\(\int \frac {(e x)^{3/2} (c+d x)}{\sqrt {a x^2+b x^3}} \, dx\) [334]

Optimal result

Integrand size = 28, antiderivative size = 131 \[ \int \frac {(e x)^{3/2} (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {(4 b c-3 a d) e^2 \sqrt {a x^2+b x^3}}{4 b^2 \sqrt {e x}}+\frac {d e \sqrt {e x} \sqrt {a x^2+b x^3}}{2 b}-\frac {a (4 b c-3 a d) e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a x^2+b x^3}}\right )}{4 b^{5/2}} \] Output:

1/4*(-3*a*d+4*b*c)*e^2*(b*x^3+a*x^2)^(1/2)/b^2/(e*x)^(1/2)+1/2*d*e*(e*x)^( 
1/2)*(b*x^3+a*x^2)^(1/2)/b-1/4*a*(-3*a*d+4*b*c)*e^(3/2)*arctanh(b^(1/2)*(e 
*x)^(3/2)/e^(3/2)/(b*x^3+a*x^2)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.85 \[ \int \frac {(e x)^{3/2} (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {(e x)^{3/2} \left (\sqrt {b} (a+b x) (4 b c-3 a d+2 b d x)+\frac {2 a (-4 b c+3 a d) \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{\sqrt {x}}\right )}{4 b^{5/2} \sqrt {x^2 (a+b x)}} \] Input:

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

Output:

((e*x)^(3/2)*(Sqrt[b]*(a + b*x)*(4*b*c - 3*a*d + 2*b*d*x) + (2*a*(-4*b*c + 
 3*a*d)*Sqrt[a + b*x]*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x]) 
])/Sqrt[x]))/(4*b^(5/2)*Sqrt[x^2*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1945, 1930, 1937, 1935, 219}

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.

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

\(\Big \downarrow \) 1945

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1937

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

\(\Big \downarrow \) 1935

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(d*e*Sqrt[e*x]*Sqrt[a*x^2 + b*x^3])/(2*b) + ((4*b*c - 3*a*d)*((e^2*Sqrt[a* 
x^2 + b*x^3])/(b*Sqrt[e*x]) - (a*e*Sqrt[e*x]*ArcTanh[(Sqrt[b]*x^(3/2))/Sqr 
t[a*x^2 + b*x^3]])/(b^(3/2)*Sqrt[x])))/(4*b)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1930

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p 
+ 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1)))   I 
nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, 
x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
 

rule 1935

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp 
[-2/(n - j)   Subst[Int[1/(1 - a*x^2), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], 
 x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]
 

rule 1937

Int[((c_)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] 
:> Simp[c^IntPart[m]*((c*x)^FracPart[m]/x^FracPart[m])   Int[x^m*(a*x^j + b 
*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && IntegerQ[p + 1/2] && 
NeQ[n, j] && EqQ[Simplify[m + j*p + 1], 0]
 

rule 1945

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* 
p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1))   Int[(e* 
x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, 
x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p 
*(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {\left (-2 b d x +3 a d -4 b c \right ) x^{2} \left (b x +a \right ) e^{2}}{4 b^{2} \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}+\frac {a \left (3 a d -4 b c \right ) \ln \left (\frac {\frac {1}{2} a e +b e x}{\sqrt {b e}}+\sqrt {b e \,x^{2}+a e x}\right ) e^{2} x \sqrt {e x \left (b x +a \right )}}{8 b^{2} \sqrt {b e}\, \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}\)	\(128\)
default	\(-\frac {x \left (b x +a \right ) e \sqrt {e x}\, \left (-4 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, b d x -3 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a^{2} d e +4 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a b c e +6 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a d -8 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, b c \right )}{8 \sqrt {b \,x^{3}+a \,x^{2}}\, \sqrt {e x \left (b x +a \right )}\, b^{2} \sqrt {b e}}\)	\(186\)

Input:

int((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4*(-2*b*d*x+3*a*d-4*b*c)*x^2*(b*x+a)/b^2*e^2/(x^2*(b*x+a))^(1/2)/(e*x)^ 
(1/2)+1/8*a*(3*a*d-4*b*c)/b^2*ln((1/2*a*e+b*e*x)/(b*e)^(1/2)+(b*e*x^2+a*e* 
x)^(1/2))/(b*e)^(1/2)*e^2/(x^2*(b*x+a))^(1/2)*x*(e*x*(b*x+a))^(1/2)/(e*x)^ 
(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.73 \[ \int \frac {(e x)^{3/2} (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\left [-\frac {{\left (4 \, a b c - 3 \, a^{2} d\right )} e x \sqrt {\frac {e}{b}} \log \left (\frac {2 \, b e x^{2} + a e x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {e x} b \sqrt {\frac {e}{b}}}{x}\right ) - 2 \, {\left (2 \, b d e x + {\left (4 \, b c - 3 \, a d\right )} e\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{8 \, b^{2} x}, \frac {{\left (4 \, a b c - 3 \, a^{2} d\right )} e x \sqrt {-\frac {e}{b}} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {e x} b \sqrt {-\frac {e}{b}}}{b e x^{2} + a e x}\right ) + {\left (2 \, b d e x + {\left (4 \, b c - 3 \, a d\right )} e\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{4 \, b^{2} x}\right ] \] Input:

integrate((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/8*((4*a*b*c - 3*a^2*d)*e*x*sqrt(e/b)*log((2*b*e*x^2 + a*e*x + 2*sqrt(b 
*x^3 + a*x^2)*sqrt(e*x)*b*sqrt(e/b))/x) - 2*(2*b*d*e*x + (4*b*c - 3*a*d)*e 
)*sqrt(b*x^3 + a*x^2)*sqrt(e*x))/(b^2*x), 1/4*((4*a*b*c - 3*a^2*d)*e*x*sqr 
t(-e/b)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(e*x)*b*sqrt(-e/b)/(b*e*x^2 + a*e*x 
)) + (2*b*d*e*x + (4*b*c - 3*a*d)*e)*sqrt(b*x^3 + a*x^2)*sqrt(e*x))/(b^2*x 
)]
 

Sympy [F]

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

integrate((e*x)**(3/2)*(d*x+c)/(b*x**3+a*x**2)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.31 \[ \int \frac {(e x)^{3/2} (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {1}{8} \, e {\left (\frac {2 \, {\left (\sqrt {b e^{2} x + a e^{2}} \sqrt {e x} {\left (\frac {2 \, d x}{b e^{2}} + \frac {4 \, b^{2} c e^{3} - 3 \, a b d e^{3}}{b^{3} e^{5}}\right )} + \frac {{\left (4 \, a b c - 3 \, a^{2} d\right )} \log \left ({\left | -\sqrt {b e} \sqrt {e x} + \sqrt {b e^{2} x + a e^{2}} \right |}\right )}{\sqrt {b e} b^{2}}\right )} e^{2}}{{\left | e \right |} \mathrm {sgn}\left (x\right )} - \frac {{\left (4 \, a b c e^{2} \log \left (e^{2} {\left | a \right |}\right ) - 3 \, a^{2} d e^{2} \log \left (e^{2} {\left | a \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{\sqrt {b e} b^{2} {\left | e \right |}}\right )} \] Input:

integrate((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")
 

Output:

1/8*e*(2*(sqrt(b*e^2*x + a*e^2)*sqrt(e*x)*(2*d*x/(b*e^2) + (4*b^2*c*e^3 - 
3*a*b*d*e^3)/(b^3*e^5)) + (4*a*b*c - 3*a^2*d)*log(abs(-sqrt(b*e)*sqrt(e*x) 
 + sqrt(b*e^2*x + a*e^2)))/(sqrt(b*e)*b^2))*e^2/(abs(e)*sgn(x)) - (4*a*b*c 
*e^2*log(e^2*abs(a)) - 3*a^2*d*e^2*log(e^2*abs(a)))*sgn(x)/(sqrt(b*e)*b^2* 
abs(e)))
 

Mupad [F(-1)]

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

int(((e*x)^(3/2)*(c + d*x))/(a*x^2 + b*x^3)^(1/2),x)
 

Output:

int(((e*x)^(3/2)*(c + d*x))/(a*x^2 + b*x^3)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.78 \[ \int \frac {(e x)^{3/2} (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {\sqrt {e}\, e \left (-3 \sqrt {x}\, \sqrt {b x +a}\, a b d +4 \sqrt {x}\, \sqrt {b x +a}\, b^{2} c +2 \sqrt {x}\, \sqrt {b x +a}\, b^{2} d x +3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2} d -4 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a b c \right )}{4 b^{3}} \] Input:

int((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(1/2),x)
 

Output:

(sqrt(e)*e*( - 3*sqrt(x)*sqrt(a + b*x)*a*b*d + 4*sqrt(x)*sqrt(a + b*x)*b** 
2*c + 2*sqrt(x)*sqrt(a + b*x)*b**2*d*x + 3*sqrt(b)*log((sqrt(a + b*x) + sq 
rt(x)*sqrt(b))/sqrt(a))*a**2*d - 4*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sq 
rt(b))/sqrt(a))*a*b*c))/(4*b**3)