\(\int (A+B x) (c+d x)^3 \sqrt {a+b x^2} \, dx\) [148]

Optimal result

Integrand size = 24, antiderivative size = 256 \[ \int (A+B x) (c+d x)^3 \sqrt {a+b x^2} \, dx=\frac {\left (2 A b c \left (4 b c^2-3 a d^2\right )-a B d \left (6 b c^2-a d^2\right )\right ) x \sqrt {a+b x^2}}{16 b^2}+\frac {(B c+2 A d) (c+d x)^2 \left (a+b x^2\right )^{3/2}}{10 b}+\frac {B (c+d x)^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\left (8 \left (2 a d^2 (3 B c+A d)-b c^2 (B c+12 A d)\right )+3 d \left (5 a B d^2-2 b c (B c+7 A d)\right ) x\right ) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {a \left (2 A b c \left (4 b c^2-3 a d^2\right )-a B d \left (6 b c^2-a d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \] Output:

1/16*(2*A*b*c*(-3*a*d^2+4*b*c^2)-a*B*d*(-a*d^2+6*b*c^2))*x*(b*x^2+a)^(1/2) 
/b^2+1/10*(2*A*d+B*c)*(d*x+c)^2*(b*x^2+a)^(3/2)/b+1/6*B*(d*x+c)^3*(b*x^2+a 
)^(3/2)/b-1/120*(16*a*d^2*(A*d+3*B*c)-8*b*c^2*(12*A*d+B*c)+3*d*(5*a*B*d^2- 
2*b*c*(7*A*d+B*c))*x)*(b*x^2+a)^(3/2)/b^2+1/16*a*(2*A*b*c*(-3*a*d^2+4*b*c^ 
2)-a*B*d*(-a*d^2+6*b*c^2))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 1.63 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.94 \[ \int (A+B x) (c+d x)^3 \sqrt {a+b x^2} \, dx=\frac {\sqrt {b} \sqrt {a+b x^2} \left (-a^2 d^2 (96 B c+32 A d+15 B d x)+2 a b \left (A d \left (120 c^2+45 c d x+8 d^2 x^2\right )+B \left (40 c^3+45 c^2 d x+24 c d^2 x^2+5 d^3 x^3\right )\right )+4 b^2 x \left (3 A \left (10 c^3+20 c^2 d x+15 c d^2 x^2+4 d^3 x^3\right )+B x \left (20 c^3+45 c^2 d x+36 c d^2 x^2+10 d^3 x^3\right )\right )\right )-15 a \left (2 A b c \left (4 b c^2-3 a d^2\right )+a B d \left (-6 b c^2+a d^2\right )\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{240 b^{5/2}} \] Input:

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

Output:

(Sqrt[b]*Sqrt[a + b*x^2]*(-(a^2*d^2*(96*B*c + 32*A*d + 15*B*d*x)) + 2*a*b* 
(A*d*(120*c^2 + 45*c*d*x + 8*d^2*x^2) + B*(40*c^3 + 45*c^2*d*x + 24*c*d^2* 
x^2 + 5*d^3*x^3)) + 4*b^2*x*(3*A*(10*c^3 + 20*c^2*d*x + 15*c*d^2*x^2 + 4*d 
^3*x^3) + B*x*(20*c^3 + 45*c^2*d*x + 36*c*d^2*x^2 + 10*d^3*x^3))) - 15*a*( 
2*A*b*c*(4*b*c^2 - 3*a*d^2) + a*B*d*(-6*b*c^2 + a*d^2))*Log[-(Sqrt[b]*x) + 
 Sqrt[a + b*x^2]])/(240*b^(5/2))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {687, 27, 687, 27, 676, 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.

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

\(\Big \downarrow \) 687

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {\int b (c+d x) \left (10 A b c^2-7 a B d c-4 a A d^2-\left (5 a B d^2-2 b c (B c+7 A d)\right ) x\right ) \sqrt {b x^2+a}dx}{5 b}+\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (2 A d+B c)}{2 b}+\frac {B \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \int (c+d x) \left (10 A b c^2-7 a B d c-4 a A d^2-\left (5 a B d^2-2 b c (B c+7 A d)\right ) x\right ) \sqrt {b x^2+a}dx+\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (2 A d+B c)}{2 b}+\frac {B \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 \left (2 A b c \left (4 b c^2-3 a d^2\right )-a B d \left (6 b c^2-a d^2\right )\right ) \int \sqrt {b x^2+a}dx}{4 b}-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a d^2 (A d+3 B c)-b c^2 (12 A d+B c)\right )}{3 b}-\frac {d x \left (a+b x^2\right )^{3/2} \left (5 a B d^2-2 b c (7 A d+B c)\right )}{4 b}\right )+\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (2 A d+B c)}{2 b}+\frac {B \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 \left (2 A b c \left (4 b c^2-3 a d^2\right )-a B d \left (6 b c^2-a d^2\right )\right ) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a d^2 (A d+3 B c)-b c^2 (12 A d+B c)\right )}{3 b}-\frac {d x \left (a+b x^2\right )^{3/2} \left (5 a B d^2-2 b c (7 A d+B c)\right )}{4 b}\right )+\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (2 A d+B c)}{2 b}+\frac {B \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 \left (2 A b c \left (4 b c^2-3 a d^2\right )-a B d \left (6 b c^2-a d^2\right )\right ) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a d^2 (A d+3 B c)-b c^2 (12 A d+B c)\right )}{3 b}-\frac {d x \left (a+b x^2\right )^{3/2} \left (5 a B d^2-2 b c (7 A d+B c)\right )}{4 b}\right )+\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (2 A d+B c)}{2 b}+\frac {B \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right ) \left (2 A b c \left (4 b c^2-3 a d^2\right )-a B d \left (6 b c^2-a d^2\right )\right )}{4 b}-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a d^2 (A d+3 B c)-b c^2 (12 A d+B c)\right )}{3 b}-\frac {d x \left (a+b x^2\right )^{3/2} \left (5 a B d^2-2 b c (7 A d+B c)\right )}{4 b}\right )+\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (2 A d+B c)}{2 b}+\frac {B \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b}\)

Input:

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

Output:

(B*(c + d*x)^3*(a + b*x^2)^(3/2))/(6*b) + (((B*c + 2*A*d)*(c + d*x)^2*(a + 
 b*x^2)^(3/2))/5 + ((-2*(2*a*d^2*(3*B*c + A*d) - b*c^2*(B*c + 12*A*d))*(a 
+ b*x^2)^(3/2))/(3*b) - (d*(5*a*B*d^2 - 2*b*c*(B*c + 7*A*d))*x*(a + b*x^2) 
^(3/2))/(4*b) + (5*(2*A*b*c*(4*b*c^2 - 3*a*d^2) - a*B*d*(6*b*c^2 - a*d^2)) 
*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt 
[b])))/(4*b))/5)/(2*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 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]
 

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 = 1.24 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.04

method	result	size
default	\(A \,c^{3} \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+d^{2} \left (A d +3 B c \right ) \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )+3 c d \left (A d +B c \right ) \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )+\frac {c^{2} \left (3 A d +B c \right ) \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}+B \,d^{3} \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )\)	\(266\)
risch	\(-\frac {\left (-40 B \,d^{3} b^{2} x^{5}-48 A \,b^{2} d^{3} x^{4}-144 B \,b^{2} c \,d^{2} x^{4}-180 A \,b^{2} c \,d^{2} x^{3}-10 B a \,d^{3} b \,x^{3}-180 B \,b^{2} c^{2} d \,x^{3}-16 A \,d^{3} a b \,x^{2}-240 A \,b^{2} c^{2} d \,x^{2}-48 B a c \,d^{2} b \,x^{2}-80 B \,b^{2} c^{3} x^{2}-90 x A \,d^{2} a c b -120 A \,b^{2} c^{3} x +15 B \,a^{2} d^{3} x -90 B a \,c^{2} d x b +32 A \,a^{2} d^{3}-240 A a b \,c^{2} d +96 B \,a^{2} c \,d^{2}-80 B a b \,c^{3}\right ) \sqrt {b \,x^{2}+a}}{240 b^{2}}-\frac {a \left (6 A a b c \,d^{2}-8 A \,b^{2} c^{3}-a^{2} B \,d^{3}+6 B a b \,c^{2} d \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}\)	\(274\)

Input:

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

Output:

A*c^3*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))+ 
d^2*(A*d+3*B*c)*(1/5*x^2*(b*x^2+a)^(3/2)/b-2/15*a/b^2*(b*x^2+a)^(3/2))+3*c 
*d*(A*d+B*c)*(1/4*x*(b*x^2+a)^(3/2)/b-1/4*a/b*(1/2*x*(b*x^2+a)^(1/2)+1/2*a 
/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))))+1/3*c^2*(3*A*d+B*c)*(b*x^2+a)^(3/ 
2)/b+B*d^3*(1/6*x^3*(b*x^2+a)^(3/2)/b-1/2*a/b*(1/4*x*(b*x^2+a)^(3/2)/b-1/4 
*a/b*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 580, normalized size of antiderivative = 2.27 \[ \int (A+B x) (c+d x)^3 \sqrt {a+b x^2} \, dx=\left [\frac {15 \, {\left (8 \, A a b^{2} c^{3} - 6 \, B a^{2} b c^{2} d - 6 \, A a^{2} b c d^{2} + B a^{3} d^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (40 \, B b^{3} d^{3} x^{5} + 80 \, B a b^{2} c^{3} + 240 \, A a b^{2} c^{2} d - 96 \, B a^{2} b c d^{2} - 32 \, A a^{2} b d^{3} + 48 \, {\left (3 \, B b^{3} c d^{2} + A b^{3} d^{3}\right )} x^{4} + 10 \, {\left (18 \, B b^{3} c^{2} d + 18 \, A b^{3} c d^{2} + B a b^{2} d^{3}\right )} x^{3} + 16 \, {\left (5 \, B b^{3} c^{3} + 15 \, A b^{3} c^{2} d + 3 \, B a b^{2} c d^{2} + A a b^{2} d^{3}\right )} x^{2} + 15 \, {\left (8 \, A b^{3} c^{3} + 6 \, B a b^{2} c^{2} d + 6 \, A a b^{2} c d^{2} - B a^{2} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{480 \, b^{3}}, -\frac {15 \, {\left (8 \, A a b^{2} c^{3} - 6 \, B a^{2} b c^{2} d - 6 \, A a^{2} b c d^{2} + B a^{3} d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (40 \, B b^{3} d^{3} x^{5} + 80 \, B a b^{2} c^{3} + 240 \, A a b^{2} c^{2} d - 96 \, B a^{2} b c d^{2} - 32 \, A a^{2} b d^{3} + 48 \, {\left (3 \, B b^{3} c d^{2} + A b^{3} d^{3}\right )} x^{4} + 10 \, {\left (18 \, B b^{3} c^{2} d + 18 \, A b^{3} c d^{2} + B a b^{2} d^{3}\right )} x^{3} + 16 \, {\left (5 \, B b^{3} c^{3} + 15 \, A b^{3} c^{2} d + 3 \, B a b^{2} c d^{2} + A a b^{2} d^{3}\right )} x^{2} + 15 \, {\left (8 \, A b^{3} c^{3} + 6 \, B a b^{2} c^{2} d + 6 \, A a b^{2} c d^{2} - B a^{2} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b^{3}}\right ] \] Input:

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

Output:

[1/480*(15*(8*A*a*b^2*c^3 - 6*B*a^2*b*c^2*d - 6*A*a^2*b*c*d^2 + B*a^3*d^3) 
*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(40*B*b^3*d^3 
*x^5 + 80*B*a*b^2*c^3 + 240*A*a*b^2*c^2*d - 96*B*a^2*b*c*d^2 - 32*A*a^2*b* 
d^3 + 48*(3*B*b^3*c*d^2 + A*b^3*d^3)*x^4 + 10*(18*B*b^3*c^2*d + 18*A*b^3*c 
*d^2 + B*a*b^2*d^3)*x^3 + 16*(5*B*b^3*c^3 + 15*A*b^3*c^2*d + 3*B*a*b^2*c*d 
^2 + A*a*b^2*d^3)*x^2 + 15*(8*A*b^3*c^3 + 6*B*a*b^2*c^2*d + 6*A*a*b^2*c*d^ 
2 - B*a^2*b*d^3)*x)*sqrt(b*x^2 + a))/b^3, -1/240*(15*(8*A*a*b^2*c^3 - 6*B* 
a^2*b*c^2*d - 6*A*a^2*b*c*d^2 + B*a^3*d^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt 
(b*x^2 + a)) - (40*B*b^3*d^3*x^5 + 80*B*a*b^2*c^3 + 240*A*a*b^2*c^2*d - 96 
*B*a^2*b*c*d^2 - 32*A*a^2*b*d^3 + 48*(3*B*b^3*c*d^2 + A*b^3*d^3)*x^4 + 10* 
(18*B*b^3*c^2*d + 18*A*b^3*c*d^2 + B*a*b^2*d^3)*x^3 + 16*(5*B*b^3*c^3 + 15 
*A*b^3*c^2*d + 3*B*a*b^2*c*d^2 + A*a*b^2*d^3)*x^2 + 15*(8*A*b^3*c^3 + 6*B* 
a*b^2*c^2*d + 6*A*a*b^2*c*d^2 - B*a^2*b*d^3)*x)*sqrt(b*x^2 + a))/b^3]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (241) = 482\).

Time = 0.64 (sec) , antiderivative size = 493, normalized size of antiderivative = 1.93 \[ \int (A+B x) (c+d x)^3 \sqrt {a+b x^2} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {B d^{3} x^{5}}{6} + \frac {x^{4} \left (A b d^{3} + 3 B b c d^{2}\right )}{5 b} + \frac {x^{3} \cdot \left (3 A b c d^{2} + \frac {B a d^{3}}{6} + 3 B b c^{2} d\right )}{4 b} + \frac {x^{2} \left (A a d^{3} + 3 A b c^{2} d + 3 B a c d^{2} + B b c^{3} - \frac {4 a \left (A b d^{3} + 3 B b c d^{2}\right )}{5 b}\right )}{3 b} + \frac {x \left (3 A a c d^{2} + A b c^{3} + 3 B a c^{2} d - \frac {3 a \left (3 A b c d^{2} + \frac {B a d^{3}}{6} + 3 B b c^{2} d\right )}{4 b}\right )}{2 b} + \frac {3 A a c^{2} d + B a c^{3} - \frac {2 a \left (A a d^{3} + 3 A b c^{2} d + 3 B a c d^{2} + B b c^{3} - \frac {4 a \left (A b d^{3} + 3 B b c d^{2}\right )}{5 b}\right )}{3 b}}{b}\right ) + \left (A a c^{3} - \frac {a \left (3 A a c d^{2} + A b c^{3} + 3 B a c^{2} d - \frac {3 a \left (3 A b c d^{2} + \frac {B a d^{3}}{6} + 3 B b c^{2} d\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (A c^{3} x + \frac {B d^{3} x^{5}}{5} + \frac {x^{4} \left (A d^{3} + 3 B c d^{2}\right )}{4} + \frac {x^{3} \cdot \left (3 A c d^{2} + 3 B c^{2} d\right )}{3} + \frac {x^{2} \cdot \left (3 A c^{2} d + B c^{3}\right )}{2}\right ) & \text {otherwise} \end {cases} \] 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**5/6 + x**4*(A*b*d**3 + 3*B*b*c*d**2 
)/(5*b) + x**3*(3*A*b*c*d**2 + B*a*d**3/6 + 3*B*b*c**2*d)/(4*b) + x**2*(A* 
a*d**3 + 3*A*b*c**2*d + 3*B*a*c*d**2 + B*b*c**3 - 4*a*(A*b*d**3 + 3*B*b*c* 
d**2)/(5*b))/(3*b) + x*(3*A*a*c*d**2 + A*b*c**3 + 3*B*a*c**2*d - 3*a*(3*A* 
b*c*d**2 + B*a*d**3/6 + 3*B*b*c**2*d)/(4*b))/(2*b) + (3*A*a*c**2*d + B*a*c 
**3 - 2*a*(A*a*d**3 + 3*A*b*c**2*d + 3*B*a*c*d**2 + B*b*c**3 - 4*a*(A*b*d* 
*3 + 3*B*b*c*d**2)/(5*b))/(3*b))/b) + (A*a*c**3 - a*(3*A*a*c*d**2 + A*b*c* 
*3 + 3*B*a*c**2*d - 3*a*(3*A*b*c*d**2 + B*a*d**3/6 + 3*B*b*c**2*d)/(4*b))/ 
(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)), (sqrt(a)*(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), True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.18 \[ \int (A+B x) (c+d x)^3 \sqrt {a+b x^2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B d^{3} x^{3}}{6 \, b} + \frac {1}{2} \, \sqrt {b x^{2} + a} A c^{3} x - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a d^{3} x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B a^{2} d^{3} x}{16 \, b^{2}} + \frac {A a c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} + \frac {B a^{3} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B c^{3}}{3 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A c^{2} d}{b} + \frac {{\left (3 \, B c d^{2} + A d^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} x^{2}}{5 \, b} + \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} x}{4 \, b} - \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} \sqrt {b x^{2} + a} a x}{8 \, b} - \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} - \frac {2 \, {\left (3 \, B c d^{2} + A d^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} a}{15 \, b^{2}} \] Input:

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

Output:

1/6*(b*x^2 + a)^(3/2)*B*d^3*x^3/b + 1/2*sqrt(b*x^2 + a)*A*c^3*x - 1/8*(b*x 
^2 + a)^(3/2)*B*a*d^3*x/b^2 + 1/16*sqrt(b*x^2 + a)*B*a^2*d^3*x/b^2 + 1/2*A 
*a*c^3*arcsinh(b*x/sqrt(a*b))/sqrt(b) + 1/16*B*a^3*d^3*arcsinh(b*x/sqrt(a* 
b))/b^(5/2) + 1/3*(b*x^2 + a)^(3/2)*B*c^3/b + (b*x^2 + a)^(3/2)*A*c^2*d/b 
+ 1/5*(3*B*c*d^2 + A*d^3)*(b*x^2 + a)^(3/2)*x^2/b + 3/4*(B*c^2*d + A*c*d^2 
)*(b*x^2 + a)^(3/2)*x/b - 3/8*(B*c^2*d + A*c*d^2)*sqrt(b*x^2 + a)*a*x/b - 
3/8*(B*c^2*d + A*c*d^2)*a^2*arcsinh(b*x/sqrt(a*b))/b^(3/2) - 2/15*(3*B*c*d 
^2 + A*d^3)*(b*x^2 + a)^(3/2)*a/b^2
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.18 \[ \int (A+B x) (c+d x)^3 \sqrt {a+b x^2} \, dx=\frac {1}{240} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, B d^{3} x + \frac {6 \, {\left (3 \, B b^{4} c d^{2} + A b^{4} d^{3}\right )}}{b^{4}}\right )} x + \frac {5 \, {\left (18 \, B b^{4} c^{2} d + 18 \, A b^{4} c d^{2} + B a b^{3} d^{3}\right )}}{b^{4}}\right )} x + \frac {8 \, {\left (5 \, B b^{4} c^{3} + 15 \, A b^{4} c^{2} d + 3 \, B a b^{3} c d^{2} + A a b^{3} d^{3}\right )}}{b^{4}}\right )} x + \frac {15 \, {\left (8 \, A b^{4} c^{3} + 6 \, B a b^{3} c^{2} d + 6 \, A a b^{3} c d^{2} - B a^{2} b^{2} d^{3}\right )}}{b^{4}}\right )} x + \frac {16 \, {\left (5 \, B a b^{3} c^{3} + 15 \, A a b^{3} c^{2} d - 6 \, B a^{2} b^{2} c d^{2} - 2 \, A a^{2} b^{2} d^{3}\right )}}{b^{4}}\right )} - \frac {{\left (8 \, A a b^{2} c^{3} - 6 \, B a^{2} b c^{2} d - 6 \, A a^{2} b c d^{2} + B a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} \] Input:

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

Output:

1/240*sqrt(b*x^2 + a)*((2*((4*(5*B*d^3*x + 6*(3*B*b^4*c*d^2 + A*b^4*d^3)/b 
^4)*x + 5*(18*B*b^4*c^2*d + 18*A*b^4*c*d^2 + B*a*b^3*d^3)/b^4)*x + 8*(5*B* 
b^4*c^3 + 15*A*b^4*c^2*d + 3*B*a*b^3*c*d^2 + A*a*b^3*d^3)/b^4)*x + 15*(8*A 
*b^4*c^3 + 6*B*a*b^3*c^2*d + 6*A*a*b^3*c*d^2 - B*a^2*b^2*d^3)/b^4)*x + 16* 
(5*B*a*b^3*c^3 + 15*A*a*b^3*c^2*d - 6*B*a^2*b^2*c*d^2 - 2*A*a^2*b^2*d^3)/b 
^4) - 1/16*(8*A*a*b^2*c^3 - 6*B*a^2*b*c^2*d - 6*A*a^2*b*c*d^2 + B*a^3*d^3) 
*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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