\(\int \frac {(A+B x) (c+d x)^3}{\sqrt {a+b x^2}} \, dx\) [170]

Optimal result

Integrand size = 24, antiderivative size = 201 \[ \int \frac {(A+B x) (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\frac {(3 B c+4 A d) (c+d x)^2 \sqrt {a+b x^2}}{12 b}+\frac {B (c+d x)^3 \sqrt {a+b x^2}}{4 b}-\frac {\left (4 \left (4 a d^2 (3 B c+A d)-b c^2 (3 B c+16 A d)\right )+d \left (9 a B d^2-2 b c (3 B c+10 A d)\right ) x\right ) \sqrt {a+b x^2}}{24 b^2}+\frac {\left (4 A b c \left (2 b c^2-3 a d^2\right )-3 a B d \left (4 b c^2-a d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] Output:

1/12*(4*A*d+3*B*c)*(d*x+c)^2*(b*x^2+a)^(1/2)/b+1/4*B*(d*x+c)^3*(b*x^2+a)^( 
1/2)/b-1/24*(16*a*d^2*(A*d+3*B*c)-4*b*c^2*(16*A*d+3*B*c)+d*(9*a*B*d^2-2*b* 
c*(10*A*d+3*B*c))*x)*(b*x^2+a)^(1/2)/b^2+1/8*(4*A*b*c*(-3*a*d^2+2*b*c^2)-3 
*a*B*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 = 1.47 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-a d^2 (48 B c+16 A d+9 B d x)+4 A b d \left (18 c^2+9 c d x+2 d^2 x^2\right )+6 b B \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )\right )}{24 b^2}-\frac {\left (4 A b c \left (2 b c^2-3 a d^2\right )+3 a B d \left (-4 b c^2+a d^2\right )\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{5/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(-(a*d^2*(48*B*c + 16*A*d + 9*B*d*x)) + 4*A*b*d*(18*c^2 + 
 9*c*d*x + 2*d^2*x^2) + 6*b*B*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)) 
)/(24*b^2) - ((4*A*b*c*(2*b*c^2 - 3*a*d^2) + 3*a*B*d*(-4*b*c^2 + a*d^2))*L 
og[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*b^(5/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {687, 687, 27, 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.

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

\(\Big \downarrow \) 687

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

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {\int \frac {b (c+d x) \left (12 A b c^2-15 a B d c-8 a A d^2-\left (9 a B d^2-2 b c (3 B c+10 A d)\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b}+\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 (4 A d+3 B c)}{4 b}+\frac {B \sqrt {a+b x^2} (c+d x)^3}{4 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \frac {(c+d x) \left (12 A b c^2-15 a B d c-8 a A d^2-\left (9 a B d^2-2 b c (3 B c+10 A d)\right ) x\right )}{\sqrt {b x^2+a}}dx+\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 (4 A d+3 B c)}{4 b}+\frac {B \sqrt {a+b x^2} (c+d x)^3}{4 b}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 \left (4 A b c \left (2 b c^2-3 a d^2\right )-3 a B d \left (4 b c^2-a d^2\right )\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}-\frac {2 \sqrt {a+b x^2} \left (4 a d^2 (A d+3 B c)-b c^2 (16 A d+3 B c)\right )}{b}-\frac {d x \sqrt {a+b x^2} \left (9 a B d^2-2 b c (10 A d+3 B c)\right )}{2 b}\right )+\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 (4 A d+3 B c)}{4 b}+\frac {B \sqrt {a+b x^2} (c+d x)^3}{4 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 \left (4 A b c \left (2 b c^2-3 a d^2\right )-3 a B d \left (4 b c^2-a d^2\right )\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}-\frac {2 \sqrt {a+b x^2} \left (4 a d^2 (A d+3 B c)-b c^2 (16 A d+3 B c)\right )}{b}-\frac {d x \sqrt {a+b x^2} \left (9 a B d^2-2 b c (10 A d+3 B c)\right )}{2 b}\right )+\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 (4 A d+3 B c)}{4 b}+\frac {B \sqrt {a+b x^2} (c+d x)^3}{4 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (4 A b c \left (2 b c^2-3 a d^2\right )-3 a B d \left (4 b c^2-a d^2\right )\right )}{2 b^{3/2}}-\frac {2 \sqrt {a+b x^2} \left (4 a d^2 (A d+3 B c)-b c^2 (16 A d+3 B c)\right )}{b}-\frac {d x \sqrt {a+b x^2} \left (9 a B d^2-2 b c (10 A d+3 B c)\right )}{2 b}\right )+\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 (4 A d+3 B c)}{4 b}+\frac {B \sqrt {a+b x^2} (c+d x)^3}{4 b}\)

Input:

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

Output:

(B*(c + d*x)^3*Sqrt[a + b*x^2])/(4*b) + (((3*B*c + 4*A*d)*(c + d*x)^2*Sqrt 
[a + b*x^2])/3 + ((-2*(4*a*d^2*(3*B*c + A*d) - b*c^2*(3*B*c + 16*A*d))*Sqr 
t[a + b*x^2])/b - (d*(9*a*B*d^2 - 2*b*c*(3*B*c + 10*A*d))*x*Sqrt[a + b*x^2 
])/(2*b) + (3*(4*A*b*c*(2*b*c^2 - 3*a*d^2) - 3*a*B*d*(4*b*c^2 - a*d^2))*Ar 
cTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)))/3)/(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 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]
 

rule 687

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2)) 
), x] + Simp[1/(c*(m + 2*p + 2))   Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*Simp 
[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x 
] /; FreeQ[{a, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && 
 (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) &&  !(IGtQ[m, 0] && Eq 
Q[f, 0])
 

Maple [A] (verified)

Time = 1.21 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.81

method	result	size
risch	\(-\frac {\left (-6 B b \,d^{3} x^{3}-8 A b \,d^{3} x^{2}-24 B b c \,d^{2} x^{2}-36 A b c \,d^{2} x +9 B a \,d^{3} x -36 B b \,c^{2} d x +16 A \,d^{3} a -72 A b \,c^{2} d +48 a B c \,d^{2}-24 b B \,c^{3}\right ) \sqrt {b \,x^{2}+a}}{24 b^{2}}-\frac {\left (12 A a b c \,d^{2}-8 A \,b^{2} c^{3}-3 a^{2} B \,d^{3}+12 B a b \,c^{2} d \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\)	\(162\)
default	\(\frac {A \,c^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+d^{2} \left (A d +3 B c \right ) \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+3 c d \left (A d +B c \right ) \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+\frac {c^{2} \left (3 A d +B c \right ) \sqrt {b \,x^{2}+a}}{b}+B \,d^{3} \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )\)	\(211\)

Input:

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

Output:

-1/24*(-6*B*b*d^3*x^3-8*A*b*d^3*x^2-24*B*b*c*d^2*x^2-36*A*b*c*d^2*x+9*B*a* 
d^3*x-36*B*b*c^2*d*x+16*A*a*d^3-72*A*b*c^2*d+48*B*a*c*d^2-24*B*b*c^3)*(b*x 
^2+a)^(1/2)/b^2-1/8*(12*A*a*b*c*d^2-8*A*b^2*c^3-3*B*a^2*d^3+12*B*a*b*c^2*d 
)/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.86 \[ \int \frac {(A+B x) (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\left [\frac {3 \, {\left (8 \, A b^{2} c^{3} - 12 \, B a b c^{2} d - 12 \, A a b c d^{2} + 3 \, B a^{2} d^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, B b^{2} d^{3} x^{3} + 24 \, B b^{2} c^{3} + 72 \, A b^{2} c^{2} d - 48 \, B a b c d^{2} - 16 \, A a b d^{3} + 8 \, {\left (3 \, B b^{2} c d^{2} + A b^{2} d^{3}\right )} x^{2} + 9 \, {\left (4 \, B b^{2} c^{2} d + 4 \, A b^{2} c d^{2} - B a b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}, -\frac {3 \, {\left (8 \, A b^{2} c^{3} - 12 \, B a b c^{2} d - 12 \, A a b c d^{2} + 3 \, B a^{2} d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (6 \, B b^{2} d^{3} x^{3} + 24 \, B b^{2} c^{3} + 72 \, A b^{2} c^{2} d - 48 \, B a b c d^{2} - 16 \, A a b d^{3} + 8 \, {\left (3 \, B b^{2} c d^{2} + A b^{2} d^{3}\right )} x^{2} + 9 \, {\left (4 \, B b^{2} c^{2} d + 4 \, A b^{2} c d^{2} - B a b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{24 \, b^{3}}\right ] \] Input:

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

Output:

[1/48*(3*(8*A*b^2*c^3 - 12*B*a*b*c^2*d - 12*A*a*b*c*d^2 + 3*B*a^2*d^3)*sqr 
t(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(6*B*b^2*d^3*x^3 
+ 24*B*b^2*c^3 + 72*A*b^2*c^2*d - 48*B*a*b*c*d^2 - 16*A*a*b*d^3 + 8*(3*B*b 
^2*c*d^2 + A*b^2*d^3)*x^2 + 9*(4*B*b^2*c^2*d + 4*A*b^2*c*d^2 - B*a*b*d^3)* 
x)*sqrt(b*x^2 + a))/b^3, -1/24*(3*(8*A*b^2*c^3 - 12*B*a*b*c^2*d - 12*A*a*b 
*c*d^2 + 3*B*a^2*d^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (6*B*b 
^2*d^3*x^3 + 24*B*b^2*c^3 + 72*A*b^2*c^2*d - 48*B*a*b*c*d^2 - 16*A*a*b*d^3 
 + 8*(3*B*b^2*c*d^2 + A*b^2*d^3)*x^2 + 9*(4*B*b^2*c^2*d + 4*A*b^2*c*d^2 - 
B*a*b*d^3)*x)*sqrt(b*x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.36 \[ \int \frac {(A+B x) (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {B d^{3} x^{3}}{4 b} + \frac {x^{2} \left (A d^{3} + 3 B c d^{2}\right )}{3 b} + \frac {x \left (3 A c d^{2} - \frac {3 B a d^{3}}{4 b} + 3 B c^{2} d\right )}{2 b} + \frac {3 A c^{2} d + B c^{3} - \frac {2 a \left (A d^{3} + 3 B c d^{2}\right )}{3 b}}{b}\right ) + \left (A c^{3} - \frac {a \left (3 A c d^{2} - \frac {3 B a d^{3}}{4 b} + 3 B c^{2} d\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\frac {A c^{3} x + \frac {B d^{3} x^{5}}{5} + \frac {x^{4} \left (A d^{3} + 3 B c d^{2}\right )}{4} + \frac {x^{3} \cdot \left (3 A c d^{2} + 3 B c^{2} d\right )}{3} + \frac {x^{2} \cdot \left (3 A c^{2} d + B c^{3}\right )}{2}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.14 \[ \int \frac {(A+B x) (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} B d^{3} x^{3}}{4 \, b} - \frac {3 \, \sqrt {b x^{2} + a} B a d^{3} x}{8 \, b^{2}} + \frac {A c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} + \frac {3 \, B a^{2} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} + \frac {\sqrt {b x^{2} + a} B c^{3}}{b} + \frac {3 \, \sqrt {b x^{2} + a} A c^{2} d}{b} + \frac {{\left (3 \, B c d^{2} + A d^{3}\right )} \sqrt {b x^{2} + a} x^{2}}{3 \, b} + \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} \sqrt {b x^{2} + a} x}{2 \, b} - \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} - \frac {2 \, {\left (3 \, B c d^{2} + A d^{3}\right )} \sqrt {b x^{2} + a} a}{3 \, b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) (c+d x)^3}{\sqrt {a+b x^2}} \, dx=\frac {1}{24} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (\frac {3 \, B d^{3} x}{b} + \frac {4 \, {\left (3 \, B b^{3} c d^{2} + A b^{3} d^{3}\right )}}{b^{4}}\right )} x + \frac {9 \, {\left (4 \, B b^{3} c^{2} d + 4 \, A b^{3} c d^{2} - B a b^{2} d^{3}\right )}}{b^{4}}\right )} x + \frac {8 \, {\left (3 \, B b^{3} c^{3} + 9 \, A b^{3} c^{2} d - 6 \, B a b^{2} c d^{2} - 2 \, A a b^{2} d^{3}\right )}}{b^{4}}\right )} - \frac {{\left (8 \, A b^{2} c^{3} - 12 \, B a b c^{2} d - 12 \, A a b c d^{2} + 3 \, B a^{2} d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \] Input:

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

Output:

1/24*sqrt(b*x^2 + a)*((2*(3*B*d^3*x/b + 4*(3*B*b^3*c*d^2 + A*b^3*d^3)/b^4) 
*x + 9*(4*B*b^3*c^2*d + 4*A*b^3*c*d^2 - B*a*b^2*d^3)/b^4)*x + 8*(3*B*b^3*c 
^3 + 9*A*b^3*c^2*d - 6*B*a*b^2*c*d^2 - 2*A*a*b^2*d^3)/b^4) - 1/8*(8*A*b^2* 
c^3 - 12*B*a*b*c^2*d - 12*A*a*b*c*d^2 + 3*B*a^2*d^3)*log(abs(-sqrt(b)*x + 
sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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