\(\int \frac {(e x)^{3/2} (c+d x)}{(a x^2+b x^3)^{5/2}} \, dx\) [353]

Optimal result

Integrand size = 28, antiderivative size = 194 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2 c e \sqrt {e x}}{5 a \left (a x^2+b x^3\right )^{3/2}}-\frac {2 (8 b c-5 a d) (e x)^{3/2}}{15 a^2 \left (a x^2+b x^3\right )^{3/2}}-\frac {4 (8 b c-5 a d) e^2}{5 a^3 \sqrt {e x} \sqrt {a x^2+b x^3}}+\frac {16 (8 b c-5 a d) e^4 \sqrt {a x^2+b x^3}}{15 a^4 (e x)^{5/2}}-\frac {32 b (8 b c-5 a d) e^3 \sqrt {a x^2+b x^3}}{15 a^5 (e x)^{3/2}} \] Output:

-2/5*c*e*(e*x)^(1/2)/a/(b*x^3+a*x^2)^(3/2)-2/15*(-5*a*d+8*b*c)*(e*x)^(3/2) 
/a^2/(b*x^3+a*x^2)^(3/2)-4/5*(-5*a*d+8*b*c)*e^2/a^3/(e*x)^(1/2)/(b*x^3+a*x 
^2)^(1/2)+16/15*(-5*a*d+8*b*c)*e^4*(b*x^3+a*x^2)^(1/2)/a^4/(e*x)^(5/2)-32/ 
15*b*(-5*a*d+8*b*c)*e^3*(b*x^3+a*x^2)^(1/2)/a^5/(e*x)^(3/2)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.53 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 e \sqrt {e x} \left (-128 b^4 c x^4+16 a b^3 x^3 (-12 c+5 d x)+24 a^2 b^2 x^2 (-2 c+5 d x)-a^4 (3 c+5 d x)+2 a^3 b x (4 c+15 d x)\right )}{15 a^5 \left (x^2 (a+b x)\right )^{3/2}} \] Input:

Integrate[((e*x)^(3/2)*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x]
 

Output:

(2*e*Sqrt[e*x]*(-128*b^4*c*x^4 + 16*a*b^3*x^3*(-12*c + 5*d*x) + 24*a^2*b^2 
*x^2*(-2*c + 5*d*x) - a^4*(3*c + 5*d*x) + 2*a^3*b*x*(4*c + 15*d*x)))/(15*a 
^5*(x^2*(a + b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1944, 1921, 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.

\(\displaystyle \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1944

\(\displaystyle -\frac {(8 b c-5 a d) \int \frac {(e x)^{5/2}}{\left (b x^3+a x^2\right )^{5/2}}dx}{5 a e}-\frac {2 c e \sqrt {e x}}{5 a \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1921

\(\displaystyle -\frac {(8 b c-5 a d) \left (\frac {2 e^2 \int \frac {\sqrt {e x}}{\left (b x^3+a x^2\right )^{3/2}}dx}{a}+\frac {2 e (e x)^{3/2}}{3 a \left (a x^2+b x^3\right )^{3/2}}\right )}{5 a e}-\frac {2 c e \sqrt {e x}}{5 a \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1921

\(\displaystyle -\frac {(8 b c-5 a d) \left (\frac {2 e^2 \left (\frac {4 e^2 \int \frac {1}{(e x)^{3/2} \sqrt {b x^3+a x^2}}dx}{a}+\frac {2 e}{a \sqrt {e x} \sqrt {a x^2+b x^3}}\right )}{a}+\frac {2 e (e x)^{3/2}}{3 a \left (a x^2+b x^3\right )^{3/2}}\right )}{5 a e}-\frac {2 c e \sqrt {e x}}{5 a \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1922

\(\displaystyle -\frac {(8 b c-5 a d) \left (\frac {2 e^2 \left (\frac {4 e^2 \left (-\frac {2 b \int \frac {1}{\sqrt {e x} \sqrt {b x^3+a x^2}}dx}{3 a e}-\frac {2 e \sqrt {a x^2+b x^3}}{3 a (e x)^{5/2}}\right )}{a}+\frac {2 e}{a \sqrt {e x} \sqrt {a x^2+b x^3}}\right )}{a}+\frac {2 e (e x)^{3/2}}{3 a \left (a x^2+b x^3\right )^{3/2}}\right )}{5 a e}-\frac {2 c e \sqrt {e x}}{5 a \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1920

\(\displaystyle -\frac {(8 b c-5 a d) \left (\frac {2 e^2 \left (\frac {4 e^2 \left (\frac {4 b \sqrt {a x^2+b x^3}}{3 a^2 (e x)^{3/2}}-\frac {2 e \sqrt {a x^2+b x^3}}{3 a (e x)^{5/2}}\right )}{a}+\frac {2 e}{a \sqrt {e x} \sqrt {a x^2+b x^3}}\right )}{a}+\frac {2 e (e x)^{3/2}}{3 a \left (a x^2+b x^3\right )^{3/2}}\right )}{5 a e}-\frac {2 c e \sqrt {e x}}{5 a \left (a x^2+b x^3\right )^{3/2}}\)

Input:

Int[((e*x)^(3/2)*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x]
 

Output:

(-2*c*e*Sqrt[e*x])/(5*a*(a*x^2 + b*x^3)^(3/2)) - ((8*b*c - 5*a*d)*((2*e*(e 
*x)^(3/2))/(3*a*(a*x^2 + b*x^3)^(3/2)) + (2*e^2*((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))/a))/(5*a*e)
 

Defintions of rubi rules used

rule 1920

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])
 

rule 1921

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])
 

rule 1922

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])
 

rule 1944

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]
 

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.59

method	result	size
gosper	\(-\frac {2 x \left (b x +a \right ) \left (-80 x^{4} a \,b^{3} d +128 x^{4} b^{4} c -120 a^{2} b^{2} d \,x^{3}+192 a \,b^{3} c \,x^{3}-30 a^{3} b d \,x^{2}+48 a^{2} b^{2} c \,x^{2}+5 a^{4} d x -8 a^{3} b c x +3 c \,a^{4}\right ) \left (e x \right )^{\frac {3}{2}}}{15 a^{5} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\)	\(115\)
orering	\(-\frac {2 x \left (b x +a \right ) \left (-80 x^{4} a \,b^{3} d +128 x^{4} b^{4} c -120 a^{2} b^{2} d \,x^{3}+192 a \,b^{3} c \,x^{3}-30 a^{3} b d \,x^{2}+48 a^{2} b^{2} c \,x^{2}+5 a^{4} d x -8 a^{3} b c x +3 c \,a^{4}\right ) \left (e x \right )^{\frac {3}{2}}}{15 a^{5} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}}}\)	\(115\)
default	\(-\frac {2 x^{2} \left (b x +a \right ) \left (-80 x^{4} a \,b^{3} d +128 x^{4} b^{4} c -120 a^{2} b^{2} d \,x^{3}+192 a \,b^{3} c \,x^{3}-30 a^{3} b d \,x^{2}+48 a^{2} b^{2} c \,x^{2}+5 a^{4} d x -8 a^{3} b c x +3 c \,a^{4}\right ) \sqrt {e x}\, e}{15 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} a^{5}}\)	\(118\)
risch	\(-\frac {2 \left (b x +a \right ) \left (-40 a b d \,x^{2}+73 b^{2} c \,x^{2}+5 a^{2} d x -14 a b c x +3 a^{2} c \right ) e^{2}}{15 a^{5} x \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}+\frac {2 b^{2} \left (8 a b d x -11 b^{2} c x +9 a^{2} d -12 a b c \right ) x^{2} e^{2}}{3 \left (b x +a \right ) a^{5} \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}\)	\(133\)

Input:

int((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/15*x*(b*x+a)*(-80*a*b^3*d*x^4+128*b^4*c*x^4-120*a^2*b^2*d*x^3+192*a*b^3 
*c*x^3-30*a^3*b*d*x^2+48*a^2*b^2*c*x^2+5*a^4*d*x-8*a^3*b*c*x+3*a^4*c)*(e*x 
)^(3/2)/a^5/(b*x^3+a*x^2)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.73 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, a^{4} c e + 16 \, {\left (8 \, b^{4} c - 5 \, a b^{3} d\right )} e x^{4} + 24 \, {\left (8 \, a b^{3} c - 5 \, a^{2} b^{2} d\right )} e x^{3} + 6 \, {\left (8 \, a^{2} b^{2} c - 5 \, a^{3} b d\right )} e x^{2} - {\left (8 \, a^{3} b c - 5 \, a^{4} d\right )} e x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{15 \, {\left (a^{5} b^{2} x^{6} + 2 \, a^{6} b x^{5} + a^{7} x^{4}\right )}} \] Input:

integrate((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="fricas")
 

Output:

-2/15*(3*a^4*c*e + 16*(8*b^4*c - 5*a*b^3*d)*e*x^4 + 24*(8*a*b^3*c - 5*a^2* 
b^2*d)*e*x^3 + 6*(8*a^2*b^2*c - 5*a^3*b*d)*e*x^2 - (8*a^3*b*c - 5*a^4*d)*e 
*x)*sqrt(b*x^3 + a*x^2)*sqrt(e*x)/(a^5*b^2*x^6 + 2*a^6*b*x^5 + a^7*x^4)
 

Sympy [F]

\[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int \frac {\left (e x\right )^{\frac {3}{2}} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((e*x)**(3/2)*(d*x+c)/(b*x**3+a*x**2)**(5/2),x)
 

Output:

Integral((e*x)**(3/2)*(c + d*x)/(x**2*(a + b*x))**(5/2), x)
 

Maxima [F]

\[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate((d*x + c)*(e*x)^(3/2)/(b*x^3 + a*x^2)^(5/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(5/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 9.50 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.78 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {16\,e\,x^3\,\sqrt {e\,x}\,\left (5\,a\,d-8\,b\,c\right )}{5\,a^4}-\frac {2\,c\,e\,\sqrt {e\,x}}{5\,a\,b^2}-\frac {2\,e\,x\,\sqrt {e\,x}\,\left (5\,a\,d-8\,b\,c\right )}{15\,a^2\,b^2}+\frac {32\,b\,e\,x^4\,\sqrt {e\,x}\,\left (5\,a\,d-8\,b\,c\right )}{15\,a^5}+\frac {4\,e\,x^2\,\sqrt {e\,x}\,\left (5\,a\,d-8\,b\,c\right )}{5\,a^3\,b}\right )}{x^6+\frac {2\,a\,x^5}{b}+\frac {a^2\,x^4}{b^2}} \] Input:

int(((e*x)^(3/2)*(c + d*x))/(a*x^2 + b*x^3)^(5/2),x)
 

Output:

((a*x^2 + b*x^3)^(1/2)*((16*e*x^3*(e*x)^(1/2)*(5*a*d - 8*b*c))/(5*a^4) - ( 
2*c*e*(e*x)^(1/2))/(5*a*b^2) - (2*e*x*(e*x)^(1/2)*(5*a*d - 8*b*c))/(15*a^2 
*b^2) + (32*b*e*x^4*(e*x)^(1/2)*(5*a*d - 8*b*c))/(15*a^5) + (4*e*x^2*(e*x) 
^(1/2)*(5*a*d - 8*b*c))/(5*a^3*b)))/(x^6 + (2*a*x^5)/b + (a^2*x^4)/b^2)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.03 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 \sqrt {e}\, e \left (-80 \sqrt {b}\, \sqrt {b x +a}\, a^{2} b d \,x^{3}+128 \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} c \,x^{3}-80 \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} d \,x^{4}+128 \sqrt {b}\, \sqrt {b x +a}\, b^{3} c \,x^{4}-3 \sqrt {x}\, a^{4} c -5 \sqrt {x}\, a^{4} d x +8 \sqrt {x}\, a^{3} b c x +30 \sqrt {x}\, a^{3} b d \,x^{2}-48 \sqrt {x}\, a^{2} b^{2} c \,x^{2}+120 \sqrt {x}\, a^{2} b^{2} d \,x^{3}-192 \sqrt {x}\, a \,b^{3} c \,x^{3}+80 \sqrt {x}\, a \,b^{3} d \,x^{4}-128 \sqrt {x}\, b^{4} c \,x^{4}\right )}{15 \sqrt {b x +a}\, a^{5} x^{3} \left (b x +a \right )} \] Input:

int((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(5/2),x)
 

Output:

(2*sqrt(e)*e*( - 80*sqrt(b)*sqrt(a + b*x)*a**2*b*d*x**3 + 128*sqrt(b)*sqrt 
(a + b*x)*a*b**2*c*x**3 - 80*sqrt(b)*sqrt(a + b*x)*a*b**2*d*x**4 + 128*sqr 
t(b)*sqrt(a + b*x)*b**3*c*x**4 - 3*sqrt(x)*a**4*c - 5*sqrt(x)*a**4*d*x + 8 
*sqrt(x)*a**3*b*c*x + 30*sqrt(x)*a**3*b*d*x**2 - 48*sqrt(x)*a**2*b**2*c*x* 
*2 + 120*sqrt(x)*a**2*b**2*d*x**3 - 192*sqrt(x)*a*b**3*c*x**3 + 80*sqrt(x) 
*a*b**3*d*x**4 - 128*sqrt(x)*b**4*c*x**4))/(15*sqrt(a + b*x)*a**5*x**3*(a 
+ b*x))