Integrand size = 28, antiderivative size = 151 \[ \int \frac {c+d x}{\sqrt {e x} \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 c e}{5 a (e x)^{3/2} \sqrt {a x^2+b x^3}}-\frac {2 (6 b c-5 a d)}{5 a^2 \sqrt {e x} \sqrt {a x^2+b x^3}}+\frac {8 (6 b c-5 a d) e^2 \sqrt {a x^2+b x^3}}{15 a^3 (e x)^{5/2}}-\frac {16 b (6 b c-5 a d) e \sqrt {a x^2+b x^3}}{15 a^4 (e x)^{3/2}} \] Output:
-2/5*c*e/a/(e*x)^(3/2)/(b*x^3+a*x^2)^(1/2)-2/5*(-5*a*d+6*b*c)/a^2/(e*x)^(1 /2)/(b*x^3+a*x^2)^(1/2)+8/15*(-5*a*d+6*b*c)*e^2*(b*x^3+a*x^2)^(1/2)/a^3/(e *x)^(5/2)-16/15*b*(-5*a*d+6*b*c)*e*(b*x^3+a*x^2)^(1/2)/a^4/(e*x)^(3/2)
Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.54 \[ \int \frac {c+d x}{\sqrt {e x} \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 e \left (48 b^3 c x^3+8 a b^2 x^2 (3 c-5 d x)+a^3 (3 c+5 d x)-2 a^2 b x (3 c+10 d x)\right )}{15 a^4 (e x)^{3/2} \sqrt {x^2 (a+b x)}} \] Input:
Integrate[(c + d*x)/(Sqrt[e*x]*(a*x^2 + b*x^3)^(3/2)),x]
Output:
(-2*e*(48*b^3*c*x^3 + 8*a*b^2*x^2*(3*c - 5*d*x) + a^3*(3*c + 5*d*x) - 2*a^ 2*b*x*(3*c + 10*d*x)))/(15*a^4*(e*x)^(3/2)*Sqrt[x^2*(a + b*x)])
Time = 0.55 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1944, 1921, 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.
| \(\displaystyle \int \frac {c+d x}{\sqrt {e x} \left (a x^2+b x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1944 |
| \(\displaystyle -\frac {(6 b c-5 a d) \int \frac {\sqrt {e x}}{\left (b x^3+a x^2\right )^{3/2}}dx}{5 a e}-\frac {2 c e}{5 a (e x)^{3/2} \sqrt {a x^2+b x^3}}\) |
| \(\Big \downarrow \) 1921 |
| \(\displaystyle -\frac {(6 b c-5 a d) \left (\frac {4 e^2 \int \frac {1}{(e x)^{3/2} \sqrt {b x^3+a x^2}}dx}{a}+\frac {2 e}{a \sqrt {e x} \sqrt {a x^2+b x^3}}\right )}{5 a e}-\frac {2 c e}{5 a (e x)^{3/2} \sqrt {a x^2+b x^3}}\) |
| \(\Big \downarrow \) 1922 |
| \(\displaystyle -\frac {(6 b c-5 a d) \left (\frac {4 e^2 \left (-\frac {2 b \int \frac {1}{\sqrt {e x} \sqrt {b x^3+a x^2}}dx}{3 a e}-\frac {2 e \sqrt {a x^2+b x^3}}{3 a (e x)^{5/2}}\right )}{a}+\frac {2 e}{a \sqrt {e x} \sqrt {a x^2+b x^3}}\right )}{5 a e}-\frac {2 c e}{5 a (e x)^{3/2} \sqrt {a x^2+b x^3}}\) |
| \(\Big \downarrow \) 1920 |
| \(\displaystyle -\frac {(6 b c-5 a d) \left (\frac {4 e^2 \left (\frac {4 b \sqrt {a x^2+b x^3}}{3 a^2 (e x)^{3/2}}-\frac {2 e \sqrt {a x^2+b x^3}}{3 a (e x)^{5/2}}\right )}{a}+\frac {2 e}{a \sqrt {e x} \sqrt {a x^2+b x^3}}\right )}{5 a e}-\frac {2 c e}{5 a (e x)^{3/2} \sqrt {a x^2+b x^3}}\) |
Input:
Int[(c + d*x)/(Sqrt[e*x]*(a*x^2 + b*x^3)^(3/2)),x]
Output:
(-2*c*e)/(5*a*(e*x)^(3/2)*Sqrt[a*x^2 + b*x^3]) - ((6*b*c - 5*a*d)*((2*e)/( a*Sqrt[e*x]*Sqrt[a*x^2 + b*x^3]) + (4*e^2*((-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))))/a))/(5*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 = 0.42 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.60
| method | result | size |
| gosper | \(-\frac {2 x \left (b x +a \right ) \left (-40 a \,b^{2} d \,x^{3}+48 b^{3} c \,x^{3}-20 a^{2} b d \,x^{2}+24 a \,b^{2} c \,x^{2}+5 a^{3} d x -6 a^{2} b c x +3 c \,a^{3}\right )}{15 a^{4} \sqrt {e x}\, \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(91\) |
| default | \(-\frac {2 x \left (b x +a \right ) \left (-40 a \,b^{2} d \,x^{3}+48 b^{3} c \,x^{3}-20 a^{2} b d \,x^{2}+24 a \,b^{2} c \,x^{2}+5 a^{3} d x -6 a^{2} b c x +3 c \,a^{3}\right )}{15 a^{4} \sqrt {e x}\, \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(91\) |
| orering | \(-\frac {2 x \left (b x +a \right ) \left (-40 a \,b^{2} d \,x^{3}+48 b^{3} c \,x^{3}-20 a^{2} b d \,x^{2}+24 a \,b^{2} c \,x^{2}+5 a^{3} d x -6 a^{2} b c x +3 c \,a^{3}\right )}{15 a^{4} \sqrt {e x}\, \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(91\) |
| risch | \(-\frac {2 \left (b x +a \right ) \left (-25 a b d \,x^{2}+33 b^{2} c \,x^{2}+5 a^{2} d x -9 a b c x +3 a^{2} c \right )}{15 a^{4} x \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}+\frac {2 \left (a d -b c \right ) b^{2} x^{2}}{a^{4} \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}\) | \(103\) |
Input:
int((d*x+c)/(e*x)^(1/2)/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/15*x*(b*x+a)*(-40*a*b^2*d*x^3+48*b^3*c*x^3-20*a^2*b*d*x^2+24*a*b^2*c*x^ 2+5*a^3*d*x-6*a^2*b*c*x+3*a^3*c)/a^4/(e*x)^(1/2)/(b*x^3+a*x^2)^(3/2)
Time = 0.10 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.68 \[ \int \frac {c+d x}{\sqrt {e x} \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left (3 \, a^{3} c + 8 \, {\left (6 \, b^{3} c - 5 \, a b^{2} d\right )} x^{3} + 4 \, {\left (6 \, a b^{2} c - 5 \, a^{2} b d\right )} x^{2} - {\left (6 \, a^{2} b c - 5 \, a^{3} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{15 \, {\left (a^{4} b e x^{5} + a^{5} e x^{4}\right )}} \] Input:
integrate((d*x+c)/(e*x)^(1/2)/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
-2/15*(3*a^3*c + 8*(6*b^3*c - 5*a*b^2*d)*x^3 + 4*(6*a*b^2*c - 5*a^2*b*d)*x ^2 - (6*a^2*b*c - 5*a^3*d)*x)*sqrt(b*x^3 + a*x^2)*sqrt(e*x)/(a^4*b*e*x^5 + a^5*e*x^4)
\[ \int \frac {c+d x}{\sqrt {e x} \left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {c + d x}{\sqrt {e x} \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((d*x+c)/(e*x)**(1/2)/(b*x**3+a*x**2)**(3/2),x)
Output:
Integral((c + d*x)/(sqrt(e*x)*(x**2*(a + b*x))**(3/2)), x)
\[ \int \frac {c+d x}{\sqrt {e x} \left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {d x + c}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} \sqrt {e x}} \,d x } \] Input:
integrate((d*x+c)/(e*x)^(1/2)/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate((d*x + c)/((b*x^3 + a*x^2)^(3/2)*sqrt(e*x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (127) = 254\).
Time = 0.28 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.14 \[ \int \frac {c+d x}{\sqrt {e x} \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 \, \sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {{\left (33 \, a^{5} b^{8} c e^{2} - 25 \, a^{6} b^{7} d e^{2}\right )} {\left (b x + a\right )}}{a^{9} b^{2} {\left | b \right |} \mathrm {sgn}\left (x\right )} - \frac {5 \, {\left (15 \, a^{6} b^{8} c e^{2} - 11 \, a^{7} b^{7} d e^{2}\right )}}{a^{9} b^{2} {\left | b \right |} \mathrm {sgn}\left (x\right )}\right )} + \frac {15 \, {\left (3 \, a^{7} b^{8} c e^{2} - 2 \, a^{8} b^{7} d e^{2}\right )}}{a^{9} b^{2} {\left | b \right |} \mathrm {sgn}\left (x\right )}\right )}}{15 \, {\left ({\left (b x + a\right )} b e - a b e\right )}^{\frac {5}{2}}} - \frac {4 \, {\left (b^{9} c^{2} e - 2 \, a b^{8} c d e + a^{2} b^{7} d^{2} e\right )}}{{\left (\sqrt {b e} a b^{5} c e - \sqrt {b e} a^{2} b^{4} d e + \sqrt {b e} {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {{\left (b x + a\right )} b e - a b e}\right )}^{2} b^{4} c - \sqrt {b e} {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {{\left (b x + a\right )} b e - a b e}\right )}^{2} a b^{3} d\right )} a^{3} {\left | b \right |} \mathrm {sgn}\left (x\right )} \] Input:
integrate((d*x+c)/(e*x)^(1/2)/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
-2/15*sqrt(b*x + a)*((b*x + a)*((33*a^5*b^8*c*e^2 - 25*a^6*b^7*d*e^2)*(b*x + a)/(a^9*b^2*abs(b)*sgn(x)) - 5*(15*a^6*b^8*c*e^2 - 11*a^7*b^7*d*e^2)/(a ^9*b^2*abs(b)*sgn(x))) + 15*(3*a^7*b^8*c*e^2 - 2*a^8*b^7*d*e^2)/(a^9*b^2*a bs(b)*sgn(x)))/((b*x + a)*b*e - a*b*e)^(5/2) - 4*(b^9*c^2*e - 2*a*b^8*c*d* e + a^2*b^7*d^2*e)/((sqrt(b*e)*a*b^5*c*e - sqrt(b*e)*a^2*b^4*d*e + sqrt(b* e)*(sqrt(b*e)*sqrt(b*x + a) - sqrt((b*x + a)*b*e - a*b*e))^2*b^4*c - sqrt( b*e)*(sqrt(b*e)*sqrt(b*x + a) - sqrt((b*x + a)*b*e - a*b*e))^2*a*b^3*d)*a^ 3*abs(b)*sgn(x))
Time = 9.82 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.76 \[ \int \frac {c+d x}{\sqrt {e x} \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {2\,c}{5\,a\,b}-\frac {8\,x^2\,\left (5\,a\,d-6\,b\,c\right )}{15\,a^3}+\frac {x\,\left (10\,a^3\,d-12\,a^2\,b\,c\right )}{15\,a^4\,b}+\frac {x^3\,\left (96\,b^3\,c-80\,a\,b^2\,d\right )}{15\,a^4\,b}\right )}{x^4\,\sqrt {e\,x}+\frac {a\,x^3\,\sqrt {e\,x}}{b}} \] Input:
int((c + d*x)/((e*x)^(1/2)*(a*x^2 + b*x^3)^(3/2)),x)
Output:
-((a*x^2 + b*x^3)^(1/2)*((2*c)/(5*a*b) - (8*x^2*(5*a*d - 6*b*c))/(15*a^3) + (x*(10*a^3*d - 12*a^2*b*c))/(15*a^4*b) + (x^3*(96*b^3*c - 80*a*b^2*d))/( 15*a^4*b)))/(x^4*(e*x)^(1/2) + (a*x^3*(e*x)^(1/2))/b)
Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.85 \[ \int \frac {c+d x}{\sqrt {e x} \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \sqrt {e}\, \left (-40 \sqrt {b}\, \sqrt {b x +a}\, a b d \,x^{3}+48 \sqrt {b}\, \sqrt {b x +a}\, b^{2} c \,x^{3}-3 \sqrt {x}\, a^{3} c -5 \sqrt {x}\, a^{3} d x +6 \sqrt {x}\, a^{2} b c x +20 \sqrt {x}\, a^{2} b d \,x^{2}-24 \sqrt {x}\, a \,b^{2} c \,x^{2}+40 \sqrt {x}\, a \,b^{2} d \,x^{3}-48 \sqrt {x}\, b^{3} c \,x^{3}\right )}{15 \sqrt {b x +a}\, a^{4} e \,x^{3}} \] Input:
int((d*x+c)/(e*x)^(1/2)/(b*x^3+a*x^2)^(3/2),x)
Output:
(2*sqrt(e)*( - 40*sqrt(b)*sqrt(a + b*x)*a*b*d*x**3 + 48*sqrt(b)*sqrt(a + b *x)*b**2*c*x**3 - 3*sqrt(x)*a**3*c - 5*sqrt(x)*a**3*d*x + 6*sqrt(x)*a**2*b *c*x + 20*sqrt(x)*a**2*b*d*x**2 - 24*sqrt(x)*a*b**2*c*x**2 + 40*sqrt(x)*a* b**2*d*x**3 - 48*sqrt(x)*b**3*c*x**3))/(15*sqrt(a + b*x)*a**4*e*x**3)