\(\int \frac {(c+d x) (a x+b x^2)^{3/2}}{x} \, dx\) [59]

Optimal result

Integrand size = 22, antiderivative size = 159 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x} \, dx=\frac {a^2 (8 b c-3 a d) \sqrt {a x+b x^2}}{64 b^2}+\frac {7 a (8 b c-3 a d) x \sqrt {a x+b x^2}}{96 b}+\frac {1}{24} (8 b c-3 a d) x^2 \sqrt {a x+b x^2}+\frac {d \left (a x+b x^2\right )^{5/2}}{4 b x}-\frac {a^3 (8 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{64 b^{5/2}} \] Output:

1/64*a^2*(-3*a*d+8*b*c)*(b*x^2+a*x)^(1/2)/b^2+7/96*a*(-3*a*d+8*b*c)*x*(b*x 
^2+a*x)^(1/2)/b+1/24*(-3*a*d+8*b*c)*x^2*(b*x^2+a*x)^(1/2)+1/4*d*(b*x^2+a*x 
)^(5/2)/b/x-1/64*a^3*(-3*a*d+8*b*c)*arctanh(b^(1/2)*x/(b*x^2+a*x)^(1/2))/b 
^(5/2)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.14 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x} \, dx=\frac {\sqrt {x} \sqrt {a+b x} \left (\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (-9 a^3 d+6 a^2 b (4 c+d x)+16 b^3 x^2 (4 c+3 d x)+8 a b^2 x (14 c+9 d x)\right )+48 a^3 b c \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )+18 a^4 d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )\right )}{192 b^{5/2} \sqrt {x (a+b x)}} \] Input:

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

Output:

(Sqrt[x]*Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(-9*a^3*d + 6*a^2*b* 
(4*c + d*x) + 16*b^3*x^2*(4*c + 3*d*x) + 8*a*b^2*x*(14*c + 9*d*x)) + 48*a^ 
3*b*c*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x])] + 18*a^4*d*ArcT 
anh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])]))/(192*b^(5/2)*Sqrt[x*(a 
 + b*x)])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.78, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1221, 1131, 1087, 1091, 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.

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1131

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(d*(a*x + b*x^2)^(5/2))/(4*b*x) + ((8*b*c - 3*a*d)*((a*x + b*x^2)^(3/2)/3 
+ (a*(((a + 2*b*x)*Sqrt[a*x + b*x^2])/(4*b) - (a^2*ArcTanh[(Sqrt[b]*x)/Sqr 
t[a*x + b*x^2]])/(4*b^(3/2))))/2))/(8*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 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1131

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x 
] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1)))   Int[(d + e*x)^(m + 1)*(a + 
 b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b 
*d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne 
Q[m + 2*p + 1, 0] && IntegerQ[2*p]
 

rule 1221

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 
)/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c 
*f - b*g))/(c*e*(m + 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && NeQ[m + 2*p + 2, 0]
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.61

method	result	size
pseudoelliptic	\(\frac {\frac {3 a^{3} \left (a d -\frac {8 b c}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )}{64}-\frac {3 \sqrt {x \left (b x +a \right )}\, \left (-\frac {8 \left (\frac {d x}{4}+c \right ) a^{2} b^{\frac {3}{2}}}{3}-\frac {112 x \left (\frac {9 d x}{14}+c \right ) a \,b^{\frac {5}{2}}}{9}-\frac {64 x^{2} \left (\frac {3 d x}{4}+c \right ) b^{\frac {7}{2}}}{9}+\sqrt {b}\, a^{3} d \right )}{64}}{b^{\frac {5}{2}}}\)	\(97\)
risch	\(-\frac {\left (-48 b^{3} d \,x^{3}-72 a \,b^{2} d \,x^{2}-64 b^{3} c \,x^{2}-6 a^{2} b d x -112 a \,b^{2} c x +9 a^{3} d -24 c \,a^{2} b \right ) x \left (b x +a \right )}{192 b^{2} \sqrt {x \left (b x +a \right )}}+\frac {a^{3} \left (3 a d -8 b c \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{128 b^{\frac {5}{2}}}\)	\(121\)
default	\(d \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )+c \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{3}+\frac {a \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{2}\right )\)	\(164\)

Input:

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

Output:

3/64/b^(5/2)*(a^3*(a*d-8/3*b*c)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))-(x*(b 
*x+a))^(1/2)*(-8/3*(1/4*d*x+c)*a^2*b^(3/2)-112/9*x*(9/14*d*x+c)*a*b^(5/2)- 
64/9*x^2*(3/4*d*x+c)*b^(7/2)+b^(1/2)*a^3*d))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.61 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x} \, dx=\left [-\frac {3 \, {\left (8 \, a^{3} b c - 3 \, a^{4} d\right )} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (48 \, b^{4} d x^{3} + 24 \, a^{2} b^{2} c - 9 \, a^{3} b d + 8 \, {\left (8 \, b^{4} c + 9 \, a b^{3} d\right )} x^{2} + 2 \, {\left (56 \, a b^{3} c + 3 \, a^{2} b^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{384 \, b^{3}}, \frac {3 \, {\left (8 \, a^{3} b c - 3 \, a^{4} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) + {\left (48 \, b^{4} d x^{3} + 24 \, a^{2} b^{2} c - 9 \, a^{3} b d + 8 \, {\left (8 \, b^{4} c + 9 \, a b^{3} d\right )} x^{2} + 2 \, {\left (56 \, a b^{3} c + 3 \, a^{2} b^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{192 \, b^{3}}\right ] \] Input:

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

Output:

[-1/384*(3*(8*a^3*b*c - 3*a^4*d)*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a* 
x)*sqrt(b)) - 2*(48*b^4*d*x^3 + 24*a^2*b^2*c - 9*a^3*b*d + 8*(8*b^4*c + 9* 
a*b^3*d)*x^2 + 2*(56*a*b^3*c + 3*a^2*b^2*d)*x)*sqrt(b*x^2 + a*x))/b^3, 1/1 
92*(3*(8*a^3*b*c - 3*a^4*d)*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b* 
x + a)) + (48*b^4*d*x^3 + 24*a^2*b^2*c - 9*a^3*b*d + 8*(8*b^4*c + 9*a*b^3* 
d)*x^2 + 2*(56*a*b^3*c + 3*a^2*b^2*d)*x)*sqrt(b*x^2 + a*x))/b^3]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 2.04 (sec) , antiderivative size = 490, normalized size of antiderivative = 3.08 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x} \, dx =\text {Too large to display} \] Input:

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

Output:

a*c*Piecewise((-a**2*Piecewise((log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b 
*x)/sqrt(b), Ne(a**2/b, 0)), ((a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2* 
b) + x)**2), True))/(8*b) + (a/(4*b) + x/2)*sqrt(a*x + b*x**2), Ne(b, 0)), 
 (2*(a*x)**(3/2)/(3*a), Ne(a, 0)), (0, True)) + a*d*Piecewise((a**3*Piecew 
ise((log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b*x)/sqrt(b), Ne(a**2/b, 0)) 
, ((a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/(16*b** 
2) + sqrt(a*x + b*x**2)*(-a**2/(8*b**2) + a*x/(12*b) + x**2/3), Ne(b, 0)), 
 (2*(a*x)**(5/2)/(5*a**2), Ne(a, 0)), (0, True)) + b*c*Piecewise((a**3*Pie 
cewise((log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b*x)/sqrt(b), Ne(a**2/b, 
0)), ((a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/(16* 
b**2) + sqrt(a*x + b*x**2)*(-a**2/(8*b**2) + a*x/(12*b) + x**2/3), Ne(b, 0 
)), (2*(a*x)**(5/2)/(5*a**2), Ne(a, 0)), (0, True)) + b*d*Piecewise((-5*a* 
*4*Piecewise((log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b*x)/sqrt(b), Ne(a* 
*2/b, 0)), ((a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True) 
)/(128*b**3) + sqrt(a*x + b*x**2)*(5*a**3/(64*b**3) - 5*a**2*x/(96*b**2) + 
 a*x**2/(24*b) + x**3/4), Ne(b, 0)), (2*(a*x)**(7/2)/(7*a**3), Ne(a, 0)), 
(0, True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.19 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x} \, dx=\frac {1}{4} \, \sqrt {b x^{2} + a x} a c x + \frac {1}{4} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} d x - \frac {3 \, \sqrt {b x^{2} + a x} a^{2} d x}{32 \, b} - \frac {a^{3} c \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {3}{2}}} + \frac {3 \, a^{4} d \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {5}{2}}} + \frac {1}{3} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} c + \frac {\sqrt {b x^{2} + a x} a^{2} c}{8 \, b} - \frac {3 \, \sqrt {b x^{2} + a x} a^{3} d}{64 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} a d}{8 \, b} \] Input:

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

Output:

1/4*sqrt(b*x^2 + a*x)*a*c*x + 1/4*(b*x^2 + a*x)^(3/2)*d*x - 3/32*sqrt(b*x^ 
2 + a*x)*a^2*d*x/b - 1/16*a^3*c*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b 
))/b^(3/2) + 3/128*a^4*d*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5 
/2) + 1/3*(b*x^2 + a*x)^(3/2)*c + 1/8*sqrt(b*x^2 + a*x)*a^2*c/b - 3/64*sqr 
t(b*x^2 + a*x)*a^3*d/b^2 + 1/8*(b*x^2 + a*x)^(3/2)*a*d/b
 

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x} \, dx=\frac {1}{192} \, \sqrt {b x^{2} + a x} {\left (2 \, {\left (4 \, {\left (6 \, b d x + \frac {8 \, b^{4} c + 9 \, a b^{3} d}{b^{3}}\right )} x + \frac {56 \, a b^{3} c + 3 \, a^{2} b^{2} d}{b^{3}}\right )} x + \frac {3 \, {\left (8 \, a^{2} b^{2} c - 3 \, a^{3} b d\right )}}{b^{3}}\right )} + \frac {{\left (8 \, a^{3} b c - 3 \, a^{4} d\right )} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{128 \, b^{\frac {5}{2}}} \] Input:

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

Output:

1/192*sqrt(b*x^2 + a*x)*(2*(4*(6*b*d*x + (8*b^4*c + 9*a*b^3*d)/b^3)*x + (5 
6*a*b^3*c + 3*a^2*b^2*d)/b^3)*x + 3*(8*a^2*b^2*c - 3*a^3*b*d)/b^3) + 1/128 
*(8*a^3*b*c - 3*a^4*d)*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + 
 a))/b^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.11 \[ \int \frac {(c+d x) \left (a x+b x^2\right )^{3/2}}{x} \, dx=\frac {-9 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b d +24 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} c +6 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} d x +112 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} c x +72 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} d \,x^{2}+64 \sqrt {x}\, \sqrt {b x +a}\, b^{4} c \,x^{2}+48 \sqrt {x}\, \sqrt {b x +a}\, b^{4} d \,x^{3}+9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{4} d -24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3} b c}{192 b^{3}} \] Input:

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

Output:

( - 9*sqrt(x)*sqrt(a + b*x)*a**3*b*d + 24*sqrt(x)*sqrt(a + b*x)*a**2*b**2* 
c + 6*sqrt(x)*sqrt(a + b*x)*a**2*b**2*d*x + 112*sqrt(x)*sqrt(a + b*x)*a*b* 
*3*c*x + 72*sqrt(x)*sqrt(a + b*x)*a*b**3*d*x**2 + 64*sqrt(x)*sqrt(a + b*x) 
*b**4*c*x**2 + 48*sqrt(x)*sqrt(a + b*x)*b**4*d*x**3 + 9*sqrt(b)*log((sqrt( 
a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**4*d - 24*sqrt(b)*log((sqrt(a + b*x 
) + sqrt(x)*sqrt(b))/sqrt(a))*a**3*b*c)/(192*b**3)