\(\int (A+B x) (c+d x) (a+b x^2)^{3/2} \, dx\) [157]

Optimal result

Integrand size = 22, antiderivative size = 137 \[ \int (A+B x) (c+d x) \left (a+b x^2\right )^{3/2} \, dx=\frac {a (6 A b c-a B d) x \sqrt {a+b x^2}}{16 b}+\frac {(6 A b c-a B d) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {(6 (B c+A d)+5 B d x) \left (a+b x^2\right )^{5/2}}{30 b}+\frac {a^2 (6 A b c-a B d) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \] Output:

1/16*a*(6*A*b*c-B*a*d)*x*(b*x^2+a)^(1/2)/b+1/24*(6*A*b*c-B*a*d)*x*(b*x^2+a 
)^(3/2)/b+1/30*(5*B*d*x+6*A*d+6*B*c)*(b*x^2+a)^(5/2)/b+1/16*a^2*(6*A*b*c-B 
*a*d)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.04 \[ \int (A+B x) (c+d x) \left (a+b x^2\right )^{3/2} \, dx=\frac {\sqrt {b} \sqrt {a+b x^2} \left (3 a^2 (16 B c+16 A d+5 B d x)+4 b^2 x^3 (3 A (5 c+4 d x)+2 B x (6 c+5 d x))+2 a b x (B x (48 c+35 d x)+A (75 c+48 d x))\right )+15 a^2 (-6 A b c+a B d) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{240 b^{3/2}} \] Input:

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

Output:

(Sqrt[b]*Sqrt[a + b*x^2]*(3*a^2*(16*B*c + 16*A*d + 5*B*d*x) + 4*b^2*x^3*(3 
*A*(5*c + 4*d*x) + 2*B*x*(6*c + 5*d*x)) + 2*a*b*x*(B*x*(48*c + 35*d*x) + A 
*(75*c + 48*d*x))) + 15*a^2*(-6*A*b*c + a*B*d)*Log[-(Sqrt[b]*x) + Sqrt[a + 
 b*x^2]])/(240*b^(3/2))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {676, 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.

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

\(\Big \downarrow \) 676

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

\(\Big \downarrow \) 211

\(\displaystyle \frac {(6 A b c-a B d) \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {\left (a+b x^2\right )^{5/2} (A d+B c)}{5 b}+\frac {B d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {(6 A b c-a B d) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {\left (a+b x^2\right )^{5/2} (A d+B c)}{5 b}+\frac {B d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {(6 A b c-a B d) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {\left (a+b x^2\right )^{5/2} (A d+B c)}{5 b}+\frac {B d x \left (a+b x^2\right )^{5/2}}{6 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right ) (6 A b c-a B d)}{6 b}+\frac {\left (a+b x^2\right )^{5/2} (A d+B c)}{5 b}+\frac {B d x \left (a+b x^2\right )^{5/2}}{6 b}\)

Input:

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

Output:

((B*c + A*d)*(a + b*x^2)^(5/2))/(5*b) + (B*d*x*(a + b*x^2)^(5/2))/(6*b) + 
((6*A*b*c - a*B*d)*((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))/(6*b)
 

Defintions of rubi rules used

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 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 = 0.83 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {\left (40 B \,b^{2} d \,x^{5}+48 A \,b^{2} d \,x^{4}+48 x^{4} B \,b^{2} c +60 A \,b^{2} c \,x^{3}+70 a B b d \,x^{3}+96 A a b d \,x^{2}+96 B a b c \,x^{2}+150 A a b c x +15 B \,a^{2} d x +48 a^{2} A d +48 B \,a^{2} c \right ) \sqrt {b \,x^{2}+a}}{240 b}+\frac {a^{2} \left (6 A b c -B a d \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}\)	\(148\)
default	\(A c \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+\frac {\left (A d +B c \right ) \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}+B d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\)	\(153\)

Input:

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

Output:

1/240/b*(40*B*b^2*d*x^5+48*A*b^2*d*x^4+48*B*b^2*c*x^4+60*A*b^2*c*x^3+70*B* 
a*b*d*x^3+96*A*a*b*d*x^2+96*B*a*b*c*x^2+150*A*a*b*c*x+15*B*a^2*d*x+48*A*a^ 
2*d+48*B*a^2*c)*(b*x^2+a)^(1/2)+1/16*a^2*(6*A*b*c-B*a*d)/b^(3/2)*ln(b^(1/2 
)*x+(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.42 \[ \int (A+B x) (c+d x) \left (a+b x^2\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (6 \, A a^{2} b c - B a^{3} d\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (40 \, B b^{3} d x^{5} + 48 \, B a^{2} b c + 48 \, A a^{2} b d + 48 \, {\left (B b^{3} c + A b^{3} d\right )} x^{4} + 10 \, {\left (6 \, A b^{3} c + 7 \, B a b^{2} d\right )} x^{3} + 96 \, {\left (B a b^{2} c + A a b^{2} d\right )} x^{2} + 15 \, {\left (10 \, A a b^{2} c + B a^{2} b d\right )} x\right )} \sqrt {b x^{2} + a}}{480 \, b^{2}}, -\frac {15 \, {\left (6 \, A a^{2} b c - B a^{3} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (40 \, B b^{3} d x^{5} + 48 \, B a^{2} b c + 48 \, A a^{2} b d + 48 \, {\left (B b^{3} c + A b^{3} d\right )} x^{4} + 10 \, {\left (6 \, A b^{3} c + 7 \, B a b^{2} d\right )} x^{3} + 96 \, {\left (B a b^{2} c + A a b^{2} d\right )} x^{2} + 15 \, {\left (10 \, A a b^{2} c + B a^{2} b d\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b^{2}}\right ] \] Input:

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

Output:

[-1/480*(15*(6*A*a^2*b*c - B*a^3*d)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + 
a)*sqrt(b)*x - a) - 2*(40*B*b^3*d*x^5 + 48*B*a^2*b*c + 48*A*a^2*b*d + 48*( 
B*b^3*c + A*b^3*d)*x^4 + 10*(6*A*b^3*c + 7*B*a*b^2*d)*x^3 + 96*(B*a*b^2*c 
+ A*a*b^2*d)*x^2 + 15*(10*A*a*b^2*c + B*a^2*b*d)*x)*sqrt(b*x^2 + a))/b^2, 
-1/240*(15*(6*A*a^2*b*c - B*a^3*d)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + 
 a)) - (40*B*b^3*d*x^5 + 48*B*a^2*b*c + 48*A*a^2*b*d + 48*(B*b^3*c + A*b^3 
*d)*x^4 + 10*(6*A*b^3*c + 7*B*a*b^2*d)*x^3 + 96*(B*a*b^2*c + A*a*b^2*d)*x^ 
2 + 15*(10*A*a*b^2*c + B*a^2*b*d)*x)*sqrt(b*x^2 + a))/b^2]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (124) = 248\).

Time = 0.65 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.45 \[ \int (A+B x) (c+d x) \left (a+b x^2\right )^{3/2} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {B b d x^{5}}{6} + \frac {x^{4} \left (A b^{2} d + B b^{2} c\right )}{5 b} + \frac {x^{3} \left (A b^{2} c + \frac {7 B a b d}{6}\right )}{4 b} + \frac {x^{2} \cdot \left (2 A a b d + 2 B a b c - \frac {4 a \left (A b^{2} d + B b^{2} c\right )}{5 b}\right )}{3 b} + \frac {x \left (2 A a b c + B a^{2} d - \frac {3 a \left (A b^{2} c + \frac {7 B a b d}{6}\right )}{4 b}\right )}{2 b} + \frac {A a^{2} d + B a^{2} c - \frac {2 a \left (2 A a b d + 2 B a b c - \frac {4 a \left (A b^{2} d + B b^{2} c\right )}{5 b}\right )}{3 b}}{b}\right ) + \left (A a^{2} c - \frac {a \left (2 A a b c + B a^{2} d - \frac {3 a \left (A b^{2} c + \frac {7 B a b d}{6}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A c x + \frac {B d x^{3}}{3} + \frac {x^{2} \left (A d + B c\right )}{2}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(a + b*x**2)*(B*b*d*x**5/6 + x**4*(A*b**2*d + B*b**2*c)/(5* 
b) + x**3*(A*b**2*c + 7*B*a*b*d/6)/(4*b) + x**2*(2*A*a*b*d + 2*B*a*b*c - 4 
*a*(A*b**2*d + B*b**2*c)/(5*b))/(3*b) + x*(2*A*a*b*c + B*a**2*d - 3*a*(A*b 
**2*c + 7*B*a*b*d/6)/(4*b))/(2*b) + (A*a**2*d + B*a**2*c - 2*a*(2*A*a*b*d 
+ 2*B*a*b*c - 4*a*(A*b**2*d + B*b**2*c)/(5*b))/(3*b))/b) + (A*a**2*c - a*( 
2*A*a*b*c + B*a**2*d - 3*a*(A*b**2*c + 7*B*a*b*d/6)/(4*b))/(2*b))*Piecewis 
e((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/s 
qrt(b*x**2), True)), Ne(b, 0)), (a**(3/2)*(A*c*x + B*d*x**3/3 + x**2*(A*d 
+ B*c)/2), True))
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.13 \[ \int (A+B x) (c+d x) \left (a+b x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A c x + \frac {3}{8} \, \sqrt {b x^{2} + a} A a c x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B d x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a d x}{24 \, b} - \frac {\sqrt {b x^{2} + a} B a^{2} d x}{16 \, b} + \frac {3 \, A a^{2} c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - \frac {B a^{3} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B c}{5 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A d}{5 \, b} \] Input:

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

Output:

1/4*(b*x^2 + a)^(3/2)*A*c*x + 3/8*sqrt(b*x^2 + a)*A*a*c*x + 1/6*(b*x^2 + a 
)^(5/2)*B*d*x/b - 1/24*(b*x^2 + a)^(3/2)*B*a*d*x/b - 1/16*sqrt(b*x^2 + a)* 
B*a^2*d*x/b + 3/8*A*a^2*c*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 1/16*B*a^3*d*ar 
csinh(b*x/sqrt(a*b))/b^(3/2) + 1/5*(b*x^2 + a)^(5/2)*B*c/b + 1/5*(b*x^2 + 
a)^(5/2)*A*d/b
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.30 \[ \int (A+B x) (c+d x) \left (a+b x^2\right )^{3/2} \, dx=\frac {1}{240} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, B b d x + \frac {6 \, {\left (B b^{5} c + A b^{5} d\right )}}{b^{4}}\right )} x + \frac {5 \, {\left (6 \, A b^{5} c + 7 \, B a b^{4} d\right )}}{b^{4}}\right )} x + \frac {48 \, {\left (B a b^{4} c + A a b^{4} d\right )}}{b^{4}}\right )} x + \frac {15 \, {\left (10 \, A a b^{4} c + B a^{2} b^{3} d\right )}}{b^{4}}\right )} x + \frac {48 \, {\left (B a^{2} b^{3} c + A a^{2} b^{3} d\right )}}{b^{4}}\right )} - \frac {{\left (6 \, A a^{2} b c - B a^{3} d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} \] Input:

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

Output:

1/240*sqrt(b*x^2 + a)*((2*((4*(5*B*b*d*x + 6*(B*b^5*c + A*b^5*d)/b^4)*x + 
5*(6*A*b^5*c + 7*B*a*b^4*d)/b^4)*x + 48*(B*a*b^4*c + A*a*b^4*d)/b^4)*x + 1 
5*(10*A*a*b^4*c + B*a^2*b^3*d)/b^4)*x + 48*(B*a^2*b^3*c + A*a^2*b^3*d)/b^4 
) - 1/16*(6*A*a^2*b*c - B*a^3*d)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^ 
(3/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.79 \[ \int (A+B x) (c+d x) \left (a+b x^2\right )^{3/2} \, dx=\frac {48 \sqrt {b \,x^{2}+a}\, a^{3} d +150 \sqrt {b \,x^{2}+a}\, a^{2} b c x +48 \sqrt {b \,x^{2}+a}\, a^{2} b c +96 \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{2}+15 \sqrt {b \,x^{2}+a}\, a^{2} b d x +60 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{3}+96 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{2}+48 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{4}+70 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{3}+48 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{4}+40 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{5}+90 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} c -15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d}{240 b} \] Input:

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

Output:

(48*sqrt(a + b*x**2)*a**3*d + 150*sqrt(a + b*x**2)*a**2*b*c*x + 48*sqrt(a 
+ b*x**2)*a**2*b*c + 96*sqrt(a + b*x**2)*a**2*b*d*x**2 + 15*sqrt(a + b*x** 
2)*a**2*b*d*x + 60*sqrt(a + b*x**2)*a*b**2*c*x**3 + 96*sqrt(a + b*x**2)*a* 
b**2*c*x**2 + 48*sqrt(a + b*x**2)*a*b**2*d*x**4 + 70*sqrt(a + b*x**2)*a*b* 
*2*d*x**3 + 48*sqrt(a + b*x**2)*b**3*c*x**4 + 40*sqrt(a + b*x**2)*b**3*d*x 
**5 + 90*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*c - 15*s 
qrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*d)/(240*b)