\(\int \frac {(c+d x) (A+B x+C x^2+D x^3)}{\sqrt {a+b x^2}} \, dx\) [103]

Optimal result

Integrand size = 32, antiderivative size = 173 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {(b B c+A b d-a C d-a c D) \sqrt {a+b x^2}}{b^2}+\frac {(4 b (c C+B d)-3 a d D) x \sqrt {a+b x^2}}{8 b^2}+\frac {d D x^3 \sqrt {a+b x^2}}{4 b}+\frac {(C d+c D) \left (a+b x^2\right )^{3/2}}{3 b^2}+\frac {\left (8 A b^2 c-a (4 b (c C+B d)-3 a d D)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.78 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-a (16 C d+16 c D+9 d D x)+2 b \left (12 A d+6 B (2 c+d x)+x \left (6 c C+4 C d x+4 c D x+3 d D x^2\right )\right )\right )}{24 b^2}-\frac {\left (8 A b^2 c+a (-4 b (c C+B d)+3 a d D)\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{5/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(-(a*(16*C*d + 16*c*D + 9*d*D*x)) + 2*b*(12*A*d + 6*B*(2* 
c + d*x) + x*(6*c*C + 4*C*d*x + 4*c*D*x + 3*d*D*x^2))))/(24*b^2) - ((8*A*b 
^2*c + a*(-4*b*(c*C + B*d) + 3*a*d*D))*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]] 
)/(8*b^(5/2))
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.42, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2185, 2185, 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.

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {\int \frac {b d^3 (c+d x) \left (d (12 A b d-8 a C d+a c D)-\left (9 a D d^2+2 b \left (-D c^2+2 C d c-6 B d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b d^2}+\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (4 C d-5 c D)}{4 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} d \int \frac {(c+d x) \left (d (12 A b d-8 a C d+a c D)-\left (9 a D d^2+2 b \left (-D c^2+2 C d c-6 B d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx+\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (4 C d-5 c D)}{4 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {1}{3} d \left (\frac {3 d^2 \left (8 A b^2 c-a (4 b (B d+c C)-3 a d D)\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}-\frac {2 \sqrt {a+b x^2} \left (4 a d^2 (c D+C d)+b \left (-6 A d^3-6 B c d^2+c^3 (-D)+2 c^2 C d\right )\right )}{b}-\frac {d x \sqrt {a+b x^2} \left (9 a d^2 D+2 b \left (-6 B d^2+c^2 (-D)+2 c C d\right )\right )}{2 b}\right )+\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (4 C d-5 c D)}{4 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{3} d \left (\frac {3 d^2 \left (8 A b^2 c-a (4 b (B d+c C)-3 a d D)\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}-\frac {2 \sqrt {a+b x^2} \left (4 a d^2 (c D+C d)+b \left (-6 A d^3-6 B c d^2+c^3 (-D)+2 c^2 C d\right )\right )}{b}-\frac {d x \sqrt {a+b x^2} \left (9 a d^2 D+2 b \left (-6 B d^2+c^2 (-D)+2 c C d\right )\right )}{2 b}\right )+\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (4 C d-5 c D)}{4 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{3} d \left (\frac {3 d^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (8 A b^2 c-a (4 b (B d+c C)-3 a d D)\right )}{2 b^{3/2}}-\frac {2 \sqrt {a+b x^2} \left (4 a d^2 (c D+C d)+b \left (-6 A d^3-6 B c d^2+c^3 (-D)+2 c^2 C d\right )\right )}{b}-\frac {d x \sqrt {a+b x^2} \left (9 a d^2 D+2 b \left (-6 B d^2+c^2 (-D)+2 c C d\right )\right )}{2 b}\right )+\frac {1}{3} d \sqrt {a+b x^2} (c+d x)^2 (4 C d-5 c D)}{4 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^3}{4 b d^2}\)

Input:

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

Output:

(D*(c + d*x)^3*Sqrt[a + b*x^2])/(4*b*d^2) + ((d*(4*C*d - 5*c*D)*(c + d*x)^ 
2*Sqrt[a + b*x^2])/3 + (d*((-2*(4*a*d^2*(C*d + c*D) + b*(2*c^2*C*d - 6*B*c 
*d^2 - 6*A*d^3 - c^3*D))*Sqrt[a + b*x^2])/b - (d*(9*a*d^2*D + 2*b*(2*c*C*d 
 - 6*B*d^2 - c^2*D))*x*Sqrt[a + b*x^2])/(2*b) + (3*d^2*(8*A*b^2*c - a*(4*b 
*(c*C + B*d) - 3*a*d*D))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)) 
))/3)/(4*b*d^3)
 

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 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {A c \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+\frac {\left (A d +B c \right ) \sqrt {b \,x^{2}+a}}{b}+\left (B d +C c \right ) \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+\left (C d +D c \right ) \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+D d \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )\)	\(196\)

Input:

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

Output:

A*c*ln(b^(1/2)*x+(b*x^2+a)^(1/2))/b^(1/2)+(A*d+B*c)/b*(b*x^2+a)^(1/2)+(B*d 
+C*c)*(1/2*x/b*(b*x^2+a)^(1/2)-1/2*a/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)) 
)+(C*d+D*c)*(1/3*x^2/b*(b*x^2+a)^(1/2)-2/3*a/b^2*(b*x^2+a)^(1/2))+D*d*(1/4 
*x^3/b*(b*x^2+a)^(1/2)-3/4*a/b*(1/2*x/b*(b*x^2+a)^(1/2)-1/2*a/b^(3/2)*ln(b 
^(1/2)*x+(b*x^2+a)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.84 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\left [\frac {3 \, {\left (4 \, {\left (C a b - 2 \, A b^{2}\right )} c - {\left (3 \, D a^{2} - 4 \, B a b\right )} d\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, D b^{2} d x^{3} + 8 \, {\left (D b^{2} c + C b^{2} d\right )} x^{2} - 8 \, {\left (2 \, D a b - 3 \, B b^{2}\right )} c - 8 \, {\left (2 \, C a b - 3 \, A b^{2}\right )} d + 3 \, {\left (4 \, C b^{2} c - {\left (3 \, D a b - 4 \, B b^{2}\right )} d\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}, \frac {3 \, {\left (4 \, {\left (C a b - 2 \, A b^{2}\right )} c - {\left (3 \, D a^{2} - 4 \, B a b\right )} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, D b^{2} d x^{3} + 8 \, {\left (D b^{2} c + C b^{2} d\right )} x^{2} - 8 \, {\left (2 \, D a b - 3 \, B b^{2}\right )} c - 8 \, {\left (2 \, C a b - 3 \, A b^{2}\right )} d + 3 \, {\left (4 \, C b^{2} c - {\left (3 \, D a b - 4 \, B b^{2}\right )} d\right )} x\right )} \sqrt {b x^{2} + a}}{24 \, b^{3}}\right ] \] Input:

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

Output:

[1/48*(3*(4*(C*a*b - 2*A*b^2)*c - (3*D*a^2 - 4*B*a*b)*d)*sqrt(b)*log(-2*b* 
x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(6*D*b^2*d*x^3 + 8*(D*b^2*c + C 
*b^2*d)*x^2 - 8*(2*D*a*b - 3*B*b^2)*c - 8*(2*C*a*b - 3*A*b^2)*d + 3*(4*C*b 
^2*c - (3*D*a*b - 4*B*b^2)*d)*x)*sqrt(b*x^2 + a))/b^3, 1/24*(3*(4*(C*a*b - 
 2*A*b^2)*c - (3*D*a^2 - 4*B*a*b)*d)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 
 + a)) + (6*D*b^2*d*x^3 + 8*(D*b^2*c + C*b^2*d)*x^2 - 8*(2*D*a*b - 3*B*b^2 
)*c - 8*(2*C*a*b - 3*A*b^2)*d + 3*(4*C*b^2*c - (3*D*a*b - 4*B*b^2)*d)*x)*s 
qrt(b*x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.15 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {D d x^{3}}{4 b} + \frac {x^{2} \left (C d + D c\right )}{3 b} + \frac {x \left (B d + C c - \frac {3 D a d}{4 b}\right )}{2 b} + \frac {A d + B c - \frac {2 a \left (C d + D c\right )}{3 b}}{b}\right ) + \left (A c - \frac {a \left (B d + C c - \frac {3 D a d}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\frac {A c x + \frac {D d x^{5}}{5} + \frac {x^{4} \left (C d + D c\right )}{4} + \frac {x^{3} \left (B d + C c\right )}{3} + \frac {x^{2} \left (A d + B c\right )}{2}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.11 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} D d x^{3}}{4 \, b} - \frac {3 \, \sqrt {b x^{2} + a} D a d x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} {\left (D c + C d\right )} x^{2}}{3 \, b} + \frac {A c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} + \frac {3 \, D a^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} + \frac {\sqrt {b x^{2} + a} B c}{b} + \frac {\sqrt {b x^{2} + a} A d}{b} + \frac {\sqrt {b x^{2} + a} {\left (C c + B d\right )} x}{2 \, b} - \frac {{\left (C c + B d\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} - \frac {2 \, \sqrt {b x^{2} + a} {\left (D c + C d\right )} a}{3 \, b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.92 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{24} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left (\frac {3 \, D d x}{b} + \frac {4 \, {\left (D b^{3} c + C b^{3} d\right )}}{b^{4}}\right )} x + \frac {3 \, {\left (4 \, C b^{3} c - 3 \, D a b^{2} d + 4 \, B b^{3} d\right )}}{b^{4}}\right )} x - \frac {8 \, {\left (2 \, D a b^{2} c - 3 \, B b^{3} c + 2 \, C a b^{2} d - 3 \, A b^{3} d\right )}}{b^{4}}\right )} + \frac {{\left (4 \, C a b c - 8 \, A b^{2} c - 3 \, D a^{2} d + 4 \, B a b d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.43 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {24 \sqrt {b \,x^{2}+a}\, a \,b^{2} d -32 \sqrt {b \,x^{2}+a}\, a b c d -9 \sqrt {b \,x^{2}+a}\, a b \,d^{2} x +24 \sqrt {b \,x^{2}+a}\, b^{3} c +12 \sqrt {b \,x^{2}+a}\, b^{3} d x +12 \sqrt {b \,x^{2}+a}\, b^{2} c^{2} x +16 \sqrt {b \,x^{2}+a}\, b^{2} c d \,x^{2}+6 \sqrt {b \,x^{2}+a}\, b^{2} d^{2} x^{3}+9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} d^{2}+24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c -12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d -12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2}}{24 b^{3}} \] Input:

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

Output:

(24*sqrt(a + b*x**2)*a*b**2*d - 32*sqrt(a + b*x**2)*a*b*c*d - 9*sqrt(a + b 
*x**2)*a*b*d**2*x + 24*sqrt(a + b*x**2)*b**3*c + 12*sqrt(a + b*x**2)*b**3* 
d*x + 12*sqrt(a + b*x**2)*b**2*c**2*x + 16*sqrt(a + b*x**2)*b**2*c*d*x**2 
+ 6*sqrt(a + b*x**2)*b**2*d**2*x**3 + 9*sqrt(b)*log((sqrt(a + b*x**2) + sq 
rt(b)*x)/sqrt(a))*a**2*d**2 + 24*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x 
)/sqrt(a))*a*b**2*c - 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a 
))*a*b**2*d - 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b*c 
**2)/(24*b**3)