\(\int \frac {(c+d x)^3 (A+B x+C x^2+D x^3)}{\sqrt {a+b x^2}} \, dx\) [101]

Optimal result

Integrand size = 34, antiderivative size = 411 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {\left (b^2 c^2 (B c+3 A d)+a^2 d^2 (C d+3 c D)-a b \left (3 c^2 C d+3 B c d^2+A d^3+c^3 D\right )\right ) \sqrt {a+b x^2}}{b^3}+\frac {\left (8 b^2 c \left (c^2 C+3 B c d+3 A d^2\right )+5 a^2 d^3 D-6 a b d \left (3 c C d+B d^2+3 c^2 D\right )\right ) x \sqrt {a+b x^2}}{16 b^3}-\frac {d \left (5 a d^2 D-6 b \left (3 c C d+B d^2+3 c^2 D\right )\right ) x^3 \sqrt {a+b x^2}}{24 b^2}+\frac {d^3 D x^5 \sqrt {a+b x^2}}{6 b}-\frac {\left (2 a d^2 (C d+3 c D)-b \left (3 c^2 C d+3 B c d^2+A d^3+c^3 D\right )\right ) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {d^2 (C d+3 c D) \left (a+b x^2\right )^{5/2}}{5 b^3}+\frac {\left (16 A b^3 c^3-a \left (8 b^2 c \left (c^2 C+3 B c d+3 A d^2\right )+5 a^2 d^3 D-6 a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}} \] Output:

(b^2*c^2*(3*A*d+B*c)+a^2*d^2*(C*d+3*D*c)-a*b*(A*d^3+3*B*c*d^2+3*C*c^2*d+D* 
c^3))*(b*x^2+a)^(1/2)/b^3+1/16*(8*b^2*c*(3*A*d^2+3*B*c*d+C*c^2)+5*a^2*d^3* 
D-6*a*b*d*(B*d^2+3*C*c*d+3*D*c^2))*x*(b*x^2+a)^(1/2)/b^3-1/24*d*(5*a*d^2*D 
-6*b*(B*d^2+3*C*c*d+3*D*c^2))*x^3*(b*x^2+a)^(1/2)/b^2+1/6*d^3*D*x^5*(b*x^2 
+a)^(1/2)/b-1/3*(2*a*d^2*(C*d+3*D*c)-b*(A*d^3+3*B*c*d^2+3*C*c^2*d+D*c^3))* 
(b*x^2+a)^(3/2)/b^3+1/5*d^2*(C*d+3*D*c)*(b*x^2+a)^(5/2)/b^3+1/16*(16*A*b^3 
*c^3-a*(8*b^2*c*(3*A*d^2+3*B*c*d+C*c^2)+5*a^2*d^3*D-6*a*b*d*(B*d^2+3*C*c*d 
+3*D*c^2)))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 1.57 (sec) , antiderivative size = 336, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (a^2 d^2 (128 C d+384 c D+75 d D x)+4 b^2 \left (10 A d \left (18 c^2+9 c d x+2 d^2 x^2\right )+15 B \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )+x \left (10 c^3 (3 C+2 D x)+15 c^2 d x (4 C+3 D x)+9 c d^2 x^2 (5 C+4 D x)+2 d^3 x^3 (6 C+5 D x)\right )\right )-2 a b \left (80 c^3 D+15 c^2 d (16 C+9 D x)+3 c d^2 (80 B+x (45 C+32 D x))+d^3 \left (80 A+x \left (45 B+32 C x+25 D x^2\right )\right )\right )\right )}{240 b^3}-\frac {\left (8 A b^2 c \left (2 b c^2-3 a d^2\right )+a \left (-8 b^2 c^2 (c C+3 B d)-5 a^2 d^3 D+6 a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{7/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(a^2*d^2*(128*C*d + 384*c*D + 75*d*D*x) + 4*b^2*(10*A*d*( 
18*c^2 + 9*c*d*x + 2*d^2*x^2) + 15*B*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^ 
3*x^3) + x*(10*c^3*(3*C + 2*D*x) + 15*c^2*d*x*(4*C + 3*D*x) + 9*c*d^2*x^2* 
(5*C + 4*D*x) + 2*d^3*x^3*(6*C + 5*D*x))) - 2*a*b*(80*c^3*D + 15*c^2*d*(16 
*C + 9*D*x) + 3*c*d^2*(80*B + x*(45*C + 32*D*x)) + d^3*(80*A + x*(45*B + 3 
2*C*x + 25*D*x^2)))))/(240*b^3) - ((8*A*b^2*c*(2*b*c^2 - 3*a*d^2) + a*(-8* 
b^2*c^2*(c*C + 3*B*d) - 5*a^2*d^3*D + 6*a*b*d*(3*c*C*d + B*d^2 + 3*c^2*D)) 
)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*b^(7/2))
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 506, normalized size of antiderivative = 1.23, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2185, 2185, 27, 687, 27, 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.

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

\(\Big \downarrow \) 2185

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

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {\int \frac {b d^3 (c+d x)^3 \left (3 d (10 A b d-8 a C d+a c D)-\left (25 a D d^2+2 b \left (-D c^2+3 C d c-15 B d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{5 b d^2}+\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (6 C d-7 c D)}{6 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^5}{6 b d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} d \int \frac {(c+d x)^3 \left (3 d (10 A b d-8 a C d+a c D)-\left (25 a D d^2+2 b \left (-D c^2+3 C d c-15 B d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx+\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (6 C d-7 c D)}{6 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^5}{6 b d^2}\)

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {1}{5} d \left (\frac {\int \frac {3 (c+d x)^2 \left (d \left (40 A c d b^2+a \left (25 a d^2 D-2 b \left (-D c^2+13 C d c+15 B d^2\right )\right )\right )-b \left (a (32 C d+21 c D) d^2+2 b \left (-D c^3+3 C d c^2-15 B d^2 c-20 A d^3\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{4 b}-\frac {\sqrt {a+b x^2} (c+d x)^3 \left (25 a d^2 D+2 b \left (-15 B d^2+c^2 (-D)+3 c C d\right )\right )}{4 b}\right )+\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (6 C d-7 c D)}{6 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^5}{6 b d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} d \left (\frac {3 \int \frac {(c+d x)^2 \left (d \left (40 A c d b^2+a \left (25 a d^2 D-2 b \left (-D c^2+13 C d c+15 B d^2\right )\right )\right )-b \left (a (32 C d+21 c D) d^2+2 b \left (-D c^3+3 C d c^2-15 B d^2 c-20 A d^3\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{4 b}-\frac {\sqrt {a+b x^2} (c+d x)^3 \left (25 a d^2 D+2 b \left (-15 B d^2+c^2 (-D)+3 c C d\right )\right )}{4 b}\right )+\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (6 C d-7 c D)}{6 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^5}{6 b d^2}\)

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {1}{5} d \left (\frac {3 \left (\frac {\int \frac {b (c+d x) \left (d \left (40 A b d \left (3 b c^2-2 a d^2\right )+a \left (a d^2 (64 C d+117 c D)-2 b c \left (-D c^2+33 C d c+75 B d^2\right )\right )\right )+\left (75 a^2 D d^4-2 a b \left (18 D c^2+71 C d c+45 B d^2\right ) d^2-4 b^2 c \left (-D c^3+3 C d c^2-15 B d^2 c-50 A d^3\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b}-\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (21 c D+32 C d)+2 b \left (-20 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )\right )}{4 b}-\frac {\sqrt {a+b x^2} (c+d x)^3 \left (25 a d^2 D+2 b \left (-15 B d^2+c^2 (-D)+3 c C d\right )\right )}{4 b}\right )+\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (6 C d-7 c D)}{6 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^5}{6 b d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} d \left (\frac {3 \left (\frac {1}{3} \int \frac {(c+d x) \left (d \left (40 A b d \left (3 b c^2-2 a d^2\right )+a \left (a d^2 (64 C d+117 c D)-2 b c \left (-D c^2+33 C d c+75 B d^2\right )\right )\right )+\left (75 a^2 D d^4-2 a b \left (18 D c^2+71 C d c+45 B d^2\right ) d^2-4 b^2 c \left (-D c^3+3 C d c^2-15 B d^2 c-50 A d^3\right )\right ) x\right )}{\sqrt {b x^2+a}}dx-\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (21 c D+32 C d)+2 b \left (-20 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )\right )}{4 b}-\frac {\sqrt {a+b x^2} (c+d x)^3 \left (25 a d^2 D+2 b \left (-15 B d^2+c^2 (-D)+3 c C d\right )\right )}{4 b}\right )+\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (6 C d-7 c D)}{6 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^5}{6 b d^2}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {1}{5} d \left (\frac {3 \left (\frac {1}{3} \left (\frac {15 d^2 \left (8 A b^2 c \left (2 b c^2-3 a d^2\right )-a \left (5 a^2 d^3 D-6 a b d \left (B d^2+3 c^2 D+3 c C d\right )+8 b^2 c^2 (3 B d+c C)\right )\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {d x \sqrt {a+b x^2} \left (75 a^2 d^4 D-2 a b d^2 \left (45 B d^2+18 c^2 D+71 c C d\right )-4 b^2 c \left (-50 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )}{2 b}+\frac {2 \sqrt {a+b x^2} \left (32 a^2 d^4 (3 c D+C d)-a b d^2 \left (40 A d^3+120 B c d^2+17 c^3 D+104 c^2 C d\right )-2 b^2 c^2 \left (-80 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )}{b}\right )-\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (21 c D+32 C d)+2 b \left (-20 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )\right )}{4 b}-\frac {\sqrt {a+b x^2} (c+d x)^3 \left (25 a d^2 D+2 b \left (-15 B d^2+c^2 (-D)+3 c C d\right )\right )}{4 b}\right )+\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (6 C d-7 c D)}{6 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^5}{6 b d^2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{5} d \left (\frac {3 \left (\frac {1}{3} \left (\frac {15 d^2 \left (8 A b^2 c \left (2 b c^2-3 a d^2\right )-a \left (5 a^2 d^3 D-6 a b d \left (B d^2+3 c^2 D+3 c C d\right )+8 b^2 c^2 (3 B d+c C)\right )\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {d x \sqrt {a+b x^2} \left (75 a^2 d^4 D-2 a b d^2 \left (45 B d^2+18 c^2 D+71 c C d\right )-4 b^2 c \left (-50 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )}{2 b}+\frac {2 \sqrt {a+b x^2} \left (32 a^2 d^4 (3 c D+C d)-a b d^2 \left (40 A d^3+120 B c d^2+17 c^3 D+104 c^2 C d\right )-2 b^2 c^2 \left (-80 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )}{b}\right )-\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (21 c D+32 C d)+2 b \left (-20 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )\right )}{4 b}-\frac {\sqrt {a+b x^2} (c+d x)^3 \left (25 a d^2 D+2 b \left (-15 B d^2+c^2 (-D)+3 c C d\right )\right )}{4 b}\right )+\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (6 C d-7 c D)}{6 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^5}{6 b d^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{5} d \left (\frac {3 \left (\frac {1}{3} \left (\frac {15 d^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (8 A b^2 c \left (2 b c^2-3 a d^2\right )-a \left (5 a^2 d^3 D-6 a b d \left (B d^2+3 c^2 D+3 c C d\right )+8 b^2 c^2 (3 B d+c C)\right )\right )}{2 b^{3/2}}+\frac {d x \sqrt {a+b x^2} \left (75 a^2 d^4 D-2 a b d^2 \left (45 B d^2+18 c^2 D+71 c C d\right )-4 b^2 c \left (-50 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )}{2 b}+\frac {2 \sqrt {a+b x^2} \left (32 a^2 d^4 (3 c D+C d)-a b d^2 \left (40 A d^3+120 B c d^2+17 c^3 D+104 c^2 C d\right )-2 b^2 c^2 \left (-80 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )}{b}\right )-\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 \left (a d^2 (21 c D+32 C d)+2 b \left (-20 A d^3-15 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )\right )}{4 b}-\frac {\sqrt {a+b x^2} (c+d x)^3 \left (25 a d^2 D+2 b \left (-15 B d^2+c^2 (-D)+3 c C d\right )\right )}{4 b}\right )+\frac {1}{5} d \sqrt {a+b x^2} (c+d x)^4 (6 C d-7 c D)}{6 b d^3}+\frac {D \sqrt {a+b x^2} (c+d x)^5}{6 b d^2}\)

Input:

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

Output:

(D*(c + d*x)^5*Sqrt[a + b*x^2])/(6*b*d^2) + ((d*(6*C*d - 7*c*D)*(c + d*x)^ 
4*Sqrt[a + b*x^2])/5 + (d*(-1/4*((25*a*d^2*D + 2*b*(3*c*C*d - 15*B*d^2 - c 
^2*D))*(c + d*x)^3*Sqrt[a + b*x^2])/b + (3*(-1/3*((a*d^2*(32*C*d + 21*c*D) 
 + 2*b*(3*c^2*C*d - 15*B*c*d^2 - 20*A*d^3 - c^3*D))*(c + d*x)^2*Sqrt[a + b 
*x^2]) + ((2*(32*a^2*d^4*(C*d + 3*c*D) - 2*b^2*c^2*(3*c^2*C*d - 15*B*c*d^2 
 - 80*A*d^3 - c^3*D) - a*b*d^2*(104*c^2*C*d + 120*B*c*d^2 + 40*A*d^3 + 17* 
c^3*D))*Sqrt[a + b*x^2])/b + (d*(75*a^2*d^4*D - 2*a*b*d^2*(71*c*C*d + 45*B 
*d^2 + 18*c^2*D) - 4*b^2*c*(3*c^2*C*d - 15*B*c*d^2 - 50*A*d^3 - c^3*D))*x* 
Sqrt[a + b*x^2])/(2*b) + (15*d^2*(8*A*b^2*c*(2*b*c^2 - 3*a*d^2) - a*(8*b^2 
*c^2*(c*C + 3*B*d) + 5*a^2*d^3*D - 6*a*b*d*(3*c*C*d + B*d^2 + 3*c^2*D)))*A 
rcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)))/3))/(4*b)))/5)/(6*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 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])
 

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 = 1.33 (sec) , antiderivative size = 407, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {A \,c^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+\frac {c^{2} \left (3 A d +B c \right ) \sqrt {b \,x^{2}+a}}{b}+d^{2} \left (C d +3 D c \right ) \left (\frac {x^{4} \sqrt {b \,x^{2}+a}}{5 b}-\frac {4 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )}{5 b}\right )+c \left (3 A \,d^{2}+3 B c d +C \,c^{2}\right ) \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+d \left (B \,d^{2}+3 C c d +3 D c^{2}\right ) \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+\left (A \,d^{3}+3 B c \,d^{2}+3 C \,c^{2} d +D c^{3}\right ) \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+D d^{3} \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )\)	\(407\)

Input:

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

Output:

A*c^3*ln(b^(1/2)*x+(b*x^2+a)^(1/2))/b^(1/2)+c^2*(3*A*d+B*c)/b*(b*x^2+a)^(1 
/2)+d^2*(C*d+3*D*c)*(1/5*x^4/b*(b*x^2+a)^(1/2)-4/5*a/b*(1/3*x^2/b*(b*x^2+a 
)^(1/2)-2/3*a/b^2*(b*x^2+a)^(1/2)))+c*(3*A*d^2+3*B*c*d+C*c^2)*(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)))+d*(B*d^2+3*C*c*d 
+3*D*c^2)*(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))))+(A*d^3+3*B*c*d^2+3*C*c^2*d+D*c^3 
)*(1/3*x^2/b*(b*x^2+a)^(1/2)-2/3*a/b^2*(b*x^2+a)^(1/2))+D*d^3*(1/6*x^5/b*( 
b*x^2+a)^(1/2)-5/6*a/b*(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 = 0.14 (sec) , antiderivative size = 843, normalized size of antiderivative = 2.05 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 536, normalized size of antiderivative = 1.30 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {D d^{3} x^{5}}{6 b} + \frac {x^{4} \left (C d^{3} + 3 D c d^{2}\right )}{5 b} + \frac {x^{3} \left (B d^{3} + 3 C c d^{2} - \frac {5 D a d^{3}}{6 b} + 3 D c^{2} d\right )}{4 b} + \frac {x^{2} \left (A d^{3} + 3 B c d^{2} + 3 C c^{2} d + D c^{3} - \frac {4 a \left (C d^{3} + 3 D c d^{2}\right )}{5 b}\right )}{3 b} + \frac {x \left (3 A c d^{2} + 3 B c^{2} d + C c^{3} - \frac {3 a \left (B d^{3} + 3 C c d^{2} - \frac {5 D a d^{3}}{6 b} + 3 D c^{2} d\right )}{4 b}\right )}{2 b} + \frac {3 A c^{2} d + B c^{3} - \frac {2 a \left (A d^{3} + 3 B c d^{2} + 3 C c^{2} d + D c^{3} - \frac {4 a \left (C d^{3} + 3 D c d^{2}\right )}{5 b}\right )}{3 b}}{b}\right ) + \left (A c^{3} - \frac {a \left (3 A c d^{2} + 3 B c^{2} d + C c^{3} - \frac {3 a \left (B d^{3} + 3 C c d^{2} - \frac {5 D a d^{3}}{6 b} + 3 D c^{2} d\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\frac {A c^{3} x + \frac {D d^{3} x^{7}}{7} + \frac {x^{6} \left (C d^{3} + 3 D c d^{2}\right )}{6} + \frac {x^{5} \left (B d^{3} + 3 C c d^{2} + 3 D c^{2} d\right )}{5} + \frac {x^{4} \left (A d^{3} + 3 B c d^{2} + 3 C c^{2} d + D c^{3}\right )}{4} + \frac {x^{3} \cdot \left (3 A c d^{2} + 3 B c^{2} d + C c^{3}\right )}{3} + \frac {x^{2} \cdot \left (3 A c^{2} d + B c^{3}\right )}{2}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

integrate((d*x+c)**3*(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**3*x**5/(6*b) + x**4*(C*d**3 + 3*D*c*d**2 
)/(5*b) + x**3*(B*d**3 + 3*C*c*d**2 - 5*D*a*d**3/(6*b) + 3*D*c**2*d)/(4*b) 
 + x**2*(A*d**3 + 3*B*c*d**2 + 3*C*c**2*d + D*c**3 - 4*a*(C*d**3 + 3*D*c*d 
**2)/(5*b))/(3*b) + x*(3*A*c*d**2 + 3*B*c**2*d + C*c**3 - 3*a*(B*d**3 + 3* 
C*c*d**2 - 5*D*a*d**3/(6*b) + 3*D*c**2*d)/(4*b))/(2*b) + (3*A*c**2*d + B*c 
**3 - 2*a*(A*d**3 + 3*B*c*d**2 + 3*C*c**2*d + D*c**3 - 4*a*(C*d**3 + 3*D*c 
*d**2)/(5*b))/(3*b))/b) + (A*c**3 - a*(3*A*c*d**2 + 3*B*c**2*d + C*c**3 - 
3*a*(B*d**3 + 3*C*c*d**2 - 5*D*a*d**3/(6*b) + 3*D*c**2*d)/(4*b))/(2*b))*Pi 
ecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*lo 
g(x)/sqrt(b*x**2), True)), Ne(b, 0)), ((A*c**3*x + D*d**3*x**7/7 + x**6*(C 
*d**3 + 3*D*c*d**2)/6 + x**5*(B*d**3 + 3*C*c*d**2 + 3*D*c**2*d)/5 + x**4*( 
A*d**3 + 3*B*c*d**2 + 3*C*c**2*d + D*c**3)/4 + x**3*(3*A*c*d**2 + 3*B*c**2 
*d + C*c**3)/3 + x**2*(3*A*c**2*d + B*c**3)/2)/sqrt(a), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 493, normalized size of antiderivative = 1.20 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} D d^{3} x^{5}}{6 \, b} - \frac {5 \, \sqrt {b x^{2} + a} D a d^{3} x^{3}}{24 \, b^{2}} + \frac {5 \, \sqrt {b x^{2} + a} D a^{2} d^{3} x}{16 \, b^{3}} + \frac {{\left (3 \, D c d^{2} + C d^{3}\right )} \sqrt {b x^{2} + a} x^{4}}{5 \, b} + \frac {A c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {5 \, D a^{3} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} + \frac {\sqrt {b x^{2} + a} B c^{3}}{b} + \frac {3 \, \sqrt {b x^{2} + a} A c^{2} d}{b} + \frac {{\left (3 \, D c^{2} d + 3 \, C c d^{2} + B d^{3}\right )} \sqrt {b x^{2} + a} x^{3}}{4 \, b} - \frac {4 \, {\left (3 \, D c d^{2} + C d^{3}\right )} \sqrt {b x^{2} + a} a x^{2}}{15 \, b^{2}} + \frac {{\left (D c^{3} + 3 \, C c^{2} d + 3 \, B c d^{2} + A d^{3}\right )} \sqrt {b x^{2} + a} x^{2}}{3 \, b} - \frac {3 \, {\left (3 \, D c^{2} d + 3 \, C c d^{2} + B d^{3}\right )} \sqrt {b x^{2} + a} a x}{8 \, b^{2}} + \frac {{\left (C c^{3} + 3 \, B c^{2} d + 3 \, A c d^{2}\right )} \sqrt {b x^{2} + a} x}{2 \, b} + \frac {3 \, {\left (3 \, D c^{2} d + 3 \, C c d^{2} + B d^{3}\right )} a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {{\left (C c^{3} + 3 \, B c^{2} d + 3 \, A c d^{2}\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {8 \, {\left (3 \, D c d^{2} + C d^{3}\right )} \sqrt {b x^{2} + a} a^{2}}{15 \, b^{3}} - \frac {2 \, {\left (D c^{3} + 3 \, C c^{2} d + 3 \, B c d^{2} + A d^{3}\right )} \sqrt {b x^{2} + a} a}{3 \, b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 448, normalized size of antiderivative = 1.09 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{240} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (\frac {5 \, D d^{3} x}{b} + \frac {6 \, {\left (3 \, D b^{5} c d^{2} + C b^{5} d^{3}\right )}}{b^{6}}\right )} x + \frac {5 \, {\left (18 \, D b^{5} c^{2} d + 18 \, C b^{5} c d^{2} - 5 \, D a b^{4} d^{3} + 6 \, B b^{5} d^{3}\right )}}{b^{6}}\right )} x + \frac {8 \, {\left (5 \, D b^{5} c^{3} + 15 \, C b^{5} c^{2} d - 12 \, D a b^{4} c d^{2} + 15 \, B b^{5} c d^{2} - 4 \, C a b^{4} d^{3} + 5 \, A b^{5} d^{3}\right )}}{b^{6}}\right )} x + \frac {15 \, {\left (8 \, C b^{5} c^{3} - 18 \, D a b^{4} c^{2} d + 24 \, B b^{5} c^{2} d - 18 \, C a b^{4} c d^{2} + 24 \, A b^{5} c d^{2} + 5 \, D a^{2} b^{3} d^{3} - 6 \, B a b^{4} d^{3}\right )}}{b^{6}}\right )} x - \frac {16 \, {\left (10 \, D a b^{4} c^{3} - 15 \, B b^{5} c^{3} + 30 \, C a b^{4} c^{2} d - 45 \, A b^{5} c^{2} d - 24 \, D a^{2} b^{3} c d^{2} + 30 \, B a b^{4} c d^{2} - 8 \, C a^{2} b^{3} d^{3} + 10 \, A a b^{4} d^{3}\right )}}{b^{6}}\right )} + \frac {{\left (8 \, C a b^{2} c^{3} - 16 \, A b^{3} c^{3} - 18 \, D a^{2} b c^{2} d + 24 \, B a b^{2} c^{2} d - 18 \, C a^{2} b c d^{2} + 24 \, A a b^{2} c d^{2} + 5 \, D a^{3} d^{3} - 6 \, B a^{2} b d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {7}{2}}} \] Input:

integrate((d*x+c)^3*(D*x^3+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*D*d^3*x/b + 6*(3*D*b^5*c*d^2 + C*b^5*d^3) 
/b^6)*x + 5*(18*D*b^5*c^2*d + 18*C*b^5*c*d^2 - 5*D*a*b^4*d^3 + 6*B*b^5*d^3 
)/b^6)*x + 8*(5*D*b^5*c^3 + 15*C*b^5*c^2*d - 12*D*a*b^4*c*d^2 + 15*B*b^5*c 
*d^2 - 4*C*a*b^4*d^3 + 5*A*b^5*d^3)/b^6)*x + 15*(8*C*b^5*c^3 - 18*D*a*b^4* 
c^2*d + 24*B*b^5*c^2*d - 18*C*a*b^4*c*d^2 + 24*A*b^5*c*d^2 + 5*D*a^2*b^3*d 
^3 - 6*B*a*b^4*d^3)/b^6)*x - 16*(10*D*a*b^4*c^3 - 15*B*b^5*c^3 + 30*C*a*b^ 
4*c^2*d - 45*A*b^5*c^2*d - 24*D*a^2*b^3*c*d^2 + 30*B*a*b^4*c*d^2 - 8*C*a^2 
*b^3*d^3 + 10*A*a*b^4*d^3)/b^6) + 1/16*(8*C*a*b^2*c^3 - 16*A*b^3*c^3 - 18* 
D*a^2*b*c^2*d + 24*B*a*b^2*c^2*d - 18*C*a^2*b*c*d^2 + 24*A*a*b^2*c*d^2 + 5 
*D*a^3*d^3 - 6*B*a^2*b*d^3)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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