\(\int \frac {x (c+d x)^3}{(a+b x^2)^{5/2}} \, dx\) [1274]

Optimal result

Integrand size = 20, antiderivative size = 95 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {(c+d x)^3}{3 b \left (a+b x^2\right )^{3/2}}-\frac {d \left (2 a c d-\left (b c^2-a d^2\right ) x\right )}{a b^2 \sqrt {a+b x^2}}+\frac {d^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {3 b^2 c^2 d x^3-3 a^2 d^2 (2 c+d x)-a b \left (c^3+9 c d^2 x^2+4 d^3 x^3\right )}{3 a b^2 \left (a+b x^2\right )^{3/2}}-\frac {d^3 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{5/2}} \] Input:

Integrate[(x*(c + d*x)^3)/(a + b*x^2)^(5/2),x]
 

Output:

(3*b^2*c^2*d*x^3 - 3*a^2*d^2*(2*c + d*x) - a*b*(c^3 + 9*c*d^2*x^2 + 4*d^3* 
x^3))/(3*a*b^2*(a + b*x^2)^(3/2)) - (d^3*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2 
]])/b^(5/2)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {531, 27, 495, 27, 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.

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

\(\Big \downarrow \) 531

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 495

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

Int[(x*(c + d*x)^3)/(a + b*x^2)^(5/2),x]
 

Output:

-1/3*(c + d*x)^3/(b*(a + b*x^2)^(3/2)) + (d*(-(((a*d - b*c*x)*(c + d*x))/( 
a*b*Sqrt[a + b*x^2])) + (d*(-(c*Sqrt[a + b*x^2]) + (a*d*ArcTanh[(Sqrt[b]*x 
)/Sqrt[a + b*x^2]])/Sqrt[b]))/(a*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 495

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(a*d - b*c*x)*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - 
 Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* 
d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ 
{a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(178\) vs. \(2(83)=166\).

Time = 0.35 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.88

method	result	size
default	\(-\frac {c^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+d^{3} \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )+3 c \,d^{2} \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )+3 c^{2} d \left (-\frac {x}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {a \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{2 b}\right )\)	\(179\)

Input:

int(x*(d*x+c)^3/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 335, normalized size of antiderivative = 3.53 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a b^{2} d^{3} x^{4} + 2 \, a^{2} b d^{3} x^{2} + a^{3} d^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (9 \, a b^{2} c d^{2} x^{2} + 3 \, a^{2} b d^{3} x + a b^{2} c^{3} + 6 \, a^{2} b c d^{2} - {\left (3 \, b^{3} c^{2} d - 4 \, a b^{2} d^{3}\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}}, -\frac {3 \, {\left (a b^{2} d^{3} x^{4} + 2 \, a^{2} b d^{3} x^{2} + a^{3} d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (9 \, a b^{2} c d^{2} x^{2} + 3 \, a^{2} b d^{3} x + a b^{2} c^{3} + 6 \, a^{2} b c d^{2} - {\left (3 \, b^{3} c^{2} d - 4 \, a b^{2} d^{3}\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}}\right ] \] Input:

integrate(x*(d*x+c)^3/(b*x^2+a)^(5/2),x, algorithm="fricas")
 

Output:

[1/6*(3*(a*b^2*d^3*x^4 + 2*a^2*b*d^3*x^2 + a^3*d^3)*sqrt(b)*log(-2*b*x^2 - 
 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(9*a*b^2*c*d^2*x^2 + 3*a^2*b*d^3*x + 
 a*b^2*c^3 + 6*a^2*b*c*d^2 - (3*b^3*c^2*d - 4*a*b^2*d^3)*x^3)*sqrt(b*x^2 + 
 a))/(a*b^5*x^4 + 2*a^2*b^4*x^2 + a^3*b^3), -1/3*(3*(a*b^2*d^3*x^4 + 2*a^2 
*b*d^3*x^2 + a^3*d^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (9*a*b 
^2*c*d^2*x^2 + 3*a^2*b*d^3*x + a*b^2*c^3 + 6*a^2*b*c*d^2 - (3*b^3*c^2*d - 
4*a*b^2*d^3)*x^3)*sqrt(b*x^2 + a))/(a*b^5*x^4 + 2*a^2*b^4*x^2 + a^3*b^3)]
 

Sympy [F]

\[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x \left (c + d x\right )^{3}}{\left (a + b x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x*(d*x+c)**3/(b*x**2+a)**(5/2),x)
 

Output:

Integral(x*(c + d*x)**3/(a + b*x**2)**(5/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (83) = 166\).

Time = 0.03 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.80 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {1}{3} \, d^{3} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} - \frac {3 \, c d^{2} x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {c^{2} d x}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {c^{2} d x}{\sqrt {b x^{2} + a} a b} - \frac {d^{3} x}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {c^{3}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {2 \, a c d^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} \] Input:

integrate(x*(d*x+c)^3/(b*x^2+a)^(5/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.25 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {d^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {5}{2}}} - \frac {{\left (\frac {3 \, a d^{3}}{b^{2}} + {\left (\frac {9 \, c d^{2}}{b} - \frac {{\left (3 \, b^{4} c^{2} d - 4 \, a b^{3} d^{3}\right )} x}{a b^{4}}\right )} x\right )} x + \frac {a b^{3} c^{3} + 6 \, a^{2} b^{2} c d^{2}}{a b^{4}}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} \] Input:

integrate(x*(d*x+c)^3/(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x\,{\left (c+d\,x\right )}^3}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:

int((x*(c + d*x)^3)/(a + b*x^2)^(5/2),x)
 

Output:

int((x*(c + d*x)^3)/(a + b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.94 \[ \int \frac {x (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-6 \sqrt {b \,x^{2}+a}\, a^{2} b c \,d^{2}-3 \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{3} x -\sqrt {b \,x^{2}+a}\, a \,b^{2} c^{3}-9 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,d^{2} x^{2}-4 \sqrt {b \,x^{2}+a}\, a \,b^{2} d^{3} x^{3}+3 \sqrt {b \,x^{2}+a}\, b^{3} c^{2} d \,x^{3}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d^{3}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b \,d^{3} x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d^{3} x^{4}+3 \sqrt {b}\, a^{2} b \,c^{2} d +6 \sqrt {b}\, a \,b^{2} c^{2} d \,x^{2}+3 \sqrt {b}\, b^{3} c^{2} d \,x^{4}}{3 a \,b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int(x*(d*x+c)^3/(b*x^2+a)^(5/2),x)
 

Output:

( - 6*sqrt(a + b*x**2)*a**2*b*c*d**2 - 3*sqrt(a + b*x**2)*a**2*b*d**3*x - 
sqrt(a + b*x**2)*a*b**2*c**3 - 9*sqrt(a + b*x**2)*a*b**2*c*d**2*x**2 - 4*s 
qrt(a + b*x**2)*a*b**2*d**3*x**3 + 3*sqrt(a + b*x**2)*b**3*c**2*d*x**3 + 3 
*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*d**3 + 6*sqrt(b) 
*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*d**3*x**2 + 3*sqrt(b)* 
log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*d**3*x**4 + 3*sqrt(b)*a 
**2*b*c**2*d + 6*sqrt(b)*a*b**2*c**2*d*x**2 + 3*sqrt(b)*b**3*c**2*d*x**4)/ 
(3*a*b**3*(a**2 + 2*a*b*x**2 + b**2*x**4))