\(\int \frac {(c+d x)^3 (A+B x+C x^2+D x^3)}{(a+b x^2)^{3/2}} \, dx\) [108]

Optimal result

Integrand size = 34, antiderivative size = 408 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {b^2 c^2 (B c+3 A d)+a^2 d^2 (C d+3 c D)-a b \left (3 c^2 C d+3 B c d^2+A d^3+c^3 D\right )}{b^3 \sqrt {a+b x^2}}+\frac {\left (A b^2 c \left (b c^2-3 a d^2\right )-a \left (b^2 c^2 (c C+3 B d)+a^2 d^3 D-a b d \left (3 c C d+B d^2+3 c^2 D\right )\right )\right ) x}{a b^3 \sqrt {a+b x^2}}-\frac {\left (2 a d^2 (C d+3 c D)-b \left (3 c^2 C d+3 B c d^2+A d^3+c^3 D\right )\right ) \sqrt {a+b x^2}}{b^3}-\frac {d \left (7 a d^2 D-4 b \left (3 c C d+B d^2+3 c^2 D\right )\right ) x \sqrt {a+b x^2}}{8 b^3}+\frac {d^3 D x^3 \sqrt {a+b x^2}}{4 b^2}+\frac {d^2 (C d+3 c D) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {\left (8 b^2 c \left (c^2 C+3 B c d+3 A d^2\right )+15 a^2 d^3 D-12 a b d \left (3 c C d+B d^2+3 c^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \] 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^3/(b*x^2+a)^(1/2)+(A*b^2*c*(-3*a*d^2+b*c^2)-a*(b^2*c^2*(3*B*d+C*c 
)+a^2*d^3*D-a*b*d*(B*d^2+3*C*c*d+3*D*c^2)))*x/a/b^3/(b*x^2+a)^(1/2)-(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)^(1/2)/b^3-1/ 
8*d*(7*a*d^2*D-4*b*(B*d^2+3*C*c*d+3*D*c^2))*x*(b*x^2+a)^(1/2)/b^3+1/4*d^3* 
D*x^3*(b*x^2+a)^(1/2)/b^2+1/3*d^2*(C*d+3*D*c)*(b*x^2+a)^(3/2)/b^3+1/8*(8*b 
^2*c*(3*A*d^2+3*B*c*d+C*c^2)+15*a^2*d^3*D-12*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 = 2.13 (sec) , antiderivative size = 332, normalized size of antiderivative = 0.81 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {24 A b^3 c^3 x-a^3 d^2 (64 C d+192 c D+45 d D x)+2 a b^2 \left (12 A d \left (-3 c^2-3 c d x+d^2 x^2\right )-6 B \left (2 c^3+6 c^2 d x-6 c d^2 x^2-d^3 x^3\right )+x \left (-12 c^3 (C-D x)+18 c^2 d x (2 C+D x)+6 c d^2 x^2 (3 C+2 D x)+d^3 x^3 (4 C+3 D x)\right )\right )+a^2 b \left (48 c^3 D+36 c^2 d (4 C+3 D x)+12 c d^2 (12 B+x (9 C-8 D x))+d^3 \left (48 A+x \left (36 B-32 C x-15 D x^2\right )\right )\right )}{24 a b^3 \sqrt {a+b x^2}}-\frac {\left (8 b^2 c \left (c^2 C+3 B c d+3 A d^2\right )+15 a^2 d^3 D-12 a b d \left (3 c C d+B d^2+3 c^2 D\right )\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{7/2}} \] Input:

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

Output:

(24*A*b^3*c^3*x - a^3*d^2*(64*C*d + 192*c*D + 45*d*D*x) + 2*a*b^2*(12*A*d* 
(-3*c^2 - 3*c*d*x + d^2*x^2) - 6*B*(2*c^3 + 6*c^2*d*x - 6*c*d^2*x^2 - d^3* 
x^3) + x*(-12*c^3*(C - D*x) + 18*c^2*d*x*(2*C + D*x) + 6*c*d^2*x^2*(3*C + 
2*D*x) + d^3*x^3*(4*C + 3*D*x))) + a^2*b*(48*c^3*D + 36*c^2*d*(4*C + 3*D*x 
) + 12*c*d^2*(12*B + x*(9*C - 8*D*x)) + d^3*(48*A + x*(36*B - 32*C*x - 15* 
D*x^2))))/(24*a*b^3*Sqrt[a + b*x^2]) - ((8*b^2*c*(c^2*C + 3*B*c*d + 3*A*d^ 
2) + 15*a^2*d^3*D - 12*a*b*d*(3*c*C*d + B*d^2 + 3*c^2*D))*Log[-(Sqrt[b]*x) 
 + Sqrt[a + b*x^2]])/(8*b^(7/2))
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 353, normalized size of antiderivative = 0.87, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {2176, 25, 2185, 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.

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

\(\Big \downarrow \) 2176

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {\int \frac {d^2 (c+d x)^2 (a (4 b (c C+3 B d)-15 a d D)-b (12 A b d-16 a C d-3 a c D) x)}{\sqrt {b x^2+a}}dx}{4 b d^2}+\frac {a D \sqrt {a+b x^2} (c+d x)^3}{4 b}}{a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {(c+d x)^2 (a (4 b (c C+3 B d)-15 a d D)-b (12 A b d-16 a C d-3 a c D) x)}{\sqrt {b x^2+a}}dx}{4 b}+\frac {a D \sqrt {a+b x^2} (c+d x)^3}{4 b}}{a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {\frac {\int \frac {b (c+d x) \left (a \left (12 b \left (C c^2+3 B d c+2 A d^2\right )-a d (32 C d+51 c D)\right )-\left (24 A c d b^2+a \left (45 a d^2 D-2 b \left (3 D c^2+22 C d c+18 B d^2\right )\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b}-\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 (-3 a c D-16 a C d+12 A b d)}{4 b}+\frac {a D \sqrt {a+b x^2} (c+d x)^3}{4 b}}{a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {1}{3} \int \frac {(c+d x) \left (a \left (12 b \left (C c^2+3 B d c+2 A d^2\right )-a d (32 C d+51 c D)\right )-\left (24 A c d b^2+a \left (45 a d^2 D-b \left (6 D c^2+44 C d c+36 B d^2\right )\right )\right ) x\right )}{\sqrt {b x^2+a}}dx-\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 (-3 a c D-16 a C d+12 A b d)}{4 b}+\frac {a D \sqrt {a+b x^2} (c+d x)^3}{4 b}}{a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {3 a \left (15 a^2 d^3 D-12 a b d \left (B d^2+3 c^2 D+3 c C d\right )+8 b^2 c \left (3 A d^2+3 B c d+c^2 C\right )\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}-\frac {d x \sqrt {a+b x^2} \left (a \left (45 a d^2 D-b \left (36 B d^2+6 c^2 D+44 c C d\right )\right )+24 A b^2 c d\right )}{2 b}-\frac {2 \sqrt {a+b x^2} \left (12 A b d \left (b c^2-a d^2\right )+a \left (16 a d^2 (3 c D+C d)-b c \left (36 B d^2+3 c^2 D+28 c C d\right )\right )\right )}{b}\right )-\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 (-3 a c D-16 a C d+12 A b d)}{4 b}+\frac {a D \sqrt {a+b x^2} (c+d x)^3}{4 b}}{a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {3 a \left (15 a^2 d^3 D-12 a b d \left (B d^2+3 c^2 D+3 c C d\right )+8 b^2 c \left (3 A d^2+3 B c d+c^2 C\right )\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}-\frac {d x \sqrt {a+b x^2} \left (a \left (45 a d^2 D-b \left (36 B d^2+6 c^2 D+44 c C d\right )\right )+24 A b^2 c d\right )}{2 b}-\frac {2 \sqrt {a+b x^2} \left (12 A b d \left (b c^2-a d^2\right )+a \left (16 a d^2 (3 c D+C d)-b c \left (36 B d^2+3 c^2 D+28 c C d\right )\right )\right )}{b}\right )-\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 (-3 a c D-16 a C d+12 A b d)}{4 b}+\frac {a D \sqrt {a+b x^2} (c+d x)^3}{4 b}}{a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (15 a^2 d^3 D-12 a b d \left (B d^2+3 c^2 D+3 c C d\right )+8 b^2 c \left (3 A d^2+3 B c d+c^2 C\right )\right )}{2 b^{3/2}}-\frac {d x \sqrt {a+b x^2} \left (a \left (45 a d^2 D-b \left (36 B d^2+6 c^2 D+44 c C d\right )\right )+24 A b^2 c d\right )}{2 b}-\frac {2 \sqrt {a+b x^2} \left (12 A b d \left (b c^2-a d^2\right )+a \left (16 a d^2 (3 c D+C d)-b c \left (36 B d^2+3 c^2 D+28 c C d\right )\right )\right )}{b}\right )-\frac {1}{3} \sqrt {a+b x^2} (c+d x)^2 (-3 a c D-16 a C d+12 A b d)}{4 b}+\frac {a D \sqrt {a+b x^2} (c+d x)^3}{4 b}}{a b}-\frac {(c+d x)^3 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{a b \sqrt {a+b x^2}}\)

Input:

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

Output:

-(((a*(B - (a*D)/b) - (A*b - a*C)*x)*(c + d*x)^3)/(a*b*Sqrt[a + b*x^2])) + 
 ((a*D*(c + d*x)^3*Sqrt[a + b*x^2])/(4*b) + (-1/3*((12*A*b*d - 16*a*C*d - 
3*a*c*D)*(c + d*x)^2*Sqrt[a + b*x^2]) + ((-2*(12*A*b*d*(b*c^2 - a*d^2) + a 
*(16*a*d^2*(C*d + 3*c*D) - b*c*(28*c*C*d + 36*B*d^2 + 3*c^2*D)))*Sqrt[a + 
b*x^2])/b - (d*(24*A*b^2*c*d + a*(45*a*d^2*D - b*(44*c*C*d + 36*B*d^2 + 6* 
c^2*D)))*x*Sqrt[a + b*x^2])/(2*b) + (3*a*(8*b^2*c*(c^2*C + 3*B*c*d + 3*A*d 
^2) + 15*a^2*d^3*D - 12*a*b*d*(3*c*C*d + B*d^2 + 3*c^2*D))*ArcTanh[(Sqrt[b 
]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)))/3)/(4*b))/(a*b)
 

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 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 2176

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a + b*x^2, x], R = Coeff[PolynomialRema 
inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 
, x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p 
 + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p 
 + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + 
b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && R 
ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 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 = 1.35 (sec) , antiderivative size = 394, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {A \,c^{3} x}{\sqrt {b \,x^{2}+a}\, a}-\frac {c^{2} \left (3 A d +B c \right )}{b \sqrt {b \,x^{2}+a}}+d^{2} \left (C d +3 D c \right ) \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )+c \left (3 A \,d^{2}+3 B c d +C \,c^{2}\right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )+d \left (B \,d^{2}+3 C c d +3 D c^{2}\right ) \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+\left (A \,d^{3}+3 B c \,d^{2}+3 C \,c^{2} d +D c^{3}\right ) \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )+D d^{3} \left (\frac {x^{5}}{4 b \sqrt {b \,x^{2}+a}}-\frac {5 a \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )\)	\(394\)

Input:

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

Output:

A*c^3/(b*x^2+a)^(1/2)/a*x-c^2*(3*A*d+B*c)/b/(b*x^2+a)^(1/2)+d^2*(C*d+3*D*c 
)*(1/3*x^4/b/(b*x^2+a)^(1/2)-4/3*a/b*(x^2/b/(b*x^2+a)^(1/2)+2*a/b^2/(b*x^2 
+a)^(1/2)))+c*(3*A*d^2+3*B*c*d+C*c^2)*(-x/b/(b*x^2+a)^(1/2)+1/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)*(1/2*x^3/b/(b*x^2+a)^ 
(1/2)-3/2*a/b*(-x/b/(b*x^2+a)^(1/2)+1/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)*(x^2/b/(b*x^2+a)^(1/2)+2*a/b^2/(b*x^ 
2+a)^(1/2))+D*d^3*(1/4*x^5/b/(b*x^2+a)^(1/2)-5/4*a/b*(1/2*x^3/b/(b*x^2+a)^ 
(1/2)-3/2*a/b*(-x/b/(b*x^2+a)^(1/2)+1/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 = 1126, normalized size of antiderivative = 2.76 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/48*(3*(8*C*a^2*b^2*c^3 - 12*(3*D*a^3*b - 2*B*a^2*b^2)*c^2*d - 12*(3*C* 
a^3*b - 2*A*a^2*b^2)*c*d^2 + 3*(5*D*a^4 - 4*B*a^3*b)*d^3 + (8*C*a*b^3*c^3 
- 12*(3*D*a^2*b^2 - 2*B*a*b^3)*c^2*d - 12*(3*C*a^2*b^2 - 2*A*a*b^3)*c*d^2 
+ 3*(5*D*a^3*b - 4*B*a^2*b^2)*d^3)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^ 
2 + a)*sqrt(b)*x - a) - 2*(6*D*a*b^3*d^3*x^5 + 8*(3*D*a*b^3*c*d^2 + C*a*b^ 
3*d^3)*x^4 + 24*(2*D*a^2*b^2 - B*a*b^3)*c^3 + 72*(2*C*a^2*b^2 - A*a*b^3)*c 
^2*d - 48*(4*D*a^3*b - 3*B*a^2*b^2)*c*d^2 - 16*(4*C*a^3*b - 3*A*a^2*b^2)*d 
^3 + 3*(12*D*a*b^3*c^2*d + 12*C*a*b^3*c*d^2 - (5*D*a^2*b^2 - 4*B*a*b^3)*d^ 
3)*x^3 + 8*(3*D*a*b^3*c^3 + 9*C*a*b^3*c^2*d - 3*(4*D*a^2*b^2 - 3*B*a*b^3)* 
c*d^2 - (4*C*a^2*b^2 - 3*A*a*b^3)*d^3)*x^2 - 3*(8*(C*a*b^3 - A*b^4)*c^3 - 
12*(3*D*a^2*b^2 - 2*B*a*b^3)*c^2*d - 12*(3*C*a^2*b^2 - 2*A*a*b^3)*c*d^2 + 
3*(5*D*a^3*b - 4*B*a^2*b^2)*d^3)*x)*sqrt(b*x^2 + a))/(a*b^5*x^2 + a^2*b^4) 
, -1/24*(3*(8*C*a^2*b^2*c^3 - 12*(3*D*a^3*b - 2*B*a^2*b^2)*c^2*d - 12*(3*C 
*a^3*b - 2*A*a^2*b^2)*c*d^2 + 3*(5*D*a^4 - 4*B*a^3*b)*d^3 + (8*C*a*b^3*c^3 
 - 12*(3*D*a^2*b^2 - 2*B*a*b^3)*c^2*d - 12*(3*C*a^2*b^2 - 2*A*a*b^3)*c*d^2 
 + 3*(5*D*a^3*b - 4*B*a^2*b^2)*d^3)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b 
*x^2 + a)) - (6*D*a*b^3*d^3*x^5 + 8*(3*D*a*b^3*c*d^2 + C*a*b^3*d^3)*x^4 + 
24*(2*D*a^2*b^2 - B*a*b^3)*c^3 + 72*(2*C*a^2*b^2 - A*a*b^3)*c^2*d - 48*(4* 
D*a^3*b - 3*B*a^2*b^2)*c*d^2 - 16*(4*C*a^3*b - 3*A*a^2*b^2)*d^3 + 3*(12*D* 
a*b^3*c^2*d + 12*C*a*b^3*c*d^2 - (5*D*a^2*b^2 - 4*B*a*b^3)*d^3)*x^3 + 8...
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 490, normalized size of antiderivative = 1.20 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {D d^{3} x^{5}}{4 \, \sqrt {b x^{2} + a} b} - \frac {5 \, D a d^{3} x^{3}}{8 \, \sqrt {b x^{2} + a} b^{2}} + \frac {A c^{3} x}{\sqrt {b x^{2} + a} a} - \frac {15 \, D a^{2} d^{3} x}{8 \, \sqrt {b x^{2} + a} b^{3}} + \frac {{\left (3 \, D c d^{2} + C d^{3}\right )} x^{4}}{3 \, \sqrt {b x^{2} + a} b} + \frac {15 \, D a^{2} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {7}{2}}} - \frac {B c^{3}}{\sqrt {b x^{2} + a} b} - \frac {3 \, A c^{2} d}{\sqrt {b x^{2} + a} b} + \frac {{\left (3 \, D c^{2} d + 3 \, C c d^{2} + B d^{3}\right )} x^{3}}{2 \, \sqrt {b x^{2} + a} b} - \frac {4 \, {\left (3 \, D c d^{2} + C d^{3}\right )} a x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {{\left (D c^{3} + 3 \, C c^{2} d + 3 \, B c d^{2} + A d^{3}\right )} x^{2}}{\sqrt {b x^{2} + a} b} + \frac {3 \, {\left (3 \, D c^{2} d + 3 \, C c d^{2} + B d^{3}\right )} a x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {{\left (C c^{3} + 3 \, B c^{2} d + 3 \, A c d^{2}\right )} x}{\sqrt {b x^{2} + a} b} - \frac {3 \, {\left (3 \, D c^{2} d + 3 \, C c d^{2} + B d^{3}\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {{\left (C c^{3} + 3 \, B c^{2} d + 3 \, A c d^{2}\right )} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} - \frac {8 \, {\left (3 \, D c d^{2} + C d^{3}\right )} a^{2}}{3 \, \sqrt {b x^{2} + a} b^{3}} + \frac {2 \, {\left (D c^{3} + 3 \, C c^{2} d + 3 \, B c d^{2} + A d^{3}\right )} a}{\sqrt {b x^{2} + a} b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 487, normalized size of antiderivative = 1.19 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left ({\left (2 \, {\left (\frac {3 \, D d^{3} x}{b} + \frac {4 \, {\left (3 \, D a b^{5} c d^{2} + C a b^{5} d^{3}\right )}}{a b^{6}}\right )} x + \frac {3 \, {\left (12 \, D a b^{5} c^{2} d + 12 \, C a b^{5} c d^{2} - 5 \, D a^{2} b^{4} d^{3} + 4 \, B a b^{5} d^{3}\right )}}{a b^{6}}\right )} x + \frac {8 \, {\left (3 \, D a b^{5} c^{3} + 9 \, C a b^{5} c^{2} d - 12 \, D a^{2} b^{4} c d^{2} + 9 \, B a b^{5} c d^{2} - 4 \, C a^{2} b^{4} d^{3} + 3 \, A a b^{5} d^{3}\right )}}{a b^{6}}\right )} x - \frac {3 \, {\left (8 \, C a b^{5} c^{3} - 8 \, A b^{6} c^{3} - 36 \, D a^{2} b^{4} c^{2} d + 24 \, B a b^{5} c^{2} d - 36 \, C a^{2} b^{4} c d^{2} + 24 \, A a b^{5} c d^{2} + 15 \, D a^{3} b^{3} d^{3} - 12 \, B a^{2} b^{4} d^{3}\right )}}{a b^{6}}\right )} x + \frac {8 \, {\left (6 \, D a^{2} b^{4} c^{3} - 3 \, B a b^{5} c^{3} + 18 \, C a^{2} b^{4} c^{2} d - 9 \, A a b^{5} c^{2} d - 24 \, D a^{3} b^{3} c d^{2} + 18 \, B a^{2} b^{4} c d^{2} - 8 \, C a^{3} b^{3} d^{3} + 6 \, A a^{2} b^{4} d^{3}\right )}}{a b^{6}}}{24 \, \sqrt {b x^{2} + a}} - \frac {{\left (8 \, C b^{2} c^{3} - 36 \, D a b c^{2} d + 24 \, B b^{2} c^{2} d - 36 \, C a b c d^{2} + 24 \, A b^{2} c d^{2} + 15 \, D a^{2} d^{3} - 12 \, B a b d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {7}{2}}} \] Input:

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

Output:

1/24*((((2*(3*D*d^3*x/b + 4*(3*D*a*b^5*c*d^2 + C*a*b^5*d^3)/(a*b^6))*x + 3 
*(12*D*a*b^5*c^2*d + 12*C*a*b^5*c*d^2 - 5*D*a^2*b^4*d^3 + 4*B*a*b^5*d^3)/( 
a*b^6))*x + 8*(3*D*a*b^5*c^3 + 9*C*a*b^5*c^2*d - 12*D*a^2*b^4*c*d^2 + 9*B* 
a*b^5*c*d^2 - 4*C*a^2*b^4*d^3 + 3*A*a*b^5*d^3)/(a*b^6))*x - 3*(8*C*a*b^5*c 
^3 - 8*A*b^6*c^3 - 36*D*a^2*b^4*c^2*d + 24*B*a*b^5*c^2*d - 36*C*a^2*b^4*c* 
d^2 + 24*A*a*b^5*c*d^2 + 15*D*a^3*b^3*d^3 - 12*B*a^2*b^4*d^3)/(a*b^6))*x + 
 8*(6*D*a^2*b^4*c^3 - 3*B*a*b^5*c^3 + 18*C*a^2*b^4*c^2*d - 9*A*a*b^5*c^2*d 
 - 24*D*a^3*b^3*c*d^2 + 18*B*a^2*b^4*c*d^2 - 8*C*a^3*b^3*d^3 + 6*A*a^2*b^4 
*d^3)/(a*b^6))/sqrt(b*x^2 + a) - 1/8*(8*C*b^2*c^3 - 36*D*a*b*c^2*d + 24*B* 
b^2*c^2*d - 36*C*a*b*c*d^2 + 24*A*b^2*c*d^2 + 15*D*a^2*d^3 - 12*B*a*b*d^3) 
*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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