\(\int \frac {\sqrt {f+g x} (a+c x^2)^2}{(d+e x)^3} \, dx\) [69]

Optimal result

Integrand size = 26, antiderivative size = 301 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {4 c \left (3 c d^2+a e^2\right ) \sqrt {f+g x}}{e^5}-\frac {\left (c d^2+a e^2\right )^2 \sqrt {f+g x}}{2 e^5 (d+e x)^2}-\frac {\left (c d^2+a e^2\right ) \left (a e^2 g-c d (16 e f-17 d g)\right ) \sqrt {f+g x}}{4 e^5 (e f-d g) (d+e x)}-\frac {2 c^2 (e f+3 d g) (f+g x)^{3/2}}{3 e^4 g^2}+\frac {2 c^2 (f+g x)^{5/2}}{5 e^3 g^2}+\frac {\left (a^2 e^4 g^2-2 a c e^2 \left (8 e^2 f^2-24 d e f g+15 d^2 g^2\right )-c^2 d^2 \left (48 e^2 f^2-112 d e f g+63 d^2 g^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{11/2} (e f-d g)^{3/2}} \] Output:

4*c*(a*e^2+3*c*d^2)*(g*x+f)^(1/2)/e^5-1/2*(a*e^2+c*d^2)^2*(g*x+f)^(1/2)/e^ 
5/(e*x+d)^2-1/4*(a*e^2+c*d^2)*(a*e^2*g-c*d*(-17*d*g+16*e*f))*(g*x+f)^(1/2) 
/e^5/(-d*g+e*f)/(e*x+d)-2/3*c^2*(3*d*g+e*f)*(g*x+f)^(3/2)/e^4/g^2+2/5*c^2* 
(g*x+f)^(5/2)/e^3/g^2+1/4*(a^2*e^4*g^2-2*a*c*e^2*(15*d^2*g^2-24*d*e*f*g+8* 
e^2*f^2)-c^2*d^2*(63*d^2*g^2-112*d*e*f*g+48*e^2*f^2))*arctanh(e^(1/2)*(g*x 
+f)^(1/2)/(-d*g+e*f)^(1/2))/e^(11/2)/(-d*g+e*f)^(3/2)
 

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=-\frac {\sqrt {f+g x} \left (15 a^2 e^4 g^2 (2 e f-d g+e g x)+30 a c e^2 g^2 \left (15 d^3 g-8 e^3 f x^2+8 d e^2 x (-3 f+g x)+d^2 e (-14 f+25 g x)\right )+c^2 \left (945 d^5 g^3+525 d^4 e g^2 (-2 f+3 g x)+8 d e^4 x (f+g x)^2 (4 f+3 g x)-8 e^5 f x^2 \left (-2 f^2+f g x+3 g^2 x^2\right )+8 d^3 e^2 g \left (13 f^2-224 f g x+63 g^2 x^2\right )+8 d^2 e^3 \left (2 f^3+25 f^2 g x-76 f g^2 x^2-9 g^3 x^3\right )\right )\right )}{60 e^5 g^2 (e f-d g) (d+e x)^2}-\frac {\left (-a^2 e^4 g^2+2 a c e^2 \left (8 e^2 f^2-24 d e f g+15 d^2 g^2\right )+c^2 d^2 \left (48 e^2 f^2-112 d e f g+63 d^2 g^2\right )\right ) \arctan \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{4 e^{11/2} (-e f+d g)^{3/2}} \] Input:

Integrate[(Sqrt[f + g*x]*(a + c*x^2)^2)/(d + e*x)^3,x]
 

Output:

-1/60*(Sqrt[f + g*x]*(15*a^2*e^4*g^2*(2*e*f - d*g + e*g*x) + 30*a*c*e^2*g^ 
2*(15*d^3*g - 8*e^3*f*x^2 + 8*d*e^2*x*(-3*f + g*x) + d^2*e*(-14*f + 25*g*x 
)) + c^2*(945*d^5*g^3 + 525*d^4*e*g^2*(-2*f + 3*g*x) + 8*d*e^4*x*(f + g*x) 
^2*(4*f + 3*g*x) - 8*e^5*f*x^2*(-2*f^2 + f*g*x + 3*g^2*x^2) + 8*d^3*e^2*g* 
(13*f^2 - 224*f*g*x + 63*g^2*x^2) + 8*d^2*e^3*(2*f^3 + 25*f^2*g*x - 76*f*g 
^2*x^2 - 9*g^3*x^3))))/(e^5*g^2*(e*f - d*g)*(d + e*x)^2) - ((-(a^2*e^4*g^2 
) + 2*a*c*e^2*(8*e^2*f^2 - 24*d*e*f*g + 15*d^2*g^2) + c^2*d^2*(48*e^2*f^2 
- 112*d*e*f*g + 63*d^2*g^2))*ArcTan[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + 
d*g]])/(4*e^(11/2)*(-(e*f) + d*g)^(3/2))
 

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.23, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {649, 25, 1580, 25, 2345, 25, 2341, 2009}

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.

\(\displaystyle \int \frac {\left (a+c x^2\right )^2 \sqrt {f+g x}}{(d+e x)^3} \, dx\)

\(\Big \downarrow \) 649

\(\displaystyle \frac {2 \int -\frac {(f+g x) \left (c f^2-2 c (f+g x) f+a g^2+c (f+g x)^2\right )^2}{(e f-d g-e (f+g x))^3}d\sqrt {f+g x}}{g^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2 \int \frac {(f+g x) \left (c f^2-2 c (f+g x) f+a g^2+c (f+g x)^2\right )^2}{(e f-d g-e (f+g x))^3}d\sqrt {f+g x}}{g^2}\)

\(\Big \downarrow \) 1580

\(\displaystyle \frac {2 \left (-\frac {\int -\frac {\left (c d^2+a e^2\right )^2 g^4+4 c^2 e^4 (f+g x)^4-4 c^2 e^3 (3 e f+d g) (f+g x)^3+4 c e^2 \left (2 a e^2 g^2+c \left (3 e^2 f^2+2 d e g f+d^2 g^2\right )\right ) (f+g x)^2-4 c e (e f+d g) \left (c e^2 f^2+c d^2 g^2+2 a e^2 g^2\right ) (f+g x)}{(e f-d g-e (f+g x))^2}d\sqrt {f+g x}}{4 e^5}-\frac {g^4 \sqrt {f+g x} \left (a e^2+c d^2\right )^2}{4 e^5 (-d g-e (f+g x)+e f)^2}\right )}{g^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (\frac {\int \frac {\left (c d^2+a e^2\right )^2 g^4+4 c^2 e^4 (f+g x)^4-4 c^2 e^3 (3 e f+d g) (f+g x)^3+4 c e^2 \left (2 a e^2 g^2+c \left (3 e^2 f^2+2 d e g f+d^2 g^2\right )\right ) (f+g x)^2-4 c e (e f+d g) \left (c e^2 f^2+c d^2 g^2+2 a e^2 g^2\right ) (f+g x)}{(e f-d g-e (f+g x))^2}d\sqrt {f+g x}}{4 e^5}-\frac {g^4 \sqrt {f+g x} \left (a e^2+c d^2\right )^2}{4 e^5 (-d g-e (f+g x)+e f)^2}\right )}{g^2}\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {2 \left (\frac {\frac {g^3 \sqrt {f+g x} \left (a e^2+c d^2\right ) \left (a e^2 g-c d (16 e f-17 d g)\right )}{2 (e f-d g) (-d g-e (f+g x)+e f)}-\frac {\int -\frac {\left (c d^2+a e^2\right ) \left (a g e^2+c d (16 e f-15 d g)\right ) g^3-8 c^2 e^3 (e f-d g) (f+g x)^3+16 c^2 e^2 (e f-d g) (e f+d g) (f+g x)^2-8 c e (e f-d g) \left (2 a e^2 g^2+c \left (e^2 f^2+2 d e g f+3 d^2 g^2\right )\right ) (f+g x)}{e f-d g-e (f+g x)}d\sqrt {f+g x}}{2 (e f-d g)}}{4 e^5}-\frac {g^4 \sqrt {f+g x} \left (a e^2+c d^2\right )^2}{4 e^5 (-d g-e (f+g x)+e f)^2}\right )}{g^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (\frac {\frac {\int \frac {\left (c d^2+a e^2\right ) \left (a g e^2+c d (16 e f-15 d g)\right ) g^3-8 c^2 e^3 (e f-d g) (f+g x)^3+16 c^2 e^2 (e f-d g) (e f+d g) (f+g x)^2-8 c e (e f-d g) \left (2 a e^2 g^2+c \left (e^2 f^2+2 d e g f+3 d^2 g^2\right )\right ) (f+g x)}{e f-d g-e (f+g x)}d\sqrt {f+g x}}{2 (e f-d g)}+\frac {g^3 \sqrt {f+g x} \left (a e^2+c d^2\right ) \left (a e^2 g-c d (16 e f-17 d g)\right )}{2 (e f-d g) (-d g-e (f+g x)+e f)}}{4 e^5}-\frac {g^4 \sqrt {f+g x} \left (a e^2+c d^2\right )^2}{4 e^5 (-d g-e (f+g x)+e f)^2}\right )}{g^2}\)

\(\Big \downarrow \) 2341

\(\displaystyle \frac {2 \left (\frac {\frac {\int \left (8 e^2 (e f-d g) (f+g x)^2 c^2-8 e (e f-d g) (e f+3 d g) (f+g x) c^2+16 \left (3 c d^2+a e^2\right ) g^2 (e f-d g) c+\frac {a^2 g^4 e^4-16 a c f^2 g^2 e^4+48 a c d f g^3 e^3-30 a c d^2 g^4 e^2-48 c^2 d^2 f^2 g^2 e^2+112 c^2 d^3 f g^3 e-63 c^2 d^4 g^4}{e f-d g-e (f+g x)}\right )d\sqrt {f+g x}}{2 (e f-d g)}+\frac {g^3 \sqrt {f+g x} \left (a e^2+c d^2\right ) \left (a e^2 g-c d (16 e f-17 d g)\right )}{2 (e f-d g) (-d g-e (f+g x)+e f)}}{4 e^5}-\frac {g^4 \sqrt {f+g x} \left (a e^2+c d^2\right )^2}{4 e^5 (-d g-e (f+g x)+e f)^2}\right )}{g^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (\frac {\frac {\frac {g^2 \left (a^2 e^4 g^2-2 a c e^2 \left (15 d^2 g^2-24 d e f g+8 e^2 f^2\right )-c^2 d^2 \left (63 d^2 g^2-112 d e f g+48 e^2 f^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e} \sqrt {e f-d g}}+16 c g^2 \sqrt {f+g x} \left (a e^2+3 c d^2\right ) (e f-d g)+\frac {8}{5} c^2 e^2 (f+g x)^{5/2} (e f-d g)-\frac {8}{3} c^2 e (f+g x)^{3/2} (e f-d g) (3 d g+e f)}{2 (e f-d g)}+\frac {g^3 \sqrt {f+g x} \left (a e^2+c d^2\right ) \left (a e^2 g-c d (16 e f-17 d g)\right )}{2 (e f-d g) (-d g-e (f+g x)+e f)}}{4 e^5}-\frac {g^4 \sqrt {f+g x} \left (a e^2+c d^2\right )^2}{4 e^5 (-d g-e (f+g x)+e f)^2}\right )}{g^2}\)

Input:

Int[(Sqrt[f + g*x]*(a + c*x^2)^2)/(d + e*x)^3,x]
 

Output:

(2*(-1/4*((c*d^2 + a*e^2)^2*g^4*Sqrt[f + g*x])/(e^5*(e*f - d*g - e*(f + g* 
x))^2) + (((c*d^2 + a*e^2)*g^3*(a*e^2*g - c*d*(16*e*f - 17*d*g))*Sqrt[f + 
g*x])/(2*(e*f - d*g)*(e*f - d*g - e*(f + g*x))) + (16*c*(3*c*d^2 + a*e^2)* 
g^2*(e*f - d*g)*Sqrt[f + g*x] - (8*c^2*e*(e*f - d*g)*(e*f + 3*d*g)*(f + g* 
x)^(3/2))/3 + (8*c^2*e^2*(e*f - d*g)*(f + g*x)^(5/2))/5 + (g^2*(a^2*e^4*g^ 
2 - 2*a*c*e^2*(8*e^2*f^2 - 24*d*e*f*g + 15*d^2*g^2) - c^2*d^2*(48*e^2*f^2 
- 112*d*e*f*g + 63*d^2*g^2))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d* 
g]])/(Sqrt[e]*Sqrt[e*f - d*g]))/(2*(e*f - d*g)))/(4*e^5)))/g^2
 

Defintions of rubi rules used

rule 25

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

rule 649

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^ 
2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1)   Subst[Int[x^(2*m + 1)*(e*f 
- d*g + g*x^2)^n*(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x 
]], x] /; FreeQ[{a, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && Integ 
erQ[m + 1/2]
 

rule 1580

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_) 
^4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d 
 + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[1/(2*e^(2*p + m/2)* 
(q + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2* 
e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b 
*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b, c, d, e 
}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2341

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.20

method	result	size
risch	\(\frac {2 c \left (3 c \,x^{2} e^{2} g^{2}-15 c d e \,g^{2} x +c \,e^{2} f g x +30 a \,e^{2} g^{2}+90 c \,d^{2} g^{2}-15 c d e f g -2 c \,e^{2} f^{2}\right ) \sqrt {g x +f}}{15 g^{2} e^{5}}-\frac {\frac {-\frac {e g \left (a^{2} e^{4} g +18 a c \,d^{2} e^{2} g -16 a c d \,e^{3} f +17 c^{2} d^{4} g -16 c^{2} d^{3} e f \right ) \left (g x +f \right )^{\frac {3}{2}}}{4 \left (d g -e f \right )}+\frac {g \left (a^{2} e^{4} g -14 a c \,d^{2} e^{2} g +16 a c d \,e^{3} f -15 c^{2} d^{4} g +16 c^{2} d^{3} e f \right ) \sqrt {g x +f}}{4}}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}-\frac {\left (a^{2} e^{4} g^{2}-30 a c \,d^{2} e^{2} g^{2}+48 a c d \,e^{3} f g -16 a c \,e^{4} f^{2}-63 c^{2} d^{4} g^{2}+112 c^{2} d^{3} e f g -48 c^{2} d^{2} e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}}{e^{5}}\)	\(361\)
derivativedivides	\(\frac {\frac {2 c \left (\frac {c \left (g x +f \right )^{\frac {5}{2}} e^{2}}{5}-c d e g \left (g x +f \right )^{\frac {3}{2}}-\frac {c \,e^{2} f \left (g x +f \right )^{\frac {3}{2}}}{3}+2 a \,e^{2} g^{2} \sqrt {g x +f}+6 c \,d^{2} g^{2} \sqrt {g x +f}\right )}{e^{5}}+\frac {2 g^{2} \left (\frac {\frac {e g \left (a^{2} e^{4} g +18 a c \,d^{2} e^{2} g -16 a c d \,e^{3} f +17 c^{2} d^{4} g -16 c^{2} d^{3} e f \right ) \left (g x +f \right )^{\frac {3}{2}}}{8 d g -8 e f}-\frac {g \left (a^{2} e^{4} g -14 a c \,d^{2} e^{2} g +16 a c d \,e^{3} f -15 c^{2} d^{4} g +16 c^{2} d^{3} e f \right ) \sqrt {g x +f}}{8}}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (a^{2} e^{4} g^{2}-30 a c \,d^{2} e^{2} g^{2}+48 a c d \,e^{3} f g -16 a c \,e^{4} f^{2}-63 c^{2} d^{4} g^{2}+112 c^{2} d^{3} e f g -48 c^{2} d^{2} e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{8 \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}\right )}{e^{5}}}{g^{2}}\)	\(365\)
default	\(\frac {\frac {2 c \left (\frac {c \left (g x +f \right )^{\frac {5}{2}} e^{2}}{5}-c d e g \left (g x +f \right )^{\frac {3}{2}}-\frac {c \,e^{2} f \left (g x +f \right )^{\frac {3}{2}}}{3}+2 a \,e^{2} g^{2} \sqrt {g x +f}+6 c \,d^{2} g^{2} \sqrt {g x +f}\right )}{e^{5}}+\frac {2 g^{2} \left (\frac {\frac {e g \left (a^{2} e^{4} g +18 a c \,d^{2} e^{2} g -16 a c d \,e^{3} f +17 c^{2} d^{4} g -16 c^{2} d^{3} e f \right ) \left (g x +f \right )^{\frac {3}{2}}}{8 d g -8 e f}-\frac {g \left (a^{2} e^{4} g -14 a c \,d^{2} e^{2} g +16 a c d \,e^{3} f -15 c^{2} d^{4} g +16 c^{2} d^{3} e f \right ) \sqrt {g x +f}}{8}}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (a^{2} e^{4} g^{2}-30 a c \,d^{2} e^{2} g^{2}+48 a c d \,e^{3} f g -16 a c \,e^{4} f^{2}-63 c^{2} d^{4} g^{2}+112 c^{2} d^{3} e f g -48 c^{2} d^{2} e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{8 \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}\right )}{e^{5}}}{g^{2}}\)	\(365\)
pseudoelliptic	\(-\frac {\sqrt {\left (d g -e f \right ) e}\, \left (\left (\left (-63 d^{5}-\frac {8}{5} d \,e^{4} x^{4}+\frac {24}{5} d^{2} e^{3} x^{3}-\frac {168}{5} d^{3} e^{2} x^{2}-105 d^{4} e x \right ) g^{3}+70 e f \left (\frac {4}{175} e^{4} x^{4}-\frac {8}{105} d \,e^{3} x^{3}+\frac {304}{525} d^{2} e^{2} x^{2}+\frac {128}{75} d^{3} e x +d^{4}\right ) g^{2}-\frac {104 e^{2} f^{2} \left (-\frac {e x}{13}+d \right ) \left (e x +d \right )^{2} g}{15}-\frac {16 e^{3} f^{3} \left (e x +d \right )^{2}}{15}\right ) c^{2}-30 e^{2} g^{2} \left (d \left (\frac {8}{15} e^{2} x^{2}+\frac {5}{3} d e x +d^{2}\right ) g -\frac {14 e f \left (\frac {4}{7} e^{2} x^{2}+\frac {12}{7} d e x +d^{2}\right )}{15}\right ) a c +e^{4} \left (\left (-e x +d \right ) g -2 e f \right ) g^{2} a^{2}\right ) \sqrt {g x +f}-\arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) g^{2} \left (\left (-63 d^{4} g^{2}+112 f g e \,d^{3}-48 d^{2} e^{2} f^{2}\right ) c^{2}-30 e^{2} a \left (d^{2} g^{2}-\frac {8}{5} d e f g +\frac {8}{15} e^{2} f^{2}\right ) c +a^{2} e^{4} g^{2}\right ) \left (e x +d \right )^{2}}{4 \sqrt {\left (d g -e f \right ) e}\, e^{5} \left (e x +d \right )^{2} \left (d g -e f \right ) g^{2}}\)	\(377\)

Input:

int((g*x+f)^(1/2)*(c*x^2+a)^2/(e*x+d)^3,x,method=_RETURNVERBOSE)
 

Output:

2/15*c*(3*c*e^2*g^2*x^2-15*c*d*e*g^2*x+c*e^2*f*g*x+30*a*e^2*g^2+90*c*d^2*g 
^2-15*c*d*e*f*g-2*c*e^2*f^2)*(g*x+f)^(1/2)/g^2/e^5-1/e^5*(2*(-1/8*e*g*(a^2 
*e^4*g+18*a*c*d^2*e^2*g-16*a*c*d*e^3*f+17*c^2*d^4*g-16*c^2*d^3*e*f)/(d*g-e 
*f)*(g*x+f)^(3/2)+1/8*g*(a^2*e^4*g-14*a*c*d^2*e^2*g+16*a*c*d*e^3*f-15*c^2* 
d^4*g+16*c^2*d^3*e*f)*(g*x+f)^(1/2))/(e*(g*x+f)+d*g-e*f)^2-1/4*(a^2*e^4*g^ 
2-30*a*c*d^2*e^2*g^2+48*a*c*d*e^3*f*g-16*a*c*e^4*f^2-63*c^2*d^4*g^2+112*c^ 
2*d^3*e*f*g-48*c^2*d^2*e^2*f^2)/(d*g-e*f)/((d*g-e*f)*e)^(1/2)*arctan(e*(g* 
x+f)^(1/2)/((d*g-e*f)*e)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 926 vs. \(2 (271) = 542\).

Time = 0.19 (sec) , antiderivative size = 1865, normalized size of antiderivative = 6.20 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\text {Too large to display} \] Input:

integrate((g*x+f)^(1/2)*(c*x^2+a)^2/(e*x+d)^3,x, algorithm="fricas")
 

Output:

[1/120*(15*(16*(3*c^2*d^4*e^2 + a*c*d^2*e^4)*f^2*g^2 - 16*(7*c^2*d^5*e + 3 
*a*c*d^3*e^3)*f*g^3 + (63*c^2*d^6 + 30*a*c*d^4*e^2 - a^2*d^2*e^4)*g^4 + (1 
6*(3*c^2*d^2*e^4 + a*c*e^6)*f^2*g^2 - 16*(7*c^2*d^3*e^3 + 3*a*c*d*e^5)*f*g 
^3 + (63*c^2*d^4*e^2 + 30*a*c*d^2*e^4 - a^2*e^6)*g^4)*x^2 + 2*(16*(3*c^2*d 
^3*e^3 + a*c*d*e^5)*f^2*g^2 - 16*(7*c^2*d^4*e^2 + 3*a*c*d^2*e^4)*f*g^3 + ( 
63*c^2*d^5*e + 30*a*c*d^3*e^3 - a^2*d*e^5)*g^4)*x)*sqrt(e^2*f - d*e*g)*log 
((e*g*x + 2*e*f - d*g - 2*sqrt(e^2*f - d*e*g)*sqrt(g*x + f))/(e*x + d)) - 
2*(16*c^2*d^2*e^5*f^4 + 88*c^2*d^3*e^4*f^3*g - 2*(577*c^2*d^4*e^3 + 210*a* 
c*d^2*e^5 - 15*a^2*e^7)*f^2*g^2 + 15*(133*c^2*d^5*e^2 + 58*a*c*d^3*e^4 - 3 
*a^2*d*e^6)*f*g^3 - 15*(63*c^2*d^6*e + 30*a*c*d^4*e^3 - a^2*d^2*e^5)*g^4 - 
 24*(c^2*e^7*f^2*g^2 - 2*c^2*d*e^6*f*g^3 + c^2*d^2*e^5*g^4)*x^4 - 8*(c^2*e 
^7*f^3*g - 11*c^2*d*e^6*f^2*g^2 + 19*c^2*d^2*e^5*f*g^3 - 9*c^2*d^3*e^4*g^4 
)*x^3 + 8*(2*c^2*e^7*f^4 + 9*c^2*d*e^6*f^3*g - 3*(29*c^2*d^2*e^5 + 10*a*c* 
e^7)*f^2*g^2 + (139*c^2*d^3*e^4 + 60*a*c*d*e^6)*f*g^3 - 3*(21*c^2*d^4*e^3 
+ 10*a*c*d^2*e^5)*g^4)*x^2 + (32*c^2*d*e^6*f^4 + 168*c^2*d^2*e^5*f^3*g - 2 
4*(83*c^2*d^3*e^4 + 30*a*c*d*e^6)*f^2*g^2 + (3367*c^2*d^4*e^3 + 1470*a*c*d 
^2*e^5 + 15*a^2*e^7)*f*g^3 - 15*(105*c^2*d^5*e^2 + 50*a*c*d^3*e^4 + a^2*d* 
e^6)*g^4)*x)*sqrt(g*x + f))/(d^2*e^8*f^2*g^2 - 2*d^3*e^7*f*g^3 + d^4*e^6*g 
^4 + (e^10*f^2*g^2 - 2*d*e^9*f*g^3 + d^2*e^8*g^4)*x^2 + 2*(d*e^9*f^2*g^2 - 
 2*d^2*e^8*f*g^3 + d^3*e^7*g^4)*x), 1/60*(15*(16*(3*c^2*d^4*e^2 + a*c*d...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\text {Timed out} \] Input:

integrate((g*x+f)**(1/2)*(c*x**2+a)**2/(e*x+d)**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((g*x+f)^(1/2)*(c*x^2+a)^2/(e*x+d)^3,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e*(d*g-e*f)>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 524, normalized size of antiderivative = 1.74 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {{\left (48 \, c^{2} d^{2} e^{2} f^{2} + 16 \, a c e^{4} f^{2} - 112 \, c^{2} d^{3} e f g - 48 \, a c d e^{3} f g + 63 \, c^{2} d^{4} g^{2} + 30 \, a c d^{2} e^{2} g^{2} - a^{2} e^{4} g^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {-e^{2} f + d e g}}\right )}{4 \, {\left (e^{6} f - d e^{5} g\right )} \sqrt {-e^{2} f + d e g}} + \frac {16 \, {\left (g x + f\right )}^{\frac {3}{2}} c^{2} d^{3} e^{2} f g + 16 \, {\left (g x + f\right )}^{\frac {3}{2}} a c d e^{4} f g - 16 \, \sqrt {g x + f} c^{2} d^{3} e^{2} f^{2} g - 16 \, \sqrt {g x + f} a c d e^{4} f^{2} g - 17 \, {\left (g x + f\right )}^{\frac {3}{2}} c^{2} d^{4} e g^{2} - 18 \, {\left (g x + f\right )}^{\frac {3}{2}} a c d^{2} e^{3} g^{2} - {\left (g x + f\right )}^{\frac {3}{2}} a^{2} e^{5} g^{2} + 31 \, \sqrt {g x + f} c^{2} d^{4} e f g^{2} + 30 \, \sqrt {g x + f} a c d^{2} e^{3} f g^{2} - \sqrt {g x + f} a^{2} e^{5} f g^{2} - 15 \, \sqrt {g x + f} c^{2} d^{5} g^{3} - 14 \, \sqrt {g x + f} a c d^{3} e^{2} g^{3} + \sqrt {g x + f} a^{2} d e^{4} g^{3}}{4 \, {\left (e^{6} f - d e^{5} g\right )} {\left ({\left (g x + f\right )} e - e f + d g\right )}^{2}} + \frac {2 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} c^{2} e^{12} g^{8} - 5 \, {\left (g x + f\right )}^{\frac {3}{2}} c^{2} e^{12} f g^{8} - 15 \, {\left (g x + f\right )}^{\frac {3}{2}} c^{2} d e^{11} g^{9} + 90 \, \sqrt {g x + f} c^{2} d^{2} e^{10} g^{10} + 30 \, \sqrt {g x + f} a c e^{12} g^{10}\right )}}{15 \, e^{15} g^{10}} \] Input:

integrate((g*x+f)^(1/2)*(c*x^2+a)^2/(e*x+d)^3,x, algorithm="giac")
 

Output:

1/4*(48*c^2*d^2*e^2*f^2 + 16*a*c*e^4*f^2 - 112*c^2*d^3*e*f*g - 48*a*c*d*e^ 
3*f*g + 63*c^2*d^4*g^2 + 30*a*c*d^2*e^2*g^2 - a^2*e^4*g^2)*arctan(sqrt(g*x 
 + f)*e/sqrt(-e^2*f + d*e*g))/((e^6*f - d*e^5*g)*sqrt(-e^2*f + d*e*g)) + 1 
/4*(16*(g*x + f)^(3/2)*c^2*d^3*e^2*f*g + 16*(g*x + f)^(3/2)*a*c*d*e^4*f*g 
- 16*sqrt(g*x + f)*c^2*d^3*e^2*f^2*g - 16*sqrt(g*x + f)*a*c*d*e^4*f^2*g - 
17*(g*x + f)^(3/2)*c^2*d^4*e*g^2 - 18*(g*x + f)^(3/2)*a*c*d^2*e^3*g^2 - (g 
*x + f)^(3/2)*a^2*e^5*g^2 + 31*sqrt(g*x + f)*c^2*d^4*e*f*g^2 + 30*sqrt(g*x 
 + f)*a*c*d^2*e^3*f*g^2 - sqrt(g*x + f)*a^2*e^5*f*g^2 - 15*sqrt(g*x + f)*c 
^2*d^5*g^3 - 14*sqrt(g*x + f)*a*c*d^3*e^2*g^3 + sqrt(g*x + f)*a^2*d*e^4*g^ 
3)/((e^6*f - d*e^5*g)*((g*x + f)*e - e*f + d*g)^2) + 2/15*(3*(g*x + f)^(5/ 
2)*c^2*e^12*g^8 - 5*(g*x + f)^(3/2)*c^2*e^12*f*g^8 - 15*(g*x + f)^(3/2)*c^ 
2*d*e^11*g^9 + 90*sqrt(g*x + f)*c^2*d^2*e^10*g^10 + 30*sqrt(g*x + f)*a*c*e 
^12*g^10)/(e^15*g^10)
 

Mupad [B] (verification not implemented)

Time = 6.19 (sec) , antiderivative size = 483, normalized size of antiderivative = 1.60 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^3} \, dx=\sqrt {f+g\,x}\,\left (\frac {12\,c^2\,f^2+4\,a\,c\,g^2}{e^3\,g^2}+\frac {3\,\left (d\,g-e\,f\right )\,\left (\frac {8\,c^2\,f}{e^3\,g^2}+\frac {6\,c^2\,\left (d\,g-e\,f\right )}{e^4\,g^2}\right )}{e}-\frac {6\,c^2\,{\left (d\,g-e\,f\right )}^2}{e^5\,g^2}\right )-{\left (f+g\,x\right )}^{3/2}\,\left (\frac {8\,c^2\,f}{3\,e^3\,g^2}+\frac {2\,c^2\,\left (d\,g-e\,f\right )}{e^4\,g^2}\right )-\frac {\sqrt {f+g\,x}\,\left (\frac {a^2\,e^4\,g^2}{4}-\frac {7\,a\,c\,d^2\,e^2\,g^2}{2}+4\,f\,a\,c\,d\,e^3\,g-\frac {15\,c^2\,d^4\,g^2}{4}+4\,f\,c^2\,d^3\,e\,g\right )-\frac {{\left (f+g\,x\right )}^{3/2}\,\left (a^2\,e^5\,g^2+18\,a\,c\,d^2\,e^3\,g^2-16\,f\,a\,c\,d\,e^4\,g+17\,c^2\,d^4\,e\,g^2-16\,f\,c^2\,d^3\,e^2\,g\right )}{4\,\left (d\,g-e\,f\right )}}{e^7\,{\left (f+g\,x\right )}^2-\left (f+g\,x\right )\,\left (2\,e^7\,f-2\,d\,e^6\,g\right )+e^7\,f^2+d^2\,e^5\,g^2-2\,d\,e^6\,f\,g}+\frac {2\,c^2\,{\left (f+g\,x\right )}^{5/2}}{5\,e^3\,g^2}-\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (-a^2\,e^4\,g^2+30\,a\,c\,d^2\,e^2\,g^2-48\,a\,c\,d\,e^3\,f\,g+16\,a\,c\,e^4\,f^2+63\,c^2\,d^4\,g^2-112\,c^2\,d^3\,e\,f\,g+48\,c^2\,d^2\,e^2\,f^2\right )}{4\,e^{11/2}\,{\left (d\,g-e\,f\right )}^{3/2}} \] Input:

int(((f + g*x)^(1/2)*(a + c*x^2)^2)/(d + e*x)^3,x)
 

Output:

(f + g*x)^(1/2)*((12*c^2*f^2 + 4*a*c*g^2)/(e^3*g^2) + (3*(d*g - e*f)*((8*c 
^2*f)/(e^3*g^2) + (6*c^2*(d*g - e*f))/(e^4*g^2)))/e - (6*c^2*(d*g - e*f)^2 
)/(e^5*g^2)) - (f + g*x)^(3/2)*((8*c^2*f)/(3*e^3*g^2) + (2*c^2*(d*g - e*f) 
)/(e^4*g^2)) - ((f + g*x)^(1/2)*((a^2*e^4*g^2)/4 - (15*c^2*d^4*g^2)/4 + 4* 
c^2*d^3*e*f*g - (7*a*c*d^2*e^2*g^2)/2 + 4*a*c*d*e^3*f*g) - ((f + g*x)^(3/2 
)*(a^2*e^5*g^2 + 17*c^2*d^4*e*g^2 + 18*a*c*d^2*e^3*g^2 - 16*c^2*d^3*e^2*f* 
g - 16*a*c*d*e^4*f*g))/(4*(d*g - e*f)))/(e^7*(f + g*x)^2 - (f + g*x)*(2*e^ 
7*f - 2*d*e^6*g) + e^7*f^2 + d^2*e^5*g^2 - 2*d*e^6*f*g) + (2*c^2*(f + g*x) 
^(5/2))/(5*e^3*g^2) - (atan((e^(1/2)*(f + g*x)^(1/2))/(d*g - e*f)^(1/2))*( 
63*c^2*d^4*g^2 - a^2*e^4*g^2 + 48*c^2*d^2*e^2*f^2 + 16*a*c*e^4*f^2 - 112*c 
^2*d^3*e*f*g + 30*a*c*d^2*e^2*g^2 - 48*a*c*d*e^3*f*g))/(4*e^(11/2)*(d*g - 
e*f)^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1926, normalized size of antiderivative = 6.40 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^3} \, dx =\text {Too large to display} \] Input:

int((g*x+f)^(1/2)*(c*x^2+a)^2/(e*x+d)^3,x)
 

Output:

(15*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f 
)))*a**2*d**2*e**4*g**4 + 30*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e 
)/(sqrt(e)*sqrt(d*g - e*f)))*a**2*d*e**5*g**4*x + 15*sqrt(e)*sqrt(d*g - e* 
f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*a**2*e**6*g**4*x**2 - 
 450*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e* 
f)))*a*c*d**4*e**2*g**4 + 720*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)* 
e)/(sqrt(e)*sqrt(d*g - e*f)))*a*c*d**3*e**3*f*g**3 - 900*sqrt(e)*sqrt(d*g 
- e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*a*c*d**3*e**3*g** 
4*x - 240*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g 
 - e*f)))*a*c*d**2*e**4*f**2*g**2 + 1440*sqrt(e)*sqrt(d*g - e*f)*atan((sqr 
t(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*a*c*d**2*e**4*f*g**3*x - 450*sqrt 
(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*a*c* 
d**2*e**4*g**4*x**2 - 480*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/( 
sqrt(e)*sqrt(d*g - e*f)))*a*c*d*e**5*f**2*g**2*x + 720*sqrt(e)*sqrt(d*g - 
e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*a*c*d*e**5*f*g**3*x 
**2 - 240*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g 
 - e*f)))*a*c*e**6*f**2*g**2*x**2 - 945*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt 
(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*c**2*d**6*g**4 + 1680*sqrt(e)*sqrt 
(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*c**2*d**5*e* 
f*g**3 - 1890*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*s...