\(\int \frac {x^3 (c+d x) (A+B x+C x^2)}{\sqrt {a+b x^2}} \, dx\) [94]

Optimal result

Integrand size = 30, antiderivative size = 234 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=-\frac {a (A b c-a c C-a B d) \sqrt {a+b x^2}}{b^3}+\frac {a (5 a C d-6 b (B c+A d)) x \sqrt {a+b x^2}}{16 b^3}-\frac {(5 a C d-6 b (B c+A d)) x^3 \sqrt {a+b x^2}}{24 b^2}+\frac {C d x^5 \sqrt {a+b x^2}}{6 b}+\frac {(A b c-2 a (c C+B d)) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {(c C+B d) \left (a+b x^2\right )^{5/2}}{5 b^3}-\frac {a^2 (5 a C d-6 b (B c+A d)) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}} \] Output:

-a*(A*b*c-B*a*d-C*a*c)*(b*x^2+a)^(1/2)/b^3+1/16*a*(5*a*C*d-6*b*(A*d+B*c))* 
x*(b*x^2+a)^(1/2)/b^3-1/24*(5*a*C*d-6*b*(A*d+B*c))*x^3*(b*x^2+a)^(1/2)/b^2 
+1/6*C*d*x^5*(b*x^2+a)^(1/2)/b+1/3*(A*b*c-2*a*(B*d+C*c))*(b*x^2+a)^(3/2)/b 
^3+1/5*(B*d+C*c)*(b*x^2+a)^(5/2)/b^3-1/16*a^2*(5*a*C*d-6*b*(A*d+B*c))*arct 
anh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.74 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (a^2 (128 c C+128 B d+75 C d x)-2 a b \left (5 A (16 c+9 d x)+x \left (45 B c+32 c C x+32 B d x+25 C d x^2\right )\right )+4 b^2 x^2 (5 A (4 c+3 d x)+x (3 B (5 c+4 d x)+2 C x (6 c+5 d x)))\right )}{240 b^3}+\frac {a^2 (5 a C d-6 b (B c+A d)) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{7/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(a^2*(128*c*C + 128*B*d + 75*C*d*x) - 2*a*b*(5*A*(16*c + 
9*d*x) + x*(45*B*c + 32*c*C*x + 32*B*d*x + 25*C*d*x^2)) + 4*b^2*x^2*(5*A*( 
4*c + 3*d*x) + x*(3*B*(5*c + 4*d*x) + 2*C*x*(6*c + 5*d*x)))))/(240*b^3) + 
(a^2*(5*a*C*d - 6*b*(B*c + A*d))*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16* 
b^(7/2))
 

Rubi [A] (verified)

Time = 2.72 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.86, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2185, 25, 2185, 25, 2185, 27, 2185, 25, 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.

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

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\int -\frac {(c+d x) \left (b d^4 (19 c C-6 B d) x^4+d^3 \left (21 b C c^2-6 A b d^2+5 a C d^2\right ) x^3+3 c C d^2 \left (3 b c^2+5 a d^2\right ) x^2+c^2 C d \left (b c^2+15 a d^2\right ) x+5 a c^3 C d^2\right )}{\sqrt {b x^2+a}}dx}{6 b d^5}+\frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\int \frac {(c+d x) \left (b d^4 (19 c C-6 B d) x^4+d^3 \left (21 b C c^2-6 A b d^2+5 a C d^2\right ) x^3+3 c C d^2 \left (3 b c^2+5 a d^2\right ) x^2+c^2 C d \left (b c^2+15 a d^2\right ) x+5 a c^3 C d^2\right )}{\sqrt {b x^2+a}}dx}{6 b d^5}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\frac {\int -\frac {(c+d x) \left (-b \left (25 a C d^2-2 b \left (52 C c^2-33 B d c+15 A d^2\right )\right ) x^3 d^7+b \left (2 b (44 c C-21 B d) c^2+a d^2 (c C-24 B d)\right ) x^2 d^6+3 a b c^2 (17 c C-8 B d) d^6+b c \left (2 b (7 c C-3 B d) c^2+a d^2 (77 c C-48 B d)\right ) x d^5\right )}{\sqrt {b x^2+a}}dx}{5 b d^4}+\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (19 c C-6 B d)}{6 b d^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (19 c C-6 B d)-\frac {\int \frac {(c+d x) \left (-b \left (25 a C d^2-2 b \left (52 C c^2-33 B d c+15 A d^2\right )\right ) x^3 d^7+b \left (2 b (44 c C-21 B d) c^2+a d^2 (c C-24 B d)\right ) x^2 d^6+3 a b c^2 (17 c C-8 B d) d^6+b c \left (2 b (7 c C-3 B d) c^2+a d^2 (77 c C-48 B d)\right ) x d^5\right )}{\sqrt {b x^2+a}}dx}{5 b d^4}}{6 b d^5}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (19 c C-6 B d)-\frac {\frac {\int \frac {3 (c+d x) \left (b^2 \left (a d^2 (43 c C-32 B d)-2 b c \left (28 C c^2-27 B d c+25 A d^2\right )\right ) x^2 d^9+a b c \left (25 a C d^2-b \left (36 C c^2-34 B d c+30 A d^2\right )\right ) d^9+b \left (25 a^2 C d^4+a b \left (7 C c^2+2 B d c-30 A d^2\right ) d^2-2 b^2 c^2 \left (8 C c^2-7 B d c+5 A d^2\right )\right ) x d^8\right )}{\sqrt {b x^2+a}}dx}{4 b d^3}+\frac {1}{4} d^5 \sqrt {a+b x^2} (c+d x)^3 \left (-25 a C d^2+30 A b d^2-66 b B c d+104 b c^2 C\right )}{5 b d^4}}{6 b d^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (19 c C-6 B d)-\frac {\frac {3 \int \frac {(c+d x) \left (b^2 \left (a d^2 (43 c C-32 B d)-2 b c \left (28 C c^2-27 B d c+25 A d^2\right )\right ) x^2 d^9+a b c \left (25 a C d^2-b \left (36 C c^2-34 B d c+30 A d^2\right )\right ) d^9+b \left (25 a^2 C d^4+a b \left (7 C c^2+2 B d c-30 A d^2\right ) d^2-2 b^2 c^2 \left (8 C c^2-7 B d c+5 A d^2\right )\right ) x d^8\right )}{\sqrt {b x^2+a}}dx}{4 b d^3}+\frac {1}{4} d^5 \sqrt {a+b x^2} (c+d x)^3 \left (-25 a C d^2+30 A b d^2-66 b B c d+104 b c^2 C\right )}{5 b d^4}}{6 b d^5}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (19 c C-6 B d)-\frac {\frac {3 \left (\frac {\int -\frac {b^2 d^{10} (c+d x) \left (a d \left (a d^2 (11 c C-64 B d)-2 b c \left (2 C c^2-3 B d c+5 A d^2\right )\right )-\left (75 a^2 C d^4-2 a b \left (11 C c^2-19 B d c+45 A d^2\right ) d^2+4 b^2 c^2 \left (2 C c^2-3 B d c+5 A d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b d^2}+\frac {1}{3} b d^8 \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (43 c C-32 B d)-2 b c \left (25 A d^2-27 B c d+28 c^2 C\right )\right )\right )}{4 b d^3}+\frac {1}{4} d^5 \sqrt {a+b x^2} (c+d x)^3 \left (-25 a C d^2+30 A b d^2-66 b B c d+104 b c^2 C\right )}{5 b d^4}}{6 b d^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (19 c C-6 B d)-\frac {\frac {3 \left (\frac {1}{3} b d^8 \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (43 c C-32 B d)-2 b c \left (25 A d^2-27 B c d+28 c^2 C\right )\right )-\frac {\int \frac {b^2 d^{10} (c+d x) \left (a d \left (a d^2 (11 c C-64 B d)-2 b c \left (2 C c^2-3 B d c+5 A d^2\right )\right )-\left (75 a^2 C d^4-2 a b \left (11 C c^2-19 B d c+45 A d^2\right ) d^2+4 b^2 c^2 \left (2 C c^2-3 B d c+5 A d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b d^2}\right )}{4 b d^3}+\frac {1}{4} d^5 \sqrt {a+b x^2} (c+d x)^3 \left (-25 a C d^2+30 A b d^2-66 b B c d+104 b c^2 C\right )}{5 b d^4}}{6 b d^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (19 c C-6 B d)-\frac {\frac {3 \left (\frac {1}{3} b d^8 \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (43 c C-32 B d)-2 b c \left (25 A d^2-27 B c d+28 c^2 C\right )\right )-\frac {1}{3} b d^8 \int \frac {(c+d x) \left (a d \left (a d^2 (11 c C-64 B d)-2 b c \left (2 C c^2-3 B d c+5 A d^2\right )\right )-\left (75 a^2 C d^4-2 a b \left (11 C c^2-19 B d c+45 A d^2\right ) d^2+4 b^2 c^2 \left (2 C c^2-3 B d c+5 A d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx\right )}{4 b d^3}+\frac {1}{4} d^5 \sqrt {a+b x^2} (c+d x)^3 \left (-25 a C d^2+30 A b d^2-66 b B c d+104 b c^2 C\right )}{5 b d^4}}{6 b d^5}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (19 c C-6 B d)-\frac {\frac {3 \left (\frac {1}{3} b d^8 \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (43 c C-32 B d)-2 b c \left (25 A d^2-27 B c d+28 c^2 C\right )\right )-\frac {1}{3} b d^8 \left (\frac {15 a^2 d^4 (5 a C d-6 b (A d+B c)) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}-\frac {d x \sqrt {a+b x^2} \left (75 a^2 C d^4-2 a b d^2 \left (45 A d^2-19 B c d+11 c^2 C\right )+4 b^2 c^2 \left (5 A d^2-3 B c d+2 c^2 C\right )\right )}{2 b}-\frac {2 \sqrt {a+b x^2} \left (32 a^2 d^4 (B d+c C)-a b c d^2 \left (40 A d^2-16 B c d+9 c^2 C\right )+2 b^2 c^3 \left (5 A d^2-3 B c d+2 c^2 C\right )\right )}{b}\right )\right )}{4 b d^3}+\frac {1}{4} d^5 \sqrt {a+b x^2} (c+d x)^3 \left (-25 a C d^2+30 A b d^2-66 b B c d+104 b c^2 C\right )}{5 b d^4}}{6 b d^5}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (19 c C-6 B d)-\frac {\frac {3 \left (\frac {1}{3} b d^8 \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (43 c C-32 B d)-2 b c \left (25 A d^2-27 B c d+28 c^2 C\right )\right )-\frac {1}{3} b d^8 \left (\frac {15 a^2 d^4 (5 a C d-6 b (A d+B c)) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}-\frac {d x \sqrt {a+b x^2} \left (75 a^2 C d^4-2 a b d^2 \left (45 A d^2-19 B c d+11 c^2 C\right )+4 b^2 c^2 \left (5 A d^2-3 B c d+2 c^2 C\right )\right )}{2 b}-\frac {2 \sqrt {a+b x^2} \left (32 a^2 d^4 (B d+c C)-a b c d^2 \left (40 A d^2-16 B c d+9 c^2 C\right )+2 b^2 c^3 \left (5 A d^2-3 B c d+2 c^2 C\right )\right )}{b}\right )\right )}{4 b d^3}+\frac {1}{4} d^5 \sqrt {a+b x^2} (c+d x)^3 \left (-25 a C d^2+30 A b d^2-66 b B c d+104 b c^2 C\right )}{5 b d^4}}{6 b d^5}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {C \sqrt {a+b x^2} (c+d x)^5}{6 b d^4}-\frac {\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (19 c C-6 B d)-\frac {\frac {3 \left (\frac {1}{3} b d^8 \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (43 c C-32 B d)-2 b c \left (25 A d^2-27 B c d+28 c^2 C\right )\right )-\frac {1}{3} b d^8 \left (\frac {15 a^2 d^4 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (5 a C d-6 b (A d+B c))}{2 b^{3/2}}-\frac {d x \sqrt {a+b x^2} \left (75 a^2 C d^4-2 a b d^2 \left (45 A d^2-19 B c d+11 c^2 C\right )+4 b^2 c^2 \left (5 A d^2-3 B c d+2 c^2 C\right )\right )}{2 b}-\frac {2 \sqrt {a+b x^2} \left (32 a^2 d^4 (B d+c C)-a b c d^2 \left (40 A d^2-16 B c d+9 c^2 C\right )+2 b^2 c^3 \left (5 A d^2-3 B c d+2 c^2 C\right )\right )}{b}\right )\right )}{4 b d^3}+\frac {1}{4} d^5 \sqrt {a+b x^2} (c+d x)^3 \left (-25 a C d^2+30 A b d^2-66 b B c d+104 b c^2 C\right )}{5 b d^4}}{6 b d^5}\)

Input:

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

Output:

(C*(c + d*x)^5*Sqrt[a + b*x^2])/(6*b*d^4) - ((d*(19*c*C - 6*B*d)*(c + d*x) 
^4*Sqrt[a + b*x^2])/5 - ((d^5*(104*b*c^2*C - 66*b*B*c*d + 30*A*b*d^2 - 25* 
a*C*d^2)*(c + d*x)^3*Sqrt[a + b*x^2])/4 + (3*((b*d^8*(a*d^2*(43*c*C - 32*B 
*d) - 2*b*c*(28*c^2*C - 27*B*c*d + 25*A*d^2))*(c + d*x)^2*Sqrt[a + b*x^2]) 
/3 - (b*d^8*((-2*(32*a^2*d^4*(c*C + B*d) + 2*b^2*c^3*(2*c^2*C - 3*B*c*d + 
5*A*d^2) - a*b*c*d^2*(9*c^2*C - 16*B*c*d + 40*A*d^2))*Sqrt[a + b*x^2])/b - 
 (d*(75*a^2*C*d^4 + 4*b^2*c^2*(2*c^2*C - 3*B*c*d + 5*A*d^2) - 2*a*b*d^2*(1 
1*c^2*C - 19*B*c*d + 45*A*d^2))*x*Sqrt[a + b*x^2])/(2*b) + (15*a^2*d^4*(5* 
a*C*d - 6*b*(B*c + A*d))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)) 
))/3))/(4*b*d^3))/(5*b*d^4))/(6*b*d^5)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 = 0.26 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {\left (-40 d C \,b^{2} x^{5}-48 B \,b^{2} d \,x^{4}-48 C \,b^{2} c \,x^{4}-60 A \,b^{2} d \,x^{3}-60 B \,b^{2} c \,x^{3}+50 a C d b \,x^{3}-80 A \,b^{2} c \,x^{2}+64 B a d b \,x^{2}+64 C a c b \,x^{2}+90 A x a d b +90 B a c x b -75 C \,a^{2} d x +160 A a b c -128 B \,a^{2} d -128 C \,a^{2} c \right ) \sqrt {b \,x^{2}+a}}{240 b^{3}}+\frac {a^{2} \left (6 A b d +6 B b c -5 a C d \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {7}{2}}}\)	\(186\)
default	\(\left (A d +B c \right ) \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+\left (B d +C c \right ) \left (\frac {x^{4} \sqrt {b \,x^{2}+a}}{5 b}-\frac {4 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )}{5 b}\right )+A c \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+d C \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )\)	\(262\)

Input:

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

Output:

-1/240*(-40*C*b^2*d*x^5-48*B*b^2*d*x^4-48*C*b^2*c*x^4-60*A*b^2*d*x^3-60*B* 
b^2*c*x^3+50*C*a*b*d*x^3-80*A*b^2*c*x^2+64*B*a*b*d*x^2+64*C*a*b*c*x^2+90*A 
*a*b*d*x+90*B*a*b*c*x-75*C*a^2*d*x+160*A*a*b*c-128*B*a^2*d-128*C*a^2*c)*(b 
*x^2+a)^(1/2)/b^3+1/16*a^2*(6*A*b*d+6*B*b*c-5*C*a*d)/b^(7/2)*ln(b^(1/2)*x+ 
(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.85 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\left [\frac {15 \, {\left (6 \, B a^{2} b c - {\left (5 \, C a^{3} - 6 \, A a^{2} b\right )} d\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (40 \, C b^{3} d x^{5} + 128 \, B a^{2} b d + 48 \, {\left (C b^{3} c + B b^{3} d\right )} x^{4} + 10 \, {\left (6 \, B b^{3} c - {\left (5 \, C a b^{2} - 6 \, A b^{3}\right )} d\right )} x^{3} - 16 \, {\left (4 \, B a b^{2} d + {\left (4 \, C a b^{2} - 5 \, A b^{3}\right )} c\right )} x^{2} + 32 \, {\left (4 \, C a^{2} b - 5 \, A a b^{2}\right )} c - 15 \, {\left (6 \, B a b^{2} c - {\left (5 \, C a^{2} b - 6 \, A a b^{2}\right )} d\right )} x\right )} \sqrt {b x^{2} + a}}{480 \, b^{4}}, -\frac {15 \, {\left (6 \, B a^{2} b c - {\left (5 \, C a^{3} - 6 \, A a^{2} b\right )} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (40 \, C b^{3} d x^{5} + 128 \, B a^{2} b d + 48 \, {\left (C b^{3} c + B b^{3} d\right )} x^{4} + 10 \, {\left (6 \, B b^{3} c - {\left (5 \, C a b^{2} - 6 \, A b^{3}\right )} d\right )} x^{3} - 16 \, {\left (4 \, B a b^{2} d + {\left (4 \, C a b^{2} - 5 \, A b^{3}\right )} c\right )} x^{2} + 32 \, {\left (4 \, C a^{2} b - 5 \, A a b^{2}\right )} c - 15 \, {\left (6 \, B a b^{2} c - {\left (5 \, C a^{2} b - 6 \, A a b^{2}\right )} d\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b^{4}}\right ] \] Input:

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

Output:

[1/480*(15*(6*B*a^2*b*c - (5*C*a^3 - 6*A*a^2*b)*d)*sqrt(b)*log(-2*b*x^2 - 
2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(40*C*b^3*d*x^5 + 128*B*a^2*b*d + 48* 
(C*b^3*c + B*b^3*d)*x^4 + 10*(6*B*b^3*c - (5*C*a*b^2 - 6*A*b^3)*d)*x^3 - 1 
6*(4*B*a*b^2*d + (4*C*a*b^2 - 5*A*b^3)*c)*x^2 + 32*(4*C*a^2*b - 5*A*a*b^2) 
*c - 15*(6*B*a*b^2*c - (5*C*a^2*b - 6*A*a*b^2)*d)*x)*sqrt(b*x^2 + a))/b^4, 
 -1/240*(15*(6*B*a^2*b*c - (5*C*a^3 - 6*A*a^2*b)*d)*sqrt(-b)*arctan(sqrt(- 
b)*x/sqrt(b*x^2 + a)) - (40*C*b^3*d*x^5 + 128*B*a^2*b*d + 48*(C*b^3*c + B* 
b^3*d)*x^4 + 10*(6*B*b^3*c - (5*C*a*b^2 - 6*A*b^3)*d)*x^3 - 16*(4*B*a*b^2* 
d + (4*C*a*b^2 - 5*A*b^3)*c)*x^2 + 32*(4*C*a^2*b - 5*A*a*b^2)*c - 15*(6*B* 
a*b^2*c - (5*C*a^2*b - 6*A*a*b^2)*d)*x)*sqrt(b*x^2 + a))/b^4]
 

Sympy [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} \frac {3 a^{2} \left (A d + B c - \frac {5 C a d}{6 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a + b x^{2}} \left (\frac {C d x^{5}}{6 b} - \frac {3 a x \left (A d + B c - \frac {5 C a d}{6 b}\right )}{8 b^{2}} - \frac {2 a \left (A c - \frac {4 a \left (B d + C c\right )}{5 b}\right )}{3 b^{2}} + \frac {x^{4} \left (B d + C c\right )}{5 b} + \frac {x^{3} \left (A d + B c - \frac {5 C a d}{6 b}\right )}{4 b} + \frac {x^{2} \left (A c - \frac {4 a \left (B d + C c\right )}{5 b}\right )}{3 b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {A c x^{4}}{4} + \frac {C d x^{7}}{7} + \frac {x^{6} \left (B d + C c\right )}{6} + \frac {x^{5} \left (A d + B c\right )}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((3*a**2*(A*d + B*c - 5*C*a*d/(6*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))/ 
(8*b**2) + sqrt(a + b*x**2)*(C*d*x**5/(6*b) - 3*a*x*(A*d + B*c - 5*C*a*d/( 
6*b))/(8*b**2) - 2*a*(A*c - 4*a*(B*d + C*c)/(5*b))/(3*b**2) + x**4*(B*d + 
C*c)/(5*b) + x**3*(A*d + B*c - 5*C*a*d/(6*b))/(4*b) + x**2*(A*c - 4*a*(B*d 
 + C*c)/(5*b))/(3*b)), Ne(b, 0)), ((A*c*x**4/4 + C*d*x**7/7 + x**6*(B*d + 
C*c)/6 + x**5*(A*d + B*c)/5)/sqrt(a), True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.11 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} C d x^{5}}{6 \, b} - \frac {5 \, \sqrt {b x^{2} + a} C a d x^{3}}{24 \, b^{2}} + \frac {\sqrt {b x^{2} + a} {\left (C c + B d\right )} x^{4}}{5 \, b} + \frac {\sqrt {b x^{2} + a} A c x^{2}}{3 \, b} + \frac {\sqrt {b x^{2} + a} {\left (B c + A d\right )} x^{3}}{4 \, b} + \frac {5 \, \sqrt {b x^{2} + a} C a^{2} d x}{16 \, b^{3}} - \frac {4 \, \sqrt {b x^{2} + a} {\left (C c + B d\right )} a x^{2}}{15 \, b^{2}} - \frac {5 \, C a^{3} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} - \frac {2 \, \sqrt {b x^{2} + a} A a c}{3 \, b^{2}} - \frac {3 \, \sqrt {b x^{2} + a} {\left (B c + A d\right )} a x}{8 \, b^{2}} + \frac {3 \, {\left (B c + A d\right )} a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} + \frac {8 \, \sqrt {b x^{2} + a} {\left (C c + B d\right )} a^{2}}{15 \, b^{3}} \] Input:

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

Output:

1/6*sqrt(b*x^2 + a)*C*d*x^5/b - 5/24*sqrt(b*x^2 + a)*C*a*d*x^3/b^2 + 1/5*s 
qrt(b*x^2 + a)*(C*c + B*d)*x^4/b + 1/3*sqrt(b*x^2 + a)*A*c*x^2/b + 1/4*sqr 
t(b*x^2 + a)*(B*c + A*d)*x^3/b + 5/16*sqrt(b*x^2 + a)*C*a^2*d*x/b^3 - 4/15 
*sqrt(b*x^2 + a)*(C*c + B*d)*a*x^2/b^2 - 5/16*C*a^3*d*arcsinh(b*x/sqrt(a*b 
))/b^(7/2) - 2/3*sqrt(b*x^2 + a)*A*a*c/b^2 - 3/8*sqrt(b*x^2 + a)*(B*c + A* 
d)*a*x/b^2 + 3/8*(B*c + A*d)*a^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) + 8/15*sqr 
t(b*x^2 + a)*(C*c + B*d)*a^2/b^3
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.95 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{240} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (\frac {5 \, C d x}{b} + \frac {6 \, {\left (C b^{5} c + B b^{5} d\right )}}{b^{6}}\right )} x + \frac {5 \, {\left (6 \, B b^{5} c - 5 \, C a b^{4} d + 6 \, A b^{5} d\right )}}{b^{6}}\right )} x - \frac {8 \, {\left (4 \, C a b^{4} c - 5 \, A b^{5} c + 4 \, B a b^{4} d\right )}}{b^{6}}\right )} x - \frac {15 \, {\left (6 \, B a b^{4} c - 5 \, C a^{2} b^{3} d + 6 \, A a b^{4} d\right )}}{b^{6}}\right )} x + \frac {32 \, {\left (4 \, C a^{2} b^{3} c - 5 \, A a b^{4} c + 4 \, B a^{2} b^{3} d\right )}}{b^{6}}\right )} - \frac {{\left (6 \, B a^{2} b c - 5 \, C a^{3} d + 6 \, A a^{2} b d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {7}{2}}} \] Input:

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

Output:

1/240*sqrt(b*x^2 + a)*((2*((4*(5*C*d*x/b + 6*(C*b^5*c + B*b^5*d)/b^6)*x + 
5*(6*B*b^5*c - 5*C*a*b^4*d + 6*A*b^5*d)/b^6)*x - 8*(4*C*a*b^4*c - 5*A*b^5* 
c + 4*B*a*b^4*d)/b^6)*x - 15*(6*B*a*b^4*c - 5*C*a^2*b^3*d + 6*A*a*b^4*d)/b 
^6)*x + 32*(4*C*a^2*b^3*c - 5*A*a*b^4*c + 4*B*a^2*b^3*d)/b^6) - 1/16*(6*B* 
a^2*b*c - 5*C*a^3*d + 6*A*a^2*b*d)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/ 
b^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.53 \[ \int \frac {x^3 (c+d x) \left (A+B x+C x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {-160 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c -90 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d x +128 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d +128 \sqrt {b \,x^{2}+a}\, a^{2} b \,c^{2}+75 \sqrt {b \,x^{2}+a}\, a^{2} b c d x +80 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,x^{2}-90 \sqrt {b \,x^{2}+a}\, a \,b^{3} c x +60 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{3}-64 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{2}-64 \sqrt {b \,x^{2}+a}\, a \,b^{2} c^{2} x^{2}-50 \sqrt {b \,x^{2}+a}\, a \,b^{2} c d \,x^{3}+60 \sqrt {b \,x^{2}+a}\, b^{4} c \,x^{3}+48 \sqrt {b \,x^{2}+a}\, b^{4} d \,x^{4}+48 \sqrt {b \,x^{2}+a}\, b^{3} c^{2} x^{4}+40 \sqrt {b \,x^{2}+a}\, b^{3} c d \,x^{5}+90 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b d -75 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} c d +90 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} c}{240 b^{4}} \] Input:

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

Output:

( - 160*sqrt(a + b*x**2)*a**2*b**2*c - 90*sqrt(a + b*x**2)*a**2*b**2*d*x + 
 128*sqrt(a + b*x**2)*a**2*b**2*d + 128*sqrt(a + b*x**2)*a**2*b*c**2 + 75* 
sqrt(a + b*x**2)*a**2*b*c*d*x + 80*sqrt(a + b*x**2)*a*b**3*c*x**2 - 90*sqr 
t(a + b*x**2)*a*b**3*c*x + 60*sqrt(a + b*x**2)*a*b**3*d*x**3 - 64*sqrt(a + 
 b*x**2)*a*b**3*d*x**2 - 64*sqrt(a + b*x**2)*a*b**2*c**2*x**2 - 50*sqrt(a 
+ b*x**2)*a*b**2*c*d*x**3 + 60*sqrt(a + b*x**2)*b**4*c*x**3 + 48*sqrt(a + 
b*x**2)*b**4*d*x**4 + 48*sqrt(a + b*x**2)*b**3*c**2*x**4 + 40*sqrt(a + b*x 
**2)*b**3*c*d*x**5 + 90*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a) 
)*a**3*b*d - 75*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*c 
*d + 90*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b**2*c)/( 
240*b**4)