\(\int \frac {(c+d x) \sqrt {a x^2+b x^3}}{\sqrt {e x}} \, dx\) [303]

Optimal result

Integrand size = 28, antiderivative size = 172 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{\sqrt {e x}} \, dx=\frac {a (2 b c-a d) \sqrt {a x^2+b x^3}}{8 b^2 \sqrt {e x}}+\frac {(2 b c-a d) \sqrt {e x} \sqrt {a x^2+b x^3}}{4 b e}+\frac {d e \left (a x^2+b x^3\right )^{3/2}}{3 b (e x)^{3/2}}-\frac {a^2 (2 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a x^2+b x^3}}\right )}{8 b^{5/2} \sqrt {e}} \] Output:

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{\sqrt {e x}} \, dx=\frac {\sqrt {x^2 (a+b x)} \left (6 a b c \sqrt {x}-3 a^2 d \sqrt {x}+12 b^2 c x^{3/2}+2 a b d x^{3/2}+8 b^2 d x^{5/2}\right )}{24 b^2 \sqrt {x} \sqrt {e x}}+\frac {a^2 (-2 b c+a d) \sqrt {x^2 (a+b x)} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{4 b^{5/2} \sqrt {x} \sqrt {e x} \sqrt {a+b x}} \] Input:

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

Output:

(Sqrt[x^2*(a + b*x)]*(6*a*b*c*Sqrt[x] - 3*a^2*d*Sqrt[x] + 12*b^2*c*x^(3/2) 
 + 2*a*b*d*x^(3/2) + 8*b^2*d*x^(5/2)))/(24*b^2*Sqrt[x]*Sqrt[e*x]) + (a^2*( 
-2*b*c + a*d)*Sqrt[x^2*(a + b*x)]*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sq 
rt[a + b*x])])/(4*b^(5/2)*Sqrt[x]*Sqrt[e*x]*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1945, 1927, 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.

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

\(\Big \downarrow \) 1945

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

\(\Big \downarrow \) 1927

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1937

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

\(\Big \downarrow \) 1935

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(d*e*(a*x^2 + b*x^3)^(3/2))/(3*b*(e*x)^(3/2)) + ((2*b*c - a*d)*((Sqrt[e*x] 
*Sqrt[a*x^2 + b*x^3])/(2*e) + (a*((e^2*Sqrt[a*x^2 + b*x^3])/(b*Sqrt[e*x]) 
- (a*e*Sqrt[e*x]*ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a*x^2 + b*x^3]])/(b^(3/2)* 
Sqrt[x])))/(4*e^2)))/(2*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 1927

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 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 = 0.57 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.84

method	result	size
risch	\(-\frac {\left (-8 b^{2} d \,x^{2}-2 a b d x -12 b^{2} c x +3 a^{2} d -6 a b c \right ) \sqrt {x^{2} \left (b x +a \right )}}{24 b^{2} \sqrt {e x}}+\frac {a^{2} \left (a d -2 b c \right ) \ln \left (\frac {\frac {1}{2} a e +b e x}{\sqrt {b e}}+\sqrt {b e \,x^{2}+a e x}\right ) \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x \left (b x +a \right )}}{16 b^{2} \sqrt {b e}\, x \left (b x +a \right ) \sqrt {e x}}\)	\(144\)
default	\(-\frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (-16 b^{2} d \,x^{2} \sqrt {b e}\, \sqrt {e x \left (b x +a \right )}-3 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a^{3} d e +6 \ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a^{2} b c e -4 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a b d x -24 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, b^{2} c x +6 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a^{2} d -12 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, a b c \right )}{48 \sqrt {e x}\, \sqrt {e x \left (b x +a \right )}\, b^{2} \sqrt {b e}}\)	\(231\)

Input:

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

Output:

-1/24*(-8*b^2*d*x^2-2*a*b*d*x-12*b^2*c*x+3*a^2*d-6*a*b*c)/b^2*(x^2*(b*x+a) 
)^(1/2)/(e*x)^(1/2)+1/16*a^2*(a*d-2*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)*(x^2*(b*x+a))^(1/2)/x/(b*x+a)*(e*x*(b 
*x+a))^(1/2)/(e*x)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.57 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{\sqrt {e x}} \, dx=\left [-\frac {3 \, {\left (2 \, a^{2} b c - a^{3} d\right )} \sqrt {b e} x \log \left (\frac {2 \, b e x^{2} + a e x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {b e} \sqrt {e x}}{x}\right ) - 2 \, {\left (8 \, b^{3} d x^{2} + 6 \, a b^{2} c - 3 \, a^{2} b d + 2 \, {\left (6 \, b^{3} c + a b^{2} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{48 \, b^{3} e x}, \frac {3 \, {\left (2 \, a^{2} b c - a^{3} d\right )} \sqrt {-b e} x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-b e} \sqrt {e x}}{b e x^{2} + a e x}\right ) + {\left (8 \, b^{3} d x^{2} + 6 \, a b^{2} c - 3 \, a^{2} b d + 2 \, {\left (6 \, b^{3} c + a b^{2} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {e x}}{24 \, b^{3} e x}\right ] \] Input:

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

Output:

[-1/48*(3*(2*a^2*b*c - a^3*d)*sqrt(b*e)*x*log((2*b*e*x^2 + a*e*x + 2*sqrt( 
b*x^3 + a*x^2)*sqrt(b*e)*sqrt(e*x))/x) - 2*(8*b^3*d*x^2 + 6*a*b^2*c - 3*a^ 
2*b*d + 2*(6*b^3*c + a*b^2*d)*x)*sqrt(b*x^3 + a*x^2)*sqrt(e*x))/(b^3*e*x), 
 1/24*(3*(2*a^2*b*c - a^3*d)*sqrt(-b*e)*x*arctan(sqrt(b*x^3 + a*x^2)*sqrt( 
-b*e)*sqrt(e*x)/(b*e*x^2 + a*e*x)) + (8*b^3*d*x^2 + 6*a*b^2*c - 3*a^2*b*d 
+ 2*(6*b^3*c + a*b^2*d)*x)*sqrt(b*x^3 + a*x^2)*sqrt(e*x))/(b^3*e*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.37 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{\sqrt {e x}} \, dx=\frac {{\left (\sqrt {{\left (b x + a\right )} b e - a b e} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )} d \mathrm {sgn}\left (x\right )}{b^{3} e} + \frac {6 \, b^{7} c e^{2} \mathrm {sgn}\left (x\right ) - 7 \, a b^{6} d e^{2} \mathrm {sgn}\left (x\right )}{b^{9} e^{3}}\right )} - \frac {3 \, {\left (2 \, a b^{7} c e^{2} \mathrm {sgn}\left (x\right ) - a^{2} b^{6} d e^{2} \mathrm {sgn}\left (x\right )\right )}}{b^{9} e^{3}}\right )} + \frac {3 \, {\left (2 \, a^{2} b c \mathrm {sgn}\left (x\right ) - a^{3} d \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {{\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} b^{2}}\right )} b}{24 \, {\left | b \right |}} - \frac {{\left (2 \, a^{2} b c \log \left (\sqrt {b e} \sqrt {a}\right ) - a^{3} d \log \left (\sqrt {b e} \sqrt {a}\right )\right )} \mathrm {sgn}\left (x\right )}{8 \, \sqrt {b e} b {\left | b \right |}} \] Input:

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

Output:

1/24*(sqrt((b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)* 
d*sgn(x)/(b^3*e) + (6*b^7*c*e^2*sgn(x) - 7*a*b^6*d*e^2*sgn(x))/(b^9*e^3)) 
- 3*(2*a*b^7*c*e^2*sgn(x) - a^2*b^6*d*e^2*sgn(x))/(b^9*e^3)) + 3*(2*a^2*b* 
c*sgn(x) - a^3*d*sgn(x))*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt((b*x + a) 
*b*e - a*b*e)))/(sqrt(b*e)*b^2))*b/abs(b) - 1/8*(2*a^2*b*c*log(sqrt(b*e)*s 
qrt(a)) - a^3*d*log(sqrt(b*e)*sqrt(a)))*sgn(x)/(sqrt(b*e)*b*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.83 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{\sqrt {e x}} \, dx=\frac {\sqrt {e}\, \left (-3 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b d +6 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} c +2 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} d x +12 \sqrt {x}\, \sqrt {b x +a}\, b^{3} c x +8 \sqrt {x}\, \sqrt {b x +a}\, b^{3} d \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3} d -6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2} b c \right )}{24 b^{3} e} \] Input:

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

Output:

(sqrt(e)*( - 3*sqrt(x)*sqrt(a + b*x)*a**2*b*d + 6*sqrt(x)*sqrt(a + b*x)*a* 
b**2*c + 2*sqrt(x)*sqrt(a + b*x)*a*b**2*d*x + 12*sqrt(x)*sqrt(a + b*x)*b** 
3*c*x + 8*sqrt(x)*sqrt(a + b*x)*b**3*d*x**2 + 3*sqrt(b)*log((sqrt(a + b*x) 
 + sqrt(x)*sqrt(b))/sqrt(a))*a**3*d - 6*sqrt(b)*log((sqrt(a + b*x) + sqrt( 
x)*sqrt(b))/sqrt(a))*a**2*b*c))/(24*b**3*e)