\(\int \frac {x (c+d x)^4}{(a+b x^2)^{3/2}} \, dx\) [1228]

Optimal result

Integrand size = 20, antiderivative size = 150 \[ \int \frac {x (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {(c+d x)^4}{b \sqrt {a+b x^2}}+\frac {8 d^2 \left (4 b c^2-a d^2\right ) \sqrt {a+b x^2}}{3 b^3}+\frac {10 c d^3 x \sqrt {a+b x^2}}{3 b^2}+\frac {4 d^2 (c+d x)^2 \sqrt {a+b x^2}}{3 b^2}+\frac {2 c d \left (2 b c^2-3 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}} \] Output:

-(d*x+c)^4/b/(b*x^2+a)^(1/2)+8/3*d^2*(-a*d^2+4*b*c^2)*(b*x^2+a)^(1/2)/b^3+ 
10/3*c*d^3*x*(b*x^2+a)^(1/2)/b^2+4/3*d^2*(d*x+c)^2*(b*x^2+a)^(1/2)/b^2+2*c 
*d*(-3*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 = 0.50 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01 \[ \int \frac {x (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-8 a^2 d^4+2 a b d^2 \left (18 c^2+9 c d x-2 d^2 x^2\right )+b^2 \left (-3 c^4-12 c^3 d x+18 c^2 d^2 x^2+6 c d^3 x^3+d^4 x^4\right )-6 \sqrt {b} c d \left (2 b c^2-3 a d^2\right ) \sqrt {a+b x^2} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{3 b^3 \sqrt {a+b x^2}} \] Input:

Integrate[(x*(c + d*x)^4)/(a + b*x^2)^(3/2),x]
 

Output:

(-8*a^2*d^4 + 2*a*b*d^2*(18*c^2 + 9*c*d*x - 2*d^2*x^2) + b^2*(-3*c^4 - 12* 
c^3*d*x + 18*c^2*d^2*x^2 + 6*c*d^3*x^3 + d^4*x^4) - 6*Sqrt[b]*c*d*(2*b*c^2 
 - 3*a*d^2)*Sqrt[a + b*x^2]*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(3*b^3*Sq 
rt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05, 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, 676, 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.

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

\(\Big \downarrow \) 531

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 497

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

\(\Big \downarrow \) 676

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

Int[(x*(c + d*x)^4)/(a + b*x^2)^(3/2),x]
 

Output:

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

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.10

method	result	size
risch	\(-\frac {d^{2} \left (-b \,x^{2} d^{2}-6 b c d x +5 a \,d^{2}-18 b \,c^{2}\right ) \sqrt {b \,x^{2}+a}}{3 b^{3}}-\frac {2 d b c \left (3 a \,d^{2}-2 b \,c^{2}\right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {-a^{2} d^{4}+6 b \,c^{2} d^{2} a -b^{2} c^{4}}{b \sqrt {b \,x^{2}+a}}+\frac {2 d^{3} c a x}{\sqrt {b \,x^{2}+a}}}{b^{2}}\)	\(165\)
default	\(-\frac {c^{4}}{b \sqrt {b \,x^{2}+a}}+d^{4} \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )+4 c \,d^{3} \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+4 d \,c^{3} \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )+6 c^{2} d^{2} \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )\)	\(227\)

Input:

int(x*(d*x+c)^4/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3*d^2*(-b*d^2*x^2-6*b*c*d*x+5*a*d^2-18*b*c^2)*(b*x^2+a)^(1/2)/b^3-1/b^2 
*(2*d*b*c*(3*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)))-(-a^2*d^4+6*a*b*c^2*d^2-b^2*c^4)/b/(b*x^2+a)^(1/2)+2*d^3* 
c*a*x/(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 382, normalized size of antiderivative = 2.55 \[ \int \frac {x (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (2 \, a b c^{3} d - 3 \, a^{2} c d^{3} + {\left (2 \, b^{2} c^{3} d - 3 \, a b c d^{3}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - {\left (b^{2} d^{4} x^{4} + 6 \, b^{2} c d^{3} x^{3} - 3 \, b^{2} c^{4} + 36 \, a b c^{2} d^{2} - 8 \, a^{2} d^{4} + 2 \, {\left (9 \, b^{2} c^{2} d^{2} - 2 \, a b d^{4}\right )} x^{2} - 6 \, {\left (2 \, b^{2} c^{3} d - 3 \, a b c d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, -\frac {6 \, {\left (2 \, a b c^{3} d - 3 \, a^{2} c d^{3} + {\left (2 \, b^{2} c^{3} d - 3 \, a b c d^{3}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (b^{2} d^{4} x^{4} + 6 \, b^{2} c d^{3} x^{3} - 3 \, b^{2} c^{4} + 36 \, a b c^{2} d^{2} - 8 \, a^{2} d^{4} + 2 \, {\left (9 \, b^{2} c^{2} d^{2} - 2 \, a b d^{4}\right )} x^{2} - 6 \, {\left (2 \, b^{2} c^{3} d - 3 \, a b c d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \] Input:

integrate(x*(d*x+c)^4/(b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

[-1/3*(3*(2*a*b*c^3*d - 3*a^2*c*d^3 + (2*b^2*c^3*d - 3*a*b*c*d^3)*x^2)*sqr 
t(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - (b^2*d^4*x^4 + 6*b^ 
2*c*d^3*x^3 - 3*b^2*c^4 + 36*a*b*c^2*d^2 - 8*a^2*d^4 + 2*(9*b^2*c^2*d^2 - 
2*a*b*d^4)*x^2 - 6*(2*b^2*c^3*d - 3*a*b*c*d^3)*x)*sqrt(b*x^2 + a))/(b^4*x^ 
2 + a*b^3), -1/3*(6*(2*a*b*c^3*d - 3*a^2*c*d^3 + (2*b^2*c^3*d - 3*a*b*c*d^ 
3)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (b^2*d^4*x^4 + 6*b^2 
*c*d^3*x^3 - 3*b^2*c^4 + 36*a*b*c^2*d^2 - 8*a^2*d^4 + 2*(9*b^2*c^2*d^2 - 2 
*a*b*d^4)*x^2 - 6*(2*b^2*c^3*d - 3*a*b*c*d^3)*x)*sqrt(b*x^2 + a))/(b^4*x^2 
 + a*b^3)]
 

Sympy [F]

\[ \int \frac {x (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x \left (c + d x\right )^{4}}{\left (a + b x^{2}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x*(d*x+c)**4/(b*x**2+a)**(3/2),x)
 

Output:

Integral(x*(c + d*x)**4/(a + b*x**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.47 \[ \int \frac {x (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {d^{4} x^{4}}{3 \, \sqrt {b x^{2} + a} b} + \frac {2 \, c d^{3} x^{3}}{\sqrt {b x^{2} + a} b} + \frac {6 \, c^{2} d^{2} x^{2}}{\sqrt {b x^{2} + a} b} - \frac {4 \, a d^{4} x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} - \frac {4 \, c^{3} d x}{\sqrt {b x^{2} + a} b} + \frac {6 \, a c d^{3} x}{\sqrt {b x^{2} + a} b^{2}} + \frac {4 \, c^{3} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} - \frac {6 \, a c d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {c^{4}}{\sqrt {b x^{2} + a} b} + \frac {12 \, a c^{2} d^{2}}{\sqrt {b x^{2} + a} b^{2}} - \frac {8 \, a^{2} d^{4}}{3 \, \sqrt {b x^{2} + a} b^{3}} \] Input:

integrate(x*(d*x+c)^4/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

1/3*d^4*x^4/(sqrt(b*x^2 + a)*b) + 2*c*d^3*x^3/(sqrt(b*x^2 + a)*b) + 6*c^2* 
d^2*x^2/(sqrt(b*x^2 + a)*b) - 4/3*a*d^4*x^2/(sqrt(b*x^2 + a)*b^2) - 4*c^3* 
d*x/(sqrt(b*x^2 + a)*b) + 6*a*c*d^3*x/(sqrt(b*x^2 + a)*b^2) + 4*c^3*d*arcs 
inh(b*x/sqrt(a*b))/b^(3/2) - 6*a*c*d^3*arcsinh(b*x/sqrt(a*b))/b^(5/2) - c^ 
4/(sqrt(b*x^2 + a)*b) + 12*a*c^2*d^2/(sqrt(b*x^2 + a)*b^2) - 8/3*a^2*d^4/( 
sqrt(b*x^2 + a)*b^3)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.10 \[ \int \frac {x (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left ({\left (\frac {d^{4} x}{b} + \frac {6 \, c d^{3}}{b}\right )} x + \frac {2 \, {\left (9 \, b^{4} c^{2} d^{2} - 2 \, a b^{3} d^{4}\right )}}{b^{5}}\right )} x - \frac {6 \, {\left (2 \, b^{4} c^{3} d - 3 \, a b^{3} c d^{3}\right )}}{b^{5}}\right )} x - \frac {3 \, b^{4} c^{4} - 36 \, a b^{3} c^{2} d^{2} + 8 \, a^{2} b^{2} d^{4}}{b^{5}}}{3 \, \sqrt {b x^{2} + a}} - \frac {2 \, {\left (2 \, b c^{3} d - 3 \, a c d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {5}{2}}} \] Input:

integrate(x*(d*x+c)^4/(b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

1/3*((((d^4*x/b + 6*c*d^3/b)*x + 2*(9*b^4*c^2*d^2 - 2*a*b^3*d^4)/b^5)*x - 
6*(2*b^4*c^3*d - 3*a*b^3*c*d^3)/b^5)*x - (3*b^4*c^4 - 36*a*b^3*c^2*d^2 + 8 
*a^2*b^2*d^4)/b^5)/sqrt(b*x^2 + a) - 2*(2*b*c^3*d - 3*a*c*d^3)*log(abs(-sq 
rt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x\,{\left (c+d\,x\right )}^4}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int((x*(c + d*x)^4)/(a + b*x^2)^(3/2),x)
 

Output:

int((x*(c + d*x)^4)/(a + b*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {x (c+d x)^4}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x \left (d x +c \right )^{4}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}}}d x \] Input:

int(x*(d*x+c)^4/(b*x^2+a)^(3/2),x)
 

Output:

int(x*(d*x+c)^4/(b*x^2+a)^(3/2),x)