathrm {sgn}\left (x\right ) - 2 \, a^{3} b^{2} d e^{7} {\left | e \right |} \mathrm {sgn}\left (x\right )\right )} x}{a^{5} b e^{2}} + \frac {3 \, {\left (2 \, a^{3} b^{2} c e^{8} {\left | e \right |} \mathrm {sgn}\left (x\right ) - a^{4} b d e^{8} {\left | e \right |} \mathrm {sgn}\left (x\right )\right )}}{a^{5} b e^{3}}\right )}}{3 \, {\left (b e^{2} x + a e^{2}\right )}^{\frac {3}{2}}} \] Input:
integrate((e*x)^(7/2)*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="giac")
Output:
-4*sqrt(b*e)*c*e^6/((a*e^2 - (sqrt(b*e)*sqrt(e*x) - sqrt(b*e^2*x + a*e^2)) ^2)*a^2*abs(e)*sgn(x)) - 2/3*sqrt(e*x)*((5*a^2*b^3*c*e^7*abs(e)*sgn(x) - 2 *a^3*b^2*d*e^7*abs(e)*sgn(x))*x/(a^5*b*e^2) + 3*(2*a^3*b^2*c*e^8*abs(e)*sg n(x) - a^4*b*d*e^8*abs(e)*sgn(x))/(a^5*b*e^3))/(b*e^2*x + a*e^2)^(3/2)
Time = 9.51 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00 \[ \int \frac {(e x)^{7/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {2\,e^3\,x\,\sqrt {e\,x}\,\left (a\,d-4\,b\,c\right )}{a^2\,b^2}-\frac {2\,c\,e^3\,\sqrt {e\,x}}{a\,b^2}+\frac {4\,e^3\,x^2\,\sqrt {e\,x}\,\left (a\,d-4\,b\,c\right )}{3\,a^3\,b}\right )}{x^4+\frac {2\,a\,x^3}{b}+\frac {a^2\,x^2}{b^2}} \] Input:
int(((e*x)^(7/2)*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x)
Output:
((a*x^2 + b*x^3)^(1/2)*((2*e^3*x*(e*x)^(1/2)*(a*d - 4*b*c))/(a^2*b^2) - (2 *c*e^3*(e*x)^(1/2))/(a*b^2) + (4*e^3*x^2*(e*x)^(1/2)*(a*d - 4*b*c))/(3*a^3 *b)))/(x^4 + (2*a*x^3)/b + (a^2*x^2)/b^2)
Time = 0.21 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.34 \[ \int \frac {(e x)^{7/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 \sqrt {e}\, e^{3} \left (-2 \sqrt {b}\, \sqrt {b x +a}\, a^{2} d x +8 \sqrt {b}\, \sqrt {b x +a}\, a b c x -2 \sqrt {b}\, \sqrt {b x +a}\, a b d \,x^{2}+8 \sqrt {b}\, \sqrt {b x +a}\, b^{2} c \,x^{2}-3 \sqrt {x}\, a^{2} b c +3 \sqrt {x}\, a^{2} b d x -12 \sqrt {x}\, a \,b^{2} c x +2 \sqrt {x}\, a \,b^{2} d \,x^{2}-8 \sqrt {x}\, b^{3} c \,x^{2}\right )}{3 \sqrt {b x +a}\, a^{3} b x \left (b x +a \right )} \] Input:
int((e*x)^(7/2)*(d*x+c)/(b*x^3+a*x^2)^(5/2),x)
Output:
(2*sqrt(e)*e**3*( - 2*sqrt(b)*sqrt(a + b*x)*a**2*d*x + 8*sqrt(b)*sqrt(a + b*x)*a*b*c*x - 2*sqrt(b)*sqrt(a + b*x)*a*b*d*x**2 + 8*sqrt(b)*sqrt(a + b*x )*b**2*c*x**2 - 3*sqrt(x)*a**2*b*c + 3*sqrt(x)*a**2*b*d*x - 12*sqrt(x)*a*b **2*c*x + 2*sqrt(x)*a*b**2*d*x**2 - 8*sqrt(x)*b**3*c*x**2))/(3*sqrt(a + b* x)*a**3*b*x*(a + b*x))