3.1.8 \(\int x^3 (A+B x) (a+b x^2)^{3/2} \, dx\) [8]

3.1.8.1 Optimal result

Integrand size = 20, antiderivative size = 150 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {3 a^3 B x \sqrt {a+b x^2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac {A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac {3 a^4 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}} \]

output

1/64*a^2*B*x*(b*x^2+a)^(3/2)/b^2+1/7*A*x^2*(b*x^2+a)^(5/2)/b+1/8*B*x^3*(b* 
x^2+a)^(5/2)/b-1/560*a*(35*B*x+32*A)*(b*x^2+a)^(5/2)/b^2+3/128*a^4*B*arcta 
nh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5/2)+3/128*a^3*B*x*(b*x^2+a)^(1/2)/b^2
 

3.1.8.2 Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.79 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {\sqrt {b} \sqrt {a+b x^2} \left (80 b^3 x^6 (8 A+7 B x)+2 a^2 b x^2 (64 A+35 B x)+8 a b^2 x^4 (128 A+105 B x)-a^3 (256 A+105 B x)\right )-105 a^4 B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{4480 b^{5/2}} \]

input

Integrate[x^3*(A + B*x)*(a + b*x^2)^(3/2),x]
 

output

(Sqrt[b]*Sqrt[a + b*x^2]*(80*b^3*x^6*(8*A + 7*B*x) + 2*a^2*b*x^2*(64*A + 3 
5*B*x) + 8*a*b^2*x^4*(128*A + 105*B*x) - a^3*(256*A + 105*B*x)) - 105*a^4* 
B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(4480*b^(5/2))
 

3.1.8.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.15, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {533, 533, 25, 27, 533, 27, 455, 211, 211, 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.

\(\displaystyle \int x^3 \left (a+b x^2\right )^{3/2} (A+B x) \, dx\)

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 533

\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {-\frac {\int -a b x (16 A+21 B x) \left (b x^2+a\right )^{3/2}dx}{7 b}-\frac {8}{7} A x^2 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\frac {\int a b x (16 A+21 B x) \left (b x^2+a\right )^{3/2}dx}{7 b}-\frac {8}{7} A x^2 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\frac {1}{7} a \int x (16 A+21 B x) \left (b x^2+a\right )^{3/2}dx-\frac {8}{7} A x^2 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\frac {1}{7} a \left (\frac {7 B x \left (a+b x^2\right )^{5/2}}{2 b}-\frac {\int 3 (7 a B-32 A b x) \left (b x^2+a\right )^{3/2}dx}{6 b}\right )-\frac {8}{7} A x^2 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\frac {1}{7} a \left (\frac {7 B x \left (a+b x^2\right )^{5/2}}{2 b}-\frac {\int (7 a B-32 A b x) \left (b x^2+a\right )^{3/2}dx}{2 b}\right )-\frac {8}{7} A x^2 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\frac {1}{7} a \left (\frac {7 B x \left (a+b x^2\right )^{5/2}}{2 b}-\frac {7 a B \int \left (b x^2+a\right )^{3/2}dx-\frac {32}{5} A \left (a+b x^2\right )^{5/2}}{2 b}\right )-\frac {8}{7} A x^2 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\frac {1}{7} a \left (\frac {7 B x \left (a+b x^2\right )^{5/2}}{2 b}-\frac {7 a B \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )-\frac {32}{5} A \left (a+b x^2\right )^{5/2}}{2 b}\right )-\frac {8}{7} A x^2 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\frac {1}{7} a \left (\frac {7 B x \left (a+b x^2\right )^{5/2}}{2 b}-\frac {7 a B \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )-\frac {32}{5} A \left (a+b x^2\right )^{5/2}}{2 b}\right )-\frac {8}{7} A x^2 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\frac {1}{7} a \left (\frac {7 B x \left (a+b x^2\right )^{5/2}}{2 b}-\frac {7 a B \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )-\frac {32}{5} A \left (a+b x^2\right )^{5/2}}{2 b}\right )-\frac {8}{7} A x^2 \left (a+b x^2\right )^{5/2}}{8 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\frac {1}{7} a \left (\frac {7 B x \left (a+b x^2\right )^{5/2}}{2 b}-\frac {7 a B \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )-\frac {32}{5} A \left (a+b x^2\right )^{5/2}}{2 b}\right )-\frac {8}{7} A x^2 \left (a+b x^2\right )^{5/2}}{8 b}\)

input

Int[x^3*(A + B*x)*(a + b*x^2)^(3/2),x]
 

output

(B*x^3*(a + b*x^2)^(5/2))/(8*b) - ((-8*A*x^2*(a + b*x^2)^(5/2))/7 + (a*((7 
*B*x*(a + b*x^2)^(5/2))/(2*b) - ((-32*A*(a + b*x^2)^(5/2))/5 + 7*a*B*((x*( 
a + b*x^2)^(3/2))/4 + (3*a*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x) 
/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/4))/(2*b)))/7)/(8*b)
 

3.1.8.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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]
 

3.1.8.4 Maple [A] (verified)

Time = 3.45 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {\left (-560 b^{3} B \,x^{7}-640 x^{6} b^{3} A -840 B a \,b^{2} x^{5}-1024 a A \,b^{2} x^{4}-70 B \,a^{2} b \,x^{3}-128 a^{2} A b \,x^{2}+105 a^{3} B x +256 a^{3} A \right ) \sqrt {b \,x^{2}+a}}{4480 b^{2}}+\frac {3 B \,a^{4} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {5}{2}}}\)	\(113\)
default	\(B \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )+A \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 b^{2}}\right )\)	\(136\)

input

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

output

-1/4480*(-560*B*b^3*x^7-640*A*b^3*x^6-840*B*a*b^2*x^5-1024*A*a*b^2*x^4-70* 
B*a^2*b*x^3-128*A*a^2*b*x^2+105*B*a^3*x+256*A*a^3)/b^2*(b*x^2+a)^(1/2)+3/1 
28*B*a^4/b^(5/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))
 

3.1.8.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.69 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\left [\frac {105 \, B a^{4} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (560 \, B b^{4} x^{7} + 640 \, A b^{4} x^{6} + 840 \, B a b^{3} x^{5} + 1024 \, A a b^{3} x^{4} + 70 \, B a^{2} b^{2} x^{3} + 128 \, A a^{2} b^{2} x^{2} - 105 \, B a^{3} b x - 256 \, A a^{3} b\right )} \sqrt {b x^{2} + a}}{8960 \, b^{3}}, -\frac {105 \, B a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (560 \, B b^{4} x^{7} + 640 \, A b^{4} x^{6} + 840 \, B a b^{3} x^{5} + 1024 \, A a b^{3} x^{4} + 70 \, B a^{2} b^{2} x^{3} + 128 \, A a^{2} b^{2} x^{2} - 105 \, B a^{3} b x - 256 \, A a^{3} b\right )} \sqrt {b x^{2} + a}}{4480 \, b^{3}}\right ] \]

input

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

output

[1/8960*(105*B*a^4*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) 
 + 2*(560*B*b^4*x^7 + 640*A*b^4*x^6 + 840*B*a*b^3*x^5 + 1024*A*a*b^3*x^4 + 
 70*B*a^2*b^2*x^3 + 128*A*a^2*b^2*x^2 - 105*B*a^3*b*x - 256*A*a^3*b)*sqrt( 
b*x^2 + a))/b^3, -1/4480*(105*B*a^4*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 
+ a)) - (560*B*b^4*x^7 + 640*A*b^4*x^6 + 840*B*a*b^3*x^5 + 1024*A*a*b^3*x^ 
4 + 70*B*a^2*b^2*x^3 + 128*A*a^2*b^2*x^2 - 105*B*a^3*b*x - 256*A*a^3*b)*sq 
rt(b*x^2 + a))/b^3]
 

3.1.8.6 Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.12 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\begin {cases} \frac {3 B a^{4} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{128 b^{2}} + \sqrt {a + b x^{2}} \left (- \frac {2 A a^{3}}{35 b^{2}} + \frac {A a^{2} x^{2}}{35 b} + \frac {8 A a x^{4}}{35} + \frac {A b x^{6}}{7} - \frac {3 B a^{3} x}{128 b^{2}} + \frac {B a^{2} x^{3}}{64 b} + \frac {3 B a x^{5}}{16} + \frac {B b x^{7}}{8}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{4}}{4} + \frac {B x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

input

integrate(x**3*(B*x+A)*(b*x**2+a)**(3/2),x)
 

output

Piecewise((3*B*a**4*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqr 
t(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(128*b**2) + sqrt(a + b*x* 
*2)*(-2*A*a**3/(35*b**2) + A*a**2*x**2/(35*b) + 8*A*a*x**4/35 + A*b*x**6/7 
 - 3*B*a**3*x/(128*b**2) + B*a**2*x**3/(64*b) + 3*B*a*x**5/16 + B*b*x**7/8 
), Ne(b, 0)), (a**(3/2)*(A*x**4/4 + B*x**5/5), True))
 

3.1.8.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.84 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x^{3}}{8 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A x^{2}}{7 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} B a^{3} x}{128 \, b^{2}} + \frac {3 \, B a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a}{35 \, b^{2}} \]

input

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

output

1/8*(b*x^2 + a)^(5/2)*B*x^3/b + 1/7*(b*x^2 + a)^(5/2)*A*x^2/b - 1/16*(b*x^ 
2 + a)^(5/2)*B*a*x/b^2 + 1/64*(b*x^2 + a)^(3/2)*B*a^2*x/b^2 + 3/128*sqrt(b 
*x^2 + a)*B*a^3*x/b^2 + 3/128*B*a^4*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 2/35* 
(b*x^2 + a)^(5/2)*A*a/b^2
 

3.1.8.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.77 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=-\frac {3 \, B a^{4} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {5}{2}}} - \frac {1}{4480} \, \sqrt {b x^{2} + a} {\left (\frac {256 \, A a^{3}}{b^{2}} + {\left (\frac {105 \, B a^{3}}{b^{2}} - 2 \, {\left (\frac {64 \, A a^{2}}{b} + {\left (\frac {35 \, B a^{2}}{b} + 4 \, {\left (128 \, A a + 5 \, {\left (21 \, B a + 2 \, {\left (7 \, B b x + 8 \, A b\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \]

input

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

output

-3/128*B*a^4*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) - 1/4480*sqrt( 
b*x^2 + a)*(256*A*a^3/b^2 + (105*B*a^3/b^2 - 2*(64*A*a^2/b + (35*B*a^2/b + 
 4*(128*A*a + 5*(21*B*a + 2*(7*B*b*x + 8*A*b)*x)*x)*x)*x)*x)*x)
 

3.1.8.9 Mupad [F(-1)]

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

input

int(x^3*(a + b*x^2)^(3/2)*(A + B*x),x)
 

output

int(x^3*(a + b*x^2)^(3/2)*(A + B*x), x)