\(\int \frac {x (c+d x)^3}{\sqrt {a+b x^2}} \, dx\) [1180]

Optimal result

Integrand size = 20, antiderivative size = 151 \[ \int \frac {x (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\frac {c \left (b c^2-2 a d^2\right ) \sqrt {a+b x^2}}{b^2}+\frac {3 d \left (4 b c^2-a d^2\right ) x \sqrt {a+b x^2}}{8 b^2}+\frac {c d^2 x^2 \sqrt {a+b x^2}}{b}+\frac {d^3 x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a d \left (4 b c^2-a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.74 \[ \int \frac {x (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (8 b c^3-16 a c d^2+12 b c^2 d x-3 a d^3 x+8 b c d^2 x^2+2 b d^3 x^3\right )}{8 b^2}-\frac {3 a d \left (-4 b c^2+a d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{5/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(8*b*c^3 - 16*a*c*d^2 + 12*b*c^2*d*x - 3*a*d^3*x + 8*b*c* 
d^2*x^2 + 2*b*d^3*x^3))/(8*b^2) - (3*a*d*(-4*b*c^2 + a*d^2)*Log[-(Sqrt[b]* 
x) + Sqrt[a + b*x^2]])/(8*b^(5/2))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {541, 2340, 27, 533, 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.

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

\(\Big \downarrow \) 541

\(\displaystyle \frac {\int \frac {x \left (4 b c^3+12 b d^2 x^2 c+3 d \left (4 b c^2-a d^2\right ) x\right )}{\sqrt {b x^2+a}}dx}{4 b}+\frac {d^3 x^3 \sqrt {a+b x^2}}{4 b}\)

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(d^3*x^3*Sqrt[a + b*x^2])/(4*b) + ((3*d*(4*b*c^2 - a*d^2)*x*Sqrt[a + b*x^2 
])/(2*b) + 4*c*d^2*x^2*Sqrt[a + b*x^2] - (-8*c*(b*c^2 - 2*a*d^2)*Sqrt[a + 
b*x^2] + (3*a*d*(4*b*c^2 - a*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sq 
rt[b])/(2*b))/(4*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 533

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* 
p + 2))   Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], 
 x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer 
Q[2*p]
 

rule 541

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x 
] + Simp[1/(b*(m + n + 2*p + 1))   Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + 
 n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) 
*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt 
Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.66

method	result	size
risch	\(-\frac {\left (-2 b \,d^{3} x^{3}-8 b c \,d^{2} x^{2}+3 a x \,d^{3}-12 b \,c^{2} d x +16 a \,d^{2} c -8 b \,c^{3}\right ) \sqrt {b \,x^{2}+a}}{8 b^{2}}+\frac {3 a d \left (a \,d^{2}-4 b \,c^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\)	\(99\)
default	\(\frac {c^{3} \sqrt {b \,x^{2}+a}}{b}+d^{3} \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+3 c \,d^{2} \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+3 c^{2} d \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\)	\(167\)

Input:

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

Output:

-1/8*(-2*b*d^3*x^3-8*b*c*d^2*x^2+3*a*d^3*x-12*b*c^2*d*x+16*a*c*d^2-8*b*c^3 
)*(b*x^2+a)^(1/2)/b^2+3/8*a*d*(a*d^2-4*b*c^2)/b^(5/2)*ln(b^(1/2)*x+(b*x^2+ 
a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.61 \[ \int \frac {x (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (4 \, a b c^{2} d - a^{2} d^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (2 \, b^{2} d^{3} x^{3} + 8 \, b^{2} c d^{2} x^{2} + 8 \, b^{2} c^{3} - 16 \, a b c d^{2} + 3 \, {\left (4 \, b^{2} c^{2} d - a b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, b^{3}}, \frac {3 \, {\left (4 \, a b c^{2} d - a^{2} d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, b^{2} d^{3} x^{3} + 8 \, b^{2} c d^{2} x^{2} + 8 \, b^{2} c^{3} - 16 \, a b c d^{2} + 3 \, {\left (4 \, b^{2} c^{2} d - a b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, b^{3}}\right ] \] Input:

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

Output:

[-1/16*(3*(4*a*b*c^2*d - a^2*d^3)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a) 
*sqrt(b)*x - a) - 2*(2*b^2*d^3*x^3 + 8*b^2*c*d^2*x^2 + 8*b^2*c^3 - 16*a*b* 
c*d^2 + 3*(4*b^2*c^2*d - a*b*d^3)*x)*sqrt(b*x^2 + a))/b^3, 1/8*(3*(4*a*b*c 
^2*d - a^2*d^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (2*b^2*d^3*x 
^3 + 8*b^2*c*d^2*x^2 + 8*b^2*c^3 - 16*a*b*c*d^2 + 3*(4*b^2*c^2*d - a*b*d^3 
)*x)*sqrt(b*x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.15 \[ \int \frac {x (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\begin {cases} - \frac {a \left (- \frac {3 a d^{3}}{4 b} + 3 c^{2} d\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {c d^{2} x^{2}}{b} + \frac {d^{3} x^{3}}{4 b} + \frac {x \left (- \frac {3 a d^{3}}{4 b} + 3 c^{2} d\right )}{2 b} + \frac {- \frac {2 a c d^{2}}{b} + c^{3}}{b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {c^{3} x^{2}}{2} + c^{2} d x^{3} + \frac {3 c d^{2} x^{4}}{4} + \frac {d^{3} x^{5}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-a*(-3*a*d**3/(4*b) + 3*c**2*d)*Piecewise((log(2*sqrt(b)*sqrt(a 
 + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(2* 
b) + sqrt(a + b*x**2)*(c*d**2*x**2/b + d**3*x**3/(4*b) + x*(-3*a*d**3/(4*b 
) + 3*c**2*d)/(2*b) + (-2*a*c*d**2/b + c**3)/b), Ne(b, 0)), ((c**3*x**2/2 
+ c**2*d*x**3 + 3*c*d**2*x**4/4 + d**3*x**5/5)/sqrt(a), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.01 \[ \int \frac {x (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} d^{3} x^{3}}{4 \, b} + \frac {\sqrt {b x^{2} + a} c d^{2} x^{2}}{b} + \frac {3 \, \sqrt {b x^{2} + a} c^{2} d x}{2 \, b} - \frac {3 \, \sqrt {b x^{2} + a} a d^{3} x}{8 \, b^{2}} - \frac {3 \, a c^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {3 \, a^{2} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} + \frac {\sqrt {b x^{2} + a} c^{3}}{b} - \frac {2 \, \sqrt {b x^{2} + a} a c d^{2}}{b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.82 \[ \int \frac {x (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\frac {1}{8} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (\frac {d^{3} x}{b} + \frac {4 \, c d^{2}}{b}\right )} x + \frac {3 \, {\left (4 \, b^{3} c^{2} d - a b^{2} d^{3}\right )}}{b^{4}}\right )} x + \frac {8 \, {\left (b^{3} c^{3} - 2 \, a b^{2} c d^{2}\right )}}{b^{4}}\right )} + \frac {3 \, {\left (4 \, a b c^{2} d - a^{2} d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.12 \[ \int \frac {x (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\frac {-16 \sqrt {b \,x^{2}+a}\, a b c \,d^{2}-3 \sqrt {b \,x^{2}+a}\, a b \,d^{3} x +8 \sqrt {b \,x^{2}+a}\, b^{2} c^{3}+12 \sqrt {b \,x^{2}+a}\, b^{2} c^{2} d x +8 \sqrt {b \,x^{2}+a}\, b^{2} c \,d^{2} x^{2}+2 \sqrt {b \,x^{2}+a}\, b^{2} d^{3} x^{3}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} d^{3}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2} d}{8 b^{3}} \] Input:

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

Output:

( - 16*sqrt(a + b*x**2)*a*b*c*d**2 - 3*sqrt(a + b*x**2)*a*b*d**3*x + 8*sqr 
t(a + b*x**2)*b**2*c**3 + 12*sqrt(a + b*x**2)*b**2*c**2*d*x + 8*sqrt(a + b 
*x**2)*b**2*c*d**2*x**2 + 2*sqrt(a + b*x**2)*b**2*d**3*x**3 + 3*sqrt(b)*lo 
g((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*d**3 - 12*sqrt(b)*log((sqrt 
(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b*c**2*d)/(8*b**3)