\(\int \frac {x (c+d x)^3}{(a+b x^2)^{3/2}} \, dx\) [1227]

Optimal result

Integrand size = 20, antiderivative size = 94 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {(c+d x)^3}{b \sqrt {a+b x^2}}+\frac {3 d^2 (4 c+d x) \sqrt {a+b x^2}}{2 b^2}+\frac {3 d \left (2 b c^2-a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.17 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-2 b c^3+12 a c d^2-6 b c^2 d x+3 a d^3 x+6 b c d^2 x^2+b d^3 x^3}{2 b^2 \sqrt {a+b x^2}}-\frac {3 \left (2 b c^2 d-a d^3\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{5/2}} \] Input:

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

Output:

(-2*b*c^3 + 12*a*c*d^2 - 6*b*c^2*d*x + 3*a*d^3*x + 6*b*c*d^2*x^2 + b*d^3*x 
^3)/(2*b^2*Sqrt[a + b*x^2]) - (3*(2*b*c^2*d - a*d^3)*Log[-(Sqrt[b]*x) + Sq 
rt[a + b*x^2]])/(2*b^(5/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {531, 27, 497, 455, 224, 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.

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

\(\Big \downarrow \) 531

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 497

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 497

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[1/(b 
*(n + 2*p + 1))   Int[(c + d*x)^(n - 2)*(a + b*x^2)^p*Simp[b*c^2*(n + 2*p + 
 1) - a*d^2*(n - 1) + 2*b*c*d*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, n 
, p}, x] && If[RationalQ[n], GtQ[n, 1], SumSimplerQ[n, -2]] && NeQ[n + 2*p 
+ 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
 

rule 531

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi 
alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a 
 + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 
2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b 
*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p 
 + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt 
Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.32

method	result	size
risch	\(\frac {d^{2} \left (d x +6 c \right ) \sqrt {b \,x^{2}+a}}{2 b^{2}}-\frac {\frac {a \,d^{3} x}{\sqrt {b \,x^{2}+a}}+3 b d \left (a \,d^{2}-2 b \,c^{2}\right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {2 c \left (3 a \,d^{2}-b \,c^{2}\right )}{\sqrt {b \,x^{2}+a}}}{2 b^{2}}\)	\(124\)
default	\(-\frac {c^{3}}{b \sqrt {b \,x^{2}+a}}+d^{3} \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+3 c \,d^{2} \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )+3 c^{2} d \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )\)	\(163\)

Input:

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

Output:

1/2*d^2*(d*x+6*c)*(b*x^2+a)^(1/2)/b^2-1/2/b^2*(a*d^3*x/(b*x^2+a)^(1/2)+3*b 
*d*(a*d^2-2*b*c^2)*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^ 
(1/2)))-2*c*(3*a*d^2-b*c^2)/(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.28 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (2 \, a b c^{2} d - a^{2} d^{3} + {\left (2 \, b^{2} c^{2} d - a b d^{3}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (b^{2} d^{3} x^{3} + 6 \, b^{2} c d^{2} x^{2} - 2 \, b^{2} c^{3} + 12 \, a b c d^{2} - 3 \, {\left (2 \, b^{2} c^{2} d - a b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{4 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, -\frac {3 \, {\left (2 \, a b c^{2} d - a^{2} d^{3} + {\left (2 \, b^{2} c^{2} d - a b d^{3}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (b^{2} d^{3} x^{3} + 6 \, b^{2} c d^{2} x^{2} - 2 \, b^{2} c^{3} + 12 \, a b c d^{2} - 3 \, {\left (2 \, b^{2} c^{2} d - a b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \] Input:

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

Output:

[-1/4*(3*(2*a*b*c^2*d - a^2*d^3 + (2*b^2*c^2*d - a*b*d^3)*x^2)*sqrt(b)*log 
(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(b^2*d^3*x^3 + 6*b^2*c*d^ 
2*x^2 - 2*b^2*c^3 + 12*a*b*c*d^2 - 3*(2*b^2*c^2*d - a*b*d^3)*x)*sqrt(b*x^2 
 + a))/(b^4*x^2 + a*b^3), -1/2*(3*(2*a*b*c^2*d - a^2*d^3 + (2*b^2*c^2*d - 
a*b*d^3)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (b^2*d^3*x^3 + 
 6*b^2*c*d^2*x^2 - 2*b^2*c^3 + 12*a*b*c*d^2 - 3*(2*b^2*c^2*d - a*b*d^3)*x) 
*sqrt(b*x^2 + a))/(b^4*x^2 + a*b^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.62 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {d^{3} x^{3}}{2 \, \sqrt {b x^{2} + a} b} + \frac {3 \, c d^{2} x^{2}}{\sqrt {b x^{2} + a} b} - \frac {3 \, c^{2} d x}{\sqrt {b x^{2} + a} b} + \frac {3 \, a d^{3} x}{2 \, \sqrt {b x^{2} + a} b^{2}} + \frac {3 \, c^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} - \frac {3 \, a d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} - \frac {c^{3}}{\sqrt {b x^{2} + a} b} + \frac {6 \, a c d^{2}}{\sqrt {b x^{2} + a} b^{2}} \] Input:

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

Output:

1/2*d^3*x^3/(sqrt(b*x^2 + a)*b) + 3*c*d^2*x^2/(sqrt(b*x^2 + a)*b) - 3*c^2* 
d*x/(sqrt(b*x^2 + a)*b) + 3/2*a*d^3*x/(sqrt(b*x^2 + a)*b^2) + 3*c^2*d*arcs 
inh(b*x/sqrt(a*b))/b^(3/2) - 3/2*a*d^3*arcsinh(b*x/sqrt(a*b))/b^(5/2) - c^ 
3/(sqrt(b*x^2 + a)*b) + 6*a*c*d^2/(sqrt(b*x^2 + a)*b^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.28 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {d^{3} x}{b} + \frac {6 \, c d^{2}}{b}\right )} x - \frac {3 \, {\left (2 \, b^{3} c^{2} d - a b^{2} d^{3}\right )}}{b^{4}}\right )} x - \frac {2 \, {\left (b^{3} c^{3} - 6 \, a b^{2} c d^{2}\right )}}{b^{4}}}{2 \, \sqrt {b x^{2} + a}} - \frac {3 \, {\left (2 \, b c^{2} d - a d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {5}{2}}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 287, normalized size of antiderivative = 3.05 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {12 \sqrt {b \,x^{2}+a}\, a b c \,d^{2}+3 \sqrt {b \,x^{2}+a}\, a b \,d^{3} x -2 \sqrt {b \,x^{2}+a}\, b^{2} c^{3}-6 \sqrt {b \,x^{2}+a}\, b^{2} c^{2} d x +6 \sqrt {b \,x^{2}+a}\, b^{2} c \,d^{2} x^{2}+\sqrt {b \,x^{2}+a}\, b^{2} d^{3} x^{3}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} d^{3}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2} d -3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,d^{3} x^{2}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} c^{2} d \,x^{2}+2 \sqrt {b}\, a^{2} d^{3}-6 \sqrt {b}\, a b \,c^{2} d +2 \sqrt {b}\, a b \,d^{3} x^{2}-6 \sqrt {b}\, b^{2} c^{2} d \,x^{2}}{2 b^{3} \left (b \,x^{2}+a \right )} \] Input:

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

Output:

(12*sqrt(a + b*x**2)*a*b*c*d**2 + 3*sqrt(a + b*x**2)*a*b*d**3*x - 2*sqrt(a 
 + b*x**2)*b**2*c**3 - 6*sqrt(a + b*x**2)*b**2*c**2*d*x + 6*sqrt(a + b*x** 
2)*b**2*c*d**2*x**2 + sqrt(a + b*x**2)*b**2*d**3*x**3 - 3*sqrt(b)*log((sqr 
t(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*d**3 + 6*sqrt(b)*log((sqrt(a + b* 
x**2) + sqrt(b)*x)/sqrt(a))*a*b*c**2*d - 3*sqrt(b)*log((sqrt(a + b*x**2) + 
 sqrt(b)*x)/sqrt(a))*a*b*d**3*x**2 + 6*sqrt(b)*log((sqrt(a + b*x**2) + sqr 
t(b)*x)/sqrt(a))*b**2*c**2*d*x**2 + 2*sqrt(b)*a**2*d**3 - 6*sqrt(b)*a*b*c* 
*2*d + 2*sqrt(b)*a*b*d**3*x**2 - 6*sqrt(b)*b**2*c**2*d*x**2)/(2*b**3*(a + 
b*x**2))