Integrand size = 28, antiderivative size = 192 \[ \int \frac {c+d x}{(e x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 c e}{7 a (e x)^{5/2} \sqrt {a x^2+b x^3}}-\frac {2 (8 b c-7 a d)}{7 a^2 (e x)^{3/2} \sqrt {a x^2+b x^3}}+\frac {12 (8 b c-7 a d) e^2 \sqrt {a x^2+b x^3}}{35 a^3 (e x)^{7/2}}-\frac {16 b (8 b c-7 a d) e \sqrt {a x^2+b x^3}}{35 a^4 (e x)^{5/2}}+\frac {32 b^2 (8 b c-7 a d) \sqrt {a x^2+b x^3}}{35 a^5 (e x)^{3/2}} \] Output:
-2/7*c*e/a/(e*x)^(5/2)/(b*x^3+a*x^2)^(1/2)-2/7*(-7*a*d+8*b*c)/a^2/(e*x)^(3 /2)/(b*x^3+a*x^2)^(1/2)+12/35*(-7*a*d+8*b*c)*e^2*(b*x^3+a*x^2)^(1/2)/a^3/( e*x)^(7/2)-16/35*b*(-7*a*d+8*b*c)*e*(b*x^3+a*x^2)^(1/2)/a^4/(e*x)^(5/2)+32 /35*b^2*(-7*a*d+8*b*c)*(b*x^3+a*x^2)^(1/2)/a^5/(e*x)^(3/2)
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.53 \[ \int \frac {c+d x}{(e x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 e \left (-128 b^4 c x^4+16 a b^3 x^3 (-4 c+7 d x)+8 a^2 b^2 x^2 (2 c+7 d x)-2 a^3 b x (4 c+7 d x)+a^4 (5 c+7 d x)\right )}{35 a^5 (e x)^{5/2} \sqrt {x^2 (a+b x)}} \] Input:
Integrate[(c + d*x)/((e*x)^(3/2)*(a*x^2 + b*x^3)^(3/2)),x]
Output:
(-2*e*(-128*b^4*c*x^4 + 16*a*b^3*x^3*(-4*c + 7*d*x) + 8*a^2*b^2*x^2*(2*c + 7*d*x) - 2*a^3*b*x*(4*c + 7*d*x) + a^4*(5*c + 7*d*x)))/(35*a^5*(e*x)^(5/2 )*Sqrt[x^2*(a + b*x)])
Time = 0.65 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1944, 1921, 1922, 1922, 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.
| \(\displaystyle \int \frac {c+d x}{(e x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1944 |
| \(\displaystyle -\frac {(8 b c-7 a d) \int \frac {1}{\sqrt {e x} \left (b x^3+a x^2\right )^{3/2}}dx}{7 a e}-\frac {2 c e}{7 a (e x)^{5/2} \sqrt {a x^2+b x^3}}\) |
| \(\Big \downarrow \) 1921 |
| \(\displaystyle -\frac {(8 b c-7 a d) \left (\frac {6 e^2 \int \frac {1}{(e x)^{5/2} \sqrt {b x^3+a x^2}}dx}{a}+\frac {2 e}{a (e x)^{3/2} \sqrt {a x^2+b x^3}}\right )}{7 a e}-\frac {2 c e}{7 a (e x)^{5/2} \sqrt {a x^2+b x^3}}\) |
| \(\Big \downarrow \) 1922 |
| \(\displaystyle -\frac {(8 b c-7 a d) \left (\frac {6 e^2 \left (-\frac {4 b \int \frac {1}{(e x)^{3/2} \sqrt {b x^3+a x^2}}dx}{5 a e}-\frac {2 e \sqrt {a x^2+b x^3}}{5 a (e x)^{7/2}}\right )}{a}+\frac {2 e}{a (e x)^{3/2} \sqrt {a x^2+b x^3}}\right )}{7 a e}-\frac {2 c e}{7 a (e x)^{5/2} \sqrt {a x^2+b x^3}}\) |
| \(\Big \downarrow \) 1922 |
| \(\displaystyle -\frac {(8 b c-7 a d) \left (\frac {6 e^2 \left (-\frac {4 b \left (-\frac {2 b \int \frac {1}{\sqrt {e x} \sqrt {b x^3+a x^2}}dx}{3 a e}-\frac {2 e \sqrt {a x^2+b x^3}}{3 a (e x)^{5/2}}\right )}{5 a e}-\frac {2 e \sqrt {a x^2+b x^3}}{5 a (e x)^{7/2}}\right )}{a}+\frac {2 e}{a (e x)^{3/2} \sqrt {a x^2+b x^3}}\right )}{7 a e}-\frac {2 c e}{7 a (e x)^{5/2} \sqrt {a x^2+b x^3}}\) |
| \(\Big \downarrow \) 1920 |
| \(\displaystyle -\frac {(8 b c-7 a d) \left (\frac {6 e^2 \left (-\frac {4 b \left (\frac {4 b \sqrt {a x^2+b x^3}}{3 a^2 (e x)^{3/2}}-\frac {2 e \sqrt {a x^2+b x^3}}{3 a (e x)^{5/2}}\right )}{5 a e}-\frac {2 e \sqrt {a x^2+b x^3}}{5 a (e x)^{7/2}}\right )}{a}+\frac {2 e}{a (e x)^{3/2} \sqrt {a x^2+b x^3}}\right )}{7 a e}-\frac {2 c e}{7 a (e x)^{5/2} \sqrt {a x^2+b x^3}}\) |
Input:
Int[(c + d*x)/((e*x)^(3/2)*(a*x^2 + b*x^3)^(3/2)),x]
Output:
(-2*c*e)/(7*a*(e*x)^(5/2)*Sqrt[a*x^2 + b*x^3]) - ((8*b*c - 7*a*d)*((2*e)/( a*(e*x)^(3/2)*Sqrt[a*x^2 + b*x^3]) + (6*e^2*((-2*e*Sqrt[a*x^2 + b*x^3])/(5 *a*(e*x)^(7/2)) - (4*b*((-2*e*Sqrt[a*x^2 + b*x^3])/(3*a*(e*x)^(5/2)) + (4* b*Sqrt[a*x^2 + b*x^3])/(3*a^2*(e*x)^(3/2))))/(5*a*e)))/a))/(7*a*e)
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[((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*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) /(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || 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 = 0.43 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.60
| method | result | size |
| gosper | \(-\frac {2 x \left (b x +a \right ) \left (112 x^{4} a \,b^{3} d -128 x^{4} b^{4} c +56 a^{2} b^{2} d \,x^{3}-64 a \,b^{3} c \,x^{3}-14 a^{3} b d \,x^{2}+16 a^{2} b^{2} c \,x^{2}+7 a^{4} d x -8 a^{3} b c x +5 c \,a^{4}\right )}{35 a^{5} \left (e x \right )^{\frac {3}{2}} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(115\) |
| orering | \(-\frac {2 x \left (b x +a \right ) \left (112 x^{4} a \,b^{3} d -128 x^{4} b^{4} c +56 a^{2} b^{2} d \,x^{3}-64 a \,b^{3} c \,x^{3}-14 a^{3} b d \,x^{2}+16 a^{2} b^{2} c \,x^{2}+7 a^{4} d x -8 a^{3} b c x +5 c \,a^{4}\right )}{35 a^{5} \left (e x \right )^{\frac {3}{2}} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(115\) |
| default | \(-\frac {2 \left (b x +a \right ) \left (112 x^{4} a \,b^{3} d -128 x^{4} b^{4} c +56 a^{2} b^{2} d \,x^{3}-64 a \,b^{3} c \,x^{3}-14 a^{3} b d \,x^{2}+16 a^{2} b^{2} c \,x^{2}+7 a^{4} d x -8 a^{3} b c x +5 c \,a^{4}\right )}{35 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \sqrt {e x}\, e \,a^{5}}\) | \(117\) |
| risch | \(-\frac {2 \left (b x +a \right ) \left (77 a \,b^{2} d \,x^{3}-93 b^{3} c \,x^{3}-21 a^{2} b d \,x^{2}+29 a \,b^{2} c \,x^{2}+7 a^{3} d x -13 a^{2} b c x +5 c \,a^{3}\right )}{35 a^{5} x^{2} e \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}-\frac {2 b^{3} \left (a d -b c \right ) x^{2}}{a^{5} e \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}\) | \(133\) |
Input:
int((d*x+c)/(e*x)^(3/2)/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/35*x*(b*x+a)*(112*a*b^3*d*x^4-128*b^4*c*x^4+56*a^2*b^2*d*x^3-64*a*b^3*c *x^3-14*a^3*b*d*x^2+16*a^2*b^2*c*x^2+7*a^4*d*x-8*a^3*b*c*x+5*a^4*c)/a^5/(e *x)^(3/2)/(b*x^3+a*x^2)^(3/2)
Time = 0.09 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.68 \[ \int \frac {c+d x}{(e x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left (5 \, a^{4} c - 16 \, {\left (8 \, b^{4} c - 7 \, a b^{3} d\right )} x^{4} - 8 \, {\left (8 \, a b^{3} c - 7 \, a^{2} b^{2} d\right )} x^{3} + 2 \, {\left (8 \, a^{2} b^{2} c - 7 \, a^{3} b d\right )} x^{2} - {\left (8 \, a^{3} b c - 7 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{35 \, {\left (a^{5} b e^{2} x^{6} + a^{6} e^{2} x^{5}\right )}} \] Input:
integrate((d*x+c)/(e*x)^(3/2)/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
-2/35*(5*a^4*c - 16*(8*b^4*c - 7*a*b^3*d)*x^4 - 8*(8*a*b^3*c - 7*a^2*b^2*d )*x^3 + 2*(8*a^2*b^2*c - 7*a^3*b*d)*x^2 - (8*a^3*b*c - 7*a^4*d)*x)*sqrt(b* x^3 + a*x^2)*sqrt(e*x)/(a^5*b*e^2*x^6 + a^6*e^2*x^5)
\[ \int \frac {c+d x}{(e x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {c + d x}{\left (e x\right )^{\frac {3}{2}} \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((d*x+c)/(e*x)**(3/2)/(b*x**3+a*x**2)**(3/2),x)
Output:
Integral((c + d*x)/((e*x)**(3/2)*(x**2*(a + b*x))**(3/2)), x)
\[ \int \frac {c+d x}{(e x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {d x + c}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((d*x+c)/(e*x)^(3/2)/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate((d*x + c)/((b*x^3 + a*x^2)^(3/2)*(e*x)^(3/2)), x)
Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (162) = 324\).
Time = 0.41 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.95 \[ \int \frac {c+d x}{(e x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (\frac {{\left ({\left (b x + a\right )} {\left ({\left (b x + a\right )} {\left (\frac {{\left (93 \, a^{9} b^{10} c e^{3} {\left | b \right |} \mathrm {sgn}\left (x\right ) - 77 \, a^{10} b^{9} d e^{3} {\left | b \right |} \mathrm {sgn}\left (x\right )\right )} {\left (b x + a\right )}}{a^{14} b^{4}} - \frac {28 \, {\left (11 \, a^{10} b^{10} c e^{3} {\left | b \right |} \mathrm {sgn}\left (x\right ) - 9 \, a^{11} b^{9} d e^{3} {\left | b \right |} \mathrm {sgn}\left (x\right )\right )}}{a^{14} b^{4}}\right )} + \frac {70 \, {\left (5 \, a^{11} b^{10} c e^{3} {\left | b \right |} \mathrm {sgn}\left (x\right ) - 4 \, a^{12} b^{9} d e^{3} {\left | b \right |} \mathrm {sgn}\left (x\right )\right )}}{a^{14} b^{4}}\right )} - \frac {35 \, {\left (4 \, a^{12} b^{10} c e^{3} {\left | b \right |} \mathrm {sgn}\left (x\right ) - 3 \, a^{13} b^{9} d e^{3} {\left | b \right |} \mathrm {sgn}\left (x\right )\right )}}{a^{14} b^{4}}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b e - a b e\right )}^{\frac {7}{2}}} + \frac {70 \, {\left (b^{11} c^{2} e - 2 \, a b^{10} c d e + a^{2} b^{9} d^{2} e\right )}}{{\left (\sqrt {b e} a b^{6} c e - \sqrt {b e} a^{2} b^{5} d e + \sqrt {b e} {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {{\left (b x + a\right )} b e - a b e}\right )}^{2} b^{5} c - \sqrt {b e} {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {{\left (b x + a\right )} b e - a b e}\right )}^{2} a b^{4} d\right )} a^{4} {\left | b \right |} \mathrm {sgn}\left (x\right )}\right )}}{35 \, e} \] Input:
integrate((d*x+c)/(e*x)^(3/2)/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
2/35*(((b*x + a)*((b*x + a)*((93*a^9*b^10*c*e^3*abs(b)*sgn(x) - 77*a^10*b^ 9*d*e^3*abs(b)*sgn(x))*(b*x + a)/(a^14*b^4) - 28*(11*a^10*b^10*c*e^3*abs(b )*sgn(x) - 9*a^11*b^9*d*e^3*abs(b)*sgn(x))/(a^14*b^4)) + 70*(5*a^11*b^10*c *e^3*abs(b)*sgn(x) - 4*a^12*b^9*d*e^3*abs(b)*sgn(x))/(a^14*b^4)) - 35*(4*a ^12*b^10*c*e^3*abs(b)*sgn(x) - 3*a^13*b^9*d*e^3*abs(b)*sgn(x))/(a^14*b^4)) *sqrt(b*x + a)/((b*x + a)*b*e - a*b*e)^(7/2) + 70*(b^11*c^2*e - 2*a*b^10*c *d*e + a^2*b^9*d^2*e)/((sqrt(b*e)*a*b^6*c*e - sqrt(b*e)*a^2*b^5*d*e + sqrt (b*e)*(sqrt(b*e)*sqrt(b*x + a) - sqrt((b*x + a)*b*e - a*b*e))^2*b^5*c - sq rt(b*e)*(sqrt(b*e)*sqrt(b*x + a) - sqrt((b*x + a)*b*e - a*b*e))^2*a*b^4*d) *a^4*abs(b)*sgn(x)))/e
Time = 9.49 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.77 \[ \int \frac {c+d x}{(e x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {2\,c}{7\,a\,b\,e}-\frac {4\,x^2\,\left (7\,a\,d-8\,b\,c\right )}{35\,a^3\,e}-\frac {x^4\,\left (256\,b^4\,c-224\,a\,b^3\,d\right )}{35\,a^5\,b\,e}+\frac {16\,b\,x^3\,\left (7\,a\,d-8\,b\,c\right )}{35\,a^4\,e}+\frac {x\,\left (14\,a^4\,d-16\,a^3\,b\,c\right )}{35\,a^5\,b\,e}\right )}{x^5\,\sqrt {e\,x}+\frac {a\,x^4\,\sqrt {e\,x}}{b}} \] Input:
int((c + d*x)/((e*x)^(3/2)*(a*x^2 + b*x^3)^(3/2)),x)
Output:
-((a*x^2 + b*x^3)^(1/2)*((2*c)/(7*a*b*e) - (4*x^2*(7*a*d - 8*b*c))/(35*a^3 *e) - (x^4*(256*b^4*c - 224*a*b^3*d))/(35*a^5*b*e) + (16*b*x^3*(7*a*d - 8* b*c))/(35*a^4*e) + (x*(14*a^4*d - 16*a^3*b*c))/(35*a^5*b*e)))/(x^5*(e*x)^( 1/2) + (a*x^4*(e*x)^(1/2))/b)
Time = 0.22 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.83 \[ \int \frac {c+d x}{(e x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \sqrt {e}\, \left (112 \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} d \,x^{4}-128 \sqrt {b}\, \sqrt {b x +a}\, b^{3} c \,x^{4}-5 \sqrt {x}\, a^{4} c -7 \sqrt {x}\, a^{4} d x +8 \sqrt {x}\, a^{3} b c x +14 \sqrt {x}\, a^{3} b d \,x^{2}-16 \sqrt {x}\, a^{2} b^{2} c \,x^{2}-56 \sqrt {x}\, a^{2} b^{2} d \,x^{3}+64 \sqrt {x}\, a \,b^{3} c \,x^{3}-112 \sqrt {x}\, a \,b^{3} d \,x^{4}+128 \sqrt {x}\, b^{4} c \,x^{4}\right )}{35 \sqrt {b x +a}\, a^{5} e^{2} x^{4}} \] Input:
int((d*x+c)/(e*x)^(3/2)/(b*x^3+a*x^2)^(3/2),x)
Output:
(2*sqrt(e)*(112*sqrt(b)*sqrt(a + b*x)*a*b**2*d*x**4 - 128*sqrt(b)*sqrt(a + b*x)*b**3*c*x**4 - 5*sqrt(x)*a**4*c - 7*sqrt(x)*a**4*d*x + 8*sqrt(x)*a**3 *b*c*x + 14*sqrt(x)*a**3*b*d*x**2 - 16*sqrt(x)*a**2*b**2*c*x**2 - 56*sqrt( x)*a**2*b**2*d*x**3 + 64*sqrt(x)*a*b**3*c*x**3 - 112*sqrt(x)*a*b**3*d*x**4 + 128*sqrt(x)*b**4*c*x**4))/(35*sqrt(a + b*x)*a**5*e**2*x**4)