\(\int (A+B x) (c+d x)^2 \sqrt {a+b x^2} \, dx\) [149]

Optimal result

Integrand size = 24, antiderivative size = 171 \[ \int (A+B x) (c+d x)^2 \sqrt {a+b x^2} \, dx=\frac {\left (4 A b c^2-2 a B c d-a A d^2\right ) x \sqrt {a+b x^2}}{8 b}+\frac {B (c+d x)^2 \left (a+b x^2\right )^{3/2}}{5 b}-\frac {\left (8 \left (a B d^2-b c (B c+5 A d)\right )-3 b d (2 B c+5 A d) x\right ) \left (a+b x^2\right )^{3/2}}{60 b^2}+\frac {a \left (4 A b c^2-2 a B c d-a A d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \] Output:

1/8*(-A*a*d^2+4*A*b*c^2-2*B*a*c*d)*x*(b*x^2+a)^(1/2)/b+1/5*B*(d*x+c)^2*(b* 
x^2+a)^(3/2)/b-1/60*(8*a*B*d^2-8*b*c*(5*A*d+B*c)-3*b*d*(5*A*d+2*B*c)*x)*(b 
*x^2+a)^(3/2)/b^2+1/8*a*(-A*a*d^2+4*A*b*c^2-2*B*a*c*d)*arctanh(b^(1/2)*x/( 
b*x^2+a)^(1/2))/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.98 \[ \int (A+B x) (c+d x)^2 \sqrt {a+b x^2} \, dx=\frac {\sqrt {a+b x^2} \left (-16 a^2 B d^2+2 b^2 x \left (5 A \left (6 c^2+8 c d x+3 d^2 x^2\right )+2 B x \left (10 c^2+15 c d x+6 d^2 x^2\right )\right )+a b \left (5 A d (16 c+3 d x)+B \left (40 c^2+30 c d x+8 d^2 x^2\right )\right )\right )+15 a \sqrt {b} \left (-4 A b c^2+2 a B c d+a A d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{120 b^2} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(-16*a^2*B*d^2 + 2*b^2*x*(5*A*(6*c^2 + 8*c*d*x + 3*d^2*x^ 
2) + 2*B*x*(10*c^2 + 15*c*d*x + 6*d^2*x^2)) + a*b*(5*A*d*(16*c + 3*d*x) + 
B*(40*c^2 + 30*c*d*x + 8*d^2*x^2))) + 15*a*Sqrt[b]*(-4*A*b*c^2 + 2*a*B*c*d 
 + a*A*d^2)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(120*b^2)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {687, 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.

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

\(\Big \downarrow \) 687

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

\(\Big \downarrow \) 676

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 = 1.07 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.99

method	result	size
default	\(A \,c^{2} \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+d \left (A d +2 B c \right ) \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )+\frac {c \left (2 A d +B c \right ) \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}+B \,d^{2} \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )\)	\(170\)
risch	\(\frac {\left (24 B \,d^{2} b^{2} x^{4}+30 A \,b^{2} d^{2} x^{3}+60 B \,b^{2} c d \,x^{3}+80 A \,b^{2} c d \,x^{2}+8 B a b \,d^{2} x^{2}+40 c^{2} x^{2} B \,b^{2}+15 A a \,d^{2} x b +60 A \,b^{2} c^{2} x +30 B a c d x b +80 A a b c d -16 a^{2} B \,d^{2}+40 B a b \,c^{2}\right ) \sqrt {b \,x^{2}+a}}{120 b^{2}}-\frac {a \left (A a \,d^{2}-4 A b \,c^{2}+2 B a c d \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {3}{2}}}\)	\(179\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 374, normalized size of antiderivative = 2.19 \[ \int (A+B x) (c+d x)^2 \sqrt {a+b x^2} \, dx=\left [-\frac {15 \, {\left (4 \, A a b c^{2} - 2 \, B a^{2} c d - A a^{2} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (24 \, B b^{2} d^{2} x^{4} + 40 \, B a b c^{2} + 80 \, A a b c d - 16 \, B a^{2} d^{2} + 30 \, {\left (2 \, B b^{2} c d + A b^{2} d^{2}\right )} x^{3} + 8 \, {\left (5 \, B b^{2} c^{2} + 10 \, A b^{2} c d + B a b d^{2}\right )} x^{2} + 15 \, {\left (4 \, A b^{2} c^{2} + 2 \, B a b c d + A a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b^{2}}, -\frac {15 \, {\left (4 \, A a b c^{2} - 2 \, B a^{2} c d - A a^{2} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (24 \, B b^{2} d^{2} x^{4} + 40 \, B a b c^{2} + 80 \, A a b c d - 16 \, B a^{2} d^{2} + 30 \, {\left (2 \, B b^{2} c d + A b^{2} d^{2}\right )} x^{3} + 8 \, {\left (5 \, B b^{2} c^{2} + 10 \, A b^{2} c d + B a b d^{2}\right )} x^{2} + 15 \, {\left (4 \, A b^{2} c^{2} + 2 \, B a b c d + A a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{120 \, b^{2}}\right ] \] Input:

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

Output:

[-1/240*(15*(4*A*a*b*c^2 - 2*B*a^2*c*d - A*a^2*d^2)*sqrt(b)*log(-2*b*x^2 + 
 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(24*B*b^2*d^2*x^4 + 40*B*a*b*c^2 + 8 
0*A*a*b*c*d - 16*B*a^2*d^2 + 30*(2*B*b^2*c*d + A*b^2*d^2)*x^3 + 8*(5*B*b^2 
*c^2 + 10*A*b^2*c*d + B*a*b*d^2)*x^2 + 15*(4*A*b^2*c^2 + 2*B*a*b*c*d + A*a 
*b*d^2)*x)*sqrt(b*x^2 + a))/b^2, -1/120*(15*(4*A*a*b*c^2 - 2*B*a^2*c*d - A 
*a^2*d^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (24*B*b^2*d^2*x^4 
+ 40*B*a*b*c^2 + 80*A*a*b*c*d - 16*B*a^2*d^2 + 30*(2*B*b^2*c*d + A*b^2*d^2 
)*x^3 + 8*(5*B*b^2*c^2 + 10*A*b^2*c*d + B*a*b*d^2)*x^2 + 15*(4*A*b^2*c^2 + 
 2*B*a*b*c*d + A*a*b*d^2)*x)*sqrt(b*x^2 + a))/b^2]
 

Sympy [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.89 \[ \int (A+B x) (c+d x)^2 \sqrt {a+b x^2} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {B d^{2} x^{4}}{5} + \frac {x^{3} \left (A b d^{2} + 2 B b c d\right )}{4 b} + \frac {x^{2} \cdot \left (2 A b c d + \frac {B a d^{2}}{5} + B b c^{2}\right )}{3 b} + \frac {x \left (A a d^{2} + A b c^{2} + 2 B a c d - \frac {3 a \left (A b d^{2} + 2 B b c d\right )}{4 b}\right )}{2 b} + \frac {2 A a c d + B a c^{2} - \frac {2 a \left (2 A b c d + \frac {B a d^{2}}{5} + B b c^{2}\right )}{3 b}}{b}\right ) + \left (A a c^{2} - \frac {a \left (A a d^{2} + A b c^{2} + 2 B a c d - \frac {3 a \left (A b d^{2} + 2 B b c d\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (A c^{2} x + \frac {B d^{2} x^{4}}{4} + \frac {x^{3} \left (A d^{2} + 2 B c d\right )}{3} + \frac {x^{2} \cdot \left (2 A c d + B c^{2}\right )}{2}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.12 \[ \int (A+B x) (c+d x)^2 \sqrt {a+b x^2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B d^{2} x^{2}}{5 \, b} + \frac {1}{2} \, \sqrt {b x^{2} + a} A c^{2} x + \frac {A a c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B c^{2}}{3 \, b} + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A c d}{3 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a d^{2}}{15 \, b^{2}} + \frac {{\left (2 \, B c d + A d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} x}{4 \, b} - \frac {{\left (2 \, B c d + A d^{2}\right )} \sqrt {b x^{2} + a} a x}{8 \, b} - \frac {{\left (2 \, B c d + A d^{2}\right )} a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.18 \[ \int (A+B x) (c+d x)^2 \sqrt {a+b x^2} \, dx=\frac {1}{120} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (4 \, B d^{2} x + \frac {5 \, {\left (2 \, B b^{3} c d + A b^{3} d^{2}\right )}}{b^{3}}\right )} x + \frac {4 \, {\left (5 \, B b^{3} c^{2} + 10 \, A b^{3} c d + B a b^{2} d^{2}\right )}}{b^{3}}\right )} x + \frac {15 \, {\left (4 \, A b^{3} c^{2} + 2 \, B a b^{2} c d + A a b^{2} d^{2}\right )}}{b^{3}}\right )} x + \frac {8 \, {\left (5 \, B a b^{2} c^{2} + 10 \, A a b^{2} c d - 2 \, B a^{2} b d^{2}\right )}}{b^{3}}\right )} - \frac {{\left (4 \, A a b c^{2} - 2 \, B a^{2} c d - A a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} \] Input:

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

Output:

1/120*sqrt(b*x^2 + a)*((2*(3*(4*B*d^2*x + 5*(2*B*b^3*c*d + A*b^3*d^2)/b^3) 
*x + 4*(5*B*b^3*c^2 + 10*A*b^3*c*d + B*a*b^2*d^2)/b^3)*x + 15*(4*A*b^3*c^2 
 + 2*B*a*b^2*c*d + A*a*b^2*d^2)/b^3)*x + 8*(5*B*a*b^2*c^2 + 10*A*a*b^2*c*d 
 - 2*B*a^2*b*d^2)/b^3) - 1/8*(4*A*a*b*c^2 - 2*B*a^2*c*d - A*a^2*d^2)*log(a 
bs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.82 \[ \int (A+B x) (c+d x)^2 \sqrt {a+b x^2} \, dx=\frac {80 \sqrt {b \,x^{2}+a}\, a^{2} b c d +15 \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{2} x -16 \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{2}+60 \sqrt {b \,x^{2}+a}\, a \,b^{2} c^{2} x +40 \sqrt {b \,x^{2}+a}\, a \,b^{2} c^{2}+80 \sqrt {b \,x^{2}+a}\, a \,b^{2} c d \,x^{2}+30 \sqrt {b \,x^{2}+a}\, a \,b^{2} c d x +30 \sqrt {b \,x^{2}+a}\, a \,b^{2} d^{2} x^{3}+8 \sqrt {b \,x^{2}+a}\, a \,b^{2} d^{2} x^{2}+40 \sqrt {b \,x^{2}+a}\, b^{3} c^{2} x^{2}+60 \sqrt {b \,x^{2}+a}\, b^{3} c d \,x^{3}+24 \sqrt {b \,x^{2}+a}\, b^{3} d^{2} x^{4}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d^{2}+60 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b \,c^{2}-30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b c d}{120 b^{2}} \] Input:

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

Output:

(80*sqrt(a + b*x**2)*a**2*b*c*d + 15*sqrt(a + b*x**2)*a**2*b*d**2*x - 16*s 
qrt(a + b*x**2)*a**2*b*d**2 + 60*sqrt(a + b*x**2)*a*b**2*c**2*x + 40*sqrt( 
a + b*x**2)*a*b**2*c**2 + 80*sqrt(a + b*x**2)*a*b**2*c*d*x**2 + 30*sqrt(a 
+ b*x**2)*a*b**2*c*d*x + 30*sqrt(a + b*x**2)*a*b**2*d**2*x**3 + 8*sqrt(a + 
 b*x**2)*a*b**2*d**2*x**2 + 40*sqrt(a + b*x**2)*b**3*c**2*x**2 + 60*sqrt(a 
 + b*x**2)*b**3*c*d*x**3 + 24*sqrt(a + b*x**2)*b**3*d**2*x**4 - 15*sqrt(b) 
*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*d**2 + 60*sqrt(b)*log((s 
qrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*c**2 - 30*sqrt(b)*log((sqrt(a 
 + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*c*d)/(120*b**2)