\(\int \frac {(f+g x)^{3/2} (a+c x^2)}{(d+e x)^2} \, dx\) [47]

Optimal result

Integrand size = 24, antiderivative size = 173 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^2} \, dx=\frac {\left (3 a e^2 g-c d (4 e f-7 d g)\right ) \sqrt {f+g x}}{e^4}-\frac {4 c d (f+g x)^{3/2}}{3 e^3}-\frac {\left (c d^2+a e^2\right ) (f+g x)^{3/2}}{e^3 (d+e x)}+\frac {2 c (f+g x)^{5/2}}{5 e^2 g}-\frac {\sqrt {e f-d g} \left (3 a e^2 g-c d (4 e f-7 d g)\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{9/2}} \] Output:

(3*a*e^2*g-c*d*(-7*d*g+4*e*f))*(g*x+f)^(1/2)/e^4-4/3*c*d*(g*x+f)^(3/2)/e^3 
-(a*e^2+c*d^2)*(g*x+f)^(3/2)/e^3/(e*x+d)+2/5*c*(g*x+f)^(5/2)/e^2/g-(-d*g+e 
*f)^(1/2)*(3*a*e^2*g-c*d*(-7*d*g+4*e*f))*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d 
*g+e*f)^(1/2))/e^(9/2)
 

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.04 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^2} \, dx=\frac {\sqrt {f+g x} \left (15 a e^2 g (-e f+3 d g+2 e g x)+c \left (105 d^3 g^2+6 e^3 x (f+g x)^2+5 d^2 e g (-19 f+14 g x)+2 d e^2 \left (3 f^2-34 f g x-7 g^2 x^2\right )\right )\right )}{15 e^4 g (d+e x)}-\frac {\sqrt {-e f+d g} \left (3 a e^2 g+c d (-4 e f+7 d g)\right ) \arctan \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{e^{9/2}} \] Input:

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

Output:

(Sqrt[f + g*x]*(15*a*e^2*g*(-(e*f) + 3*d*g + 2*e*g*x) + c*(105*d^3*g^2 + 6 
*e^3*x*(f + g*x)^2 + 5*d^2*e*g*(-19*f + 14*g*x) + 2*d*e^2*(3*f^2 - 34*f*g* 
x - 7*g^2*x^2))))/(15*e^4*g*(d + e*x)) - (Sqrt[-(e*f) + d*g]*(3*a*e^2*g + 
c*d*(-4*e*f + 7*d*g))*ArcTan[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g]])/ 
e^(9/2)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {649, 1580, 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.

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

\(\Big \downarrow \) 649

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

\(\Big \downarrow \) 1580

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

\(\Big \downarrow \) 2341

\(\displaystyle \frac {2 \left (\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right ) (e f-d g)}{2 e^4 (-d g-e (f+g x)+e f)}-\frac {\int \left (-2 c e^2 (f+g x)^2+4 c d e g (f+g x)-2 g \left (a e^2 g-c d (2 e f-3 d g)\right )+\frac {3 a f g^2 e^3-3 a d g^3 e^2-4 c d f^2 g e^2+11 c d^2 f g^2 e-7 c d^3 g^3}{e f-d g-e (f+g x)}\right )d\sqrt {f+g x}}{2 e^4}\right )}{g}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right ) (e f-d g)}{2 e^4 (-d g-e (f+g x)+e f)}-\frac {\frac {g \sqrt {e f-d g} \left (3 a e^2 g-c d (4 e f-7 d g)\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e}}-2 g \sqrt {f+g x} \left (a e^2 g-c d (2 e f-3 d g)\right )+\frac {4}{3} c d e g (f+g x)^{3/2}-\frac {2}{5} c e^2 (f+g x)^{5/2}}{2 e^4}\right )}{g}\)

Input:

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

Output:

(2*(((c*d^2 + a*e^2)*g^2*(e*f - d*g)*Sqrt[f + g*x])/(2*e^4*(e*f - d*g - e* 
(f + g*x))) - (-2*g*(a*e^2*g - c*d*(2*e*f - 3*d*g))*Sqrt[f + g*x] + (4*c*d 
*e*g*(f + g*x)^(3/2))/3 - (2*c*e^2*(f + g*x)^(5/2))/5 + (g*Sqrt[e*f - d*g] 
*(3*a*e^2*g - c*d*(4*e*f - 7*d*g))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e* 
f - d*g]])/Sqrt[e])/(2*e^4)))/g
 

Defintions of rubi rules used

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]
 

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {2 \left (3 c \,x^{2} e^{2} g^{2}-10 c d e \,g^{2} x +6 c \,e^{2} f g x +15 a \,e^{2} g^{2}+45 c \,d^{2} g^{2}-40 c d e f g +3 c \,e^{2} f^{2}\right ) \sqrt {g x +f}}{15 g \,e^{4}}-\frac {\left (2 d g -2 e f \right ) \left (\frac {\left (-\frac {1}{2} a \,e^{2} g -\frac {1}{2} c \,d^{2} g \right ) \sqrt {g x +f}}{e \left (g x +f \right )+d g -e f}+\frac {\left (3 a \,e^{2} g +7 c \,d^{2} g -4 c d e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{2 \sqrt {\left (d g -e f \right ) e}}\right )}{e^{4}}\)	\(194\)
pseudoelliptic	\(\frac {-3 g \left (a \,e^{2} g +\frac {7}{3} c \,d^{2} g -\frac {4}{3} c d e f \right ) \left (e x +d \right ) \left (d g -e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )+3 \sqrt {g x +f}\, \left (\frac {\left (2 \left (\frac {c \,x^{2}}{5}+a \right ) x \,g^{2}-f \left (-\frac {4 c \,x^{2}}{5}+a \right ) g +\frac {2 c \,f^{2} x}{5}\right ) e^{3}}{3}+\left (\left (-\frac {14 c \,x^{2}}{45}+a \right ) g^{2}-\frac {68 c f x g}{45}+\frac {2 c \,f^{2}}{15}\right ) d \,e^{2}-\frac {19 g c \,d^{2} \left (-\frac {14 g x}{19}+f \right ) e}{9}+\frac {7 c \,d^{3} g^{2}}{3}\right ) \sqrt {\left (d g -e f \right ) e}}{e^{4} \left (e x +d \right ) g \sqrt {\left (d g -e f \right ) e}}\)	\(200\)
derivativedivides	\(\frac {\frac {2 \left (\frac {c \left (g x +f \right )^{\frac {5}{2}} e^{2}}{5}-\frac {2 c d e g \left (g x +f \right )^{\frac {3}{2}}}{3}+a \,e^{2} g^{2} \sqrt {g x +f}+3 c \,d^{2} g^{2} \sqrt {g x +f}-2 \sqrt {g x +f}\, c d e f g \right )}{e^{4}}-\frac {2 g \left (\frac {\left (-\frac {1}{2} a d \,e^{2} g^{2}+\frac {1}{2} a \,e^{3} f g -\frac {1}{2} c \,d^{3} g^{2}+\frac {1}{2} c \,d^{2} e f g \right ) \sqrt {g x +f}}{e \left (g x +f \right )+d g -e f}+\frac {\left (3 a d \,e^{2} g^{2}-3 a \,e^{3} f g +7 c \,d^{3} g^{2}-11 c \,d^{2} e f g +4 c d \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{2 \sqrt {\left (d g -e f \right ) e}}\right )}{e^{4}}}{g}\)	\(235\)
default	\(\frac {\frac {2 \left (\frac {c \left (g x +f \right )^{\frac {5}{2}} e^{2}}{5}-\frac {2 c d e g \left (g x +f \right )^{\frac {3}{2}}}{3}+a \,e^{2} g^{2} \sqrt {g x +f}+3 c \,d^{2} g^{2} \sqrt {g x +f}-2 \sqrt {g x +f}\, c d e f g \right )}{e^{4}}-\frac {2 g \left (\frac {\left (-\frac {1}{2} a d \,e^{2} g^{2}+\frac {1}{2} a \,e^{3} f g -\frac {1}{2} c \,d^{3} g^{2}+\frac {1}{2} c \,d^{2} e f g \right ) \sqrt {g x +f}}{e \left (g x +f \right )+d g -e f}+\frac {\left (3 a d \,e^{2} g^{2}-3 a \,e^{3} f g +7 c \,d^{3} g^{2}-11 c \,d^{2} e f g +4 c d \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{2 \sqrt {\left (d g -e f \right ) e}}\right )}{e^{4}}}{g}\)	\(235\)

Input:

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

Output:

2/15/g*(3*c*e^2*g^2*x^2-10*c*d*e*g^2*x+6*c*e^2*f*g*x+15*a*e^2*g^2+45*c*d^2 
*g^2-40*c*d*e*f*g+3*c*e^2*f^2)*(g*x+f)^(1/2)/e^4-1/e^4*(2*d*g-2*e*f)*((-1/ 
2*a*e^2*g-1/2*c*d^2*g)*(g*x+f)^(1/2)/(e*(g*x+f)+d*g-e*f)+1/2*(3*a*e^2*g+7* 
c*d^2*g-4*c*d*e*f)/((d*g-e*f)*e)^(1/2)*arctan(e*(g*x+f)^(1/2)/((d*g-e*f)*e 
)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 539, normalized size of antiderivative = 3.12 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^2} \, dx=\left [-\frac {15 \, {\left (4 \, c d^{2} e f g - {\left (7 \, c d^{3} + 3 \, a d e^{2}\right )} g^{2} + {\left (4 \, c d e^{2} f g - {\left (7 \, c d^{2} e + 3 \, a e^{3}\right )} g^{2}\right )} x\right )} \sqrt {\frac {e f - d g}{e}} \log \left (\frac {e g x + 2 \, e f - d g - 2 \, \sqrt {g x + f} e \sqrt {\frac {e f - d g}{e}}}{e x + d}\right ) - 2 \, {\left (6 \, c e^{3} g^{2} x^{3} + 6 \, c d e^{2} f^{2} - 5 \, {\left (19 \, c d^{2} e + 3 \, a e^{3}\right )} f g + 15 \, {\left (7 \, c d^{3} + 3 \, a d e^{2}\right )} g^{2} + 2 \, {\left (6 \, c e^{3} f g - 7 \, c d e^{2} g^{2}\right )} x^{2} + 2 \, {\left (3 \, c e^{3} f^{2} - 34 \, c d e^{2} f g + 5 \, {\left (7 \, c d^{2} e + 3 \, a e^{3}\right )} g^{2}\right )} x\right )} \sqrt {g x + f}}{30 \, {\left (e^{5} g x + d e^{4} g\right )}}, \frac {15 \, {\left (4 \, c d^{2} e f g - {\left (7 \, c d^{3} + 3 \, a d e^{2}\right )} g^{2} + {\left (4 \, c d e^{2} f g - {\left (7 \, c d^{2} e + 3 \, a e^{3}\right )} g^{2}\right )} x\right )} \sqrt {-\frac {e f - d g}{e}} \arctan \left (-\frac {\sqrt {g x + f} e \sqrt {-\frac {e f - d g}{e}}}{e f - d g}\right ) + {\left (6 \, c e^{3} g^{2} x^{3} + 6 \, c d e^{2} f^{2} - 5 \, {\left (19 \, c d^{2} e + 3 \, a e^{3}\right )} f g + 15 \, {\left (7 \, c d^{3} + 3 \, a d e^{2}\right )} g^{2} + 2 \, {\left (6 \, c e^{3} f g - 7 \, c d e^{2} g^{2}\right )} x^{2} + 2 \, {\left (3 \, c e^{3} f^{2} - 34 \, c d e^{2} f g + 5 \, {\left (7 \, c d^{2} e + 3 \, a e^{3}\right )} g^{2}\right )} x\right )} \sqrt {g x + f}}{15 \, {\left (e^{5} g x + d e^{4} g\right )}}\right ] \] Input:

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

Output:

[-1/30*(15*(4*c*d^2*e*f*g - (7*c*d^3 + 3*a*d*e^2)*g^2 + (4*c*d*e^2*f*g - ( 
7*c*d^2*e + 3*a*e^3)*g^2)*x)*sqrt((e*f - d*g)/e)*log((e*g*x + 2*e*f - d*g 
- 2*sqrt(g*x + f)*e*sqrt((e*f - d*g)/e))/(e*x + d)) - 2*(6*c*e^3*g^2*x^3 + 
 6*c*d*e^2*f^2 - 5*(19*c*d^2*e + 3*a*e^3)*f*g + 15*(7*c*d^3 + 3*a*d*e^2)*g 
^2 + 2*(6*c*e^3*f*g - 7*c*d*e^2*g^2)*x^2 + 2*(3*c*e^3*f^2 - 34*c*d*e^2*f*g 
 + 5*(7*c*d^2*e + 3*a*e^3)*g^2)*x)*sqrt(g*x + f))/(e^5*g*x + d*e^4*g), 1/1 
5*(15*(4*c*d^2*e*f*g - (7*c*d^3 + 3*a*d*e^2)*g^2 + (4*c*d*e^2*f*g - (7*c*d 
^2*e + 3*a*e^3)*g^2)*x)*sqrt(-(e*f - d*g)/e)*arctan(-sqrt(g*x + f)*e*sqrt( 
-(e*f - d*g)/e)/(e*f - d*g)) + (6*c*e^3*g^2*x^3 + 6*c*d*e^2*f^2 - 5*(19*c* 
d^2*e + 3*a*e^3)*f*g + 15*(7*c*d^3 + 3*a*d*e^2)*g^2 + 2*(6*c*e^3*f*g - 7*c 
*d*e^2*g^2)*x^2 + 2*(3*c*e^3*f^2 - 34*c*d*e^2*f*g + 5*(7*c*d^2*e + 3*a*e^3 
)*g^2)*x)*sqrt(g*x + f))/(e^5*g*x + d*e^4*g)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((g*x+f)^(3/2)*(c*x^2+a)/(e*x+d)^2,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 = 269, normalized size of antiderivative = 1.55 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^2} \, dx=-\frac {{\left (4 \, c d e^{2} f^{2} - 11 \, c d^{2} e f g - 3 \, a e^{3} f g + 7 \, c d^{3} g^{2} + 3 \, a d e^{2} g^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {-e^{2} f + d e g}}\right )}{\sqrt {-e^{2} f + d e g} e^{4}} - \frac {\sqrt {g x + f} c d^{2} e f g + \sqrt {g x + f} a e^{3} f g - \sqrt {g x + f} c d^{3} g^{2} - \sqrt {g x + f} a d e^{2} g^{2}}{{\left ({\left (g x + f\right )} e - e f + d g\right )} e^{4}} + \frac {2 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} c e^{8} g^{4} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} c d e^{7} g^{5} - 30 \, \sqrt {g x + f} c d e^{7} f g^{5} + 45 \, \sqrt {g x + f} c d^{2} e^{6} g^{6} + 15 \, \sqrt {g x + f} a e^{8} g^{6}\right )}}{15 \, e^{10} g^{5}} \] Input:

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

Output:

-(4*c*d*e^2*f^2 - 11*c*d^2*e*f*g - 3*a*e^3*f*g + 7*c*d^3*g^2 + 3*a*d*e^2*g 
^2)*arctan(sqrt(g*x + f)*e/sqrt(-e^2*f + d*e*g))/(sqrt(-e^2*f + d*e*g)*e^4 
) - (sqrt(g*x + f)*c*d^2*e*f*g + sqrt(g*x + f)*a*e^3*f*g - sqrt(g*x + f)*c 
*d^3*g^2 - sqrt(g*x + f)*a*d*e^2*g^2)/(((g*x + f)*e - e*f + d*g)*e^4) + 2/ 
15*(3*(g*x + f)^(5/2)*c*e^8*g^4 - 10*(g*x + f)^(3/2)*c*d*e^7*g^5 - 30*sqrt 
(g*x + f)*c*d*e^7*f*g^5 + 45*sqrt(g*x + f)*c*d^2*e^6*g^6 + 15*sqrt(g*x + f 
)*a*e^8*g^6)/(e^10*g^5)
 

Mupad [B] (verification not implemented)

Time = 5.66 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.95 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^2} \, dx=\sqrt {f+g\,x}\,\left (\frac {2\,c\,f^2+2\,a\,g^2}{e^2\,g}+\frac {2\,\left (\frac {4\,c\,\left (d\,g-e\,f\right )}{e^3\,g}+\frac {4\,c\,f}{e^2\,g}\right )\,\left (d\,g-e\,f\right )}{e}-\frac {2\,c\,{\left (d\,g-e\,f\right )}^2}{e^4\,g}\right )-{\left (f+g\,x\right )}^{3/2}\,\left (\frac {4\,c\,\left (d\,g-e\,f\right )}{3\,e^3\,g}+\frac {4\,c\,f}{3\,e^2\,g}\right )+\frac {\sqrt {f+g\,x}\,\left (c\,d^3\,g^2-c\,f\,d^2\,e\,g+a\,d\,e^2\,g^2-a\,f\,e^3\,g\right )}{e^5\,\left (f+g\,x\right )-e^5\,f+d\,e^4\,g}+\frac {2\,c\,{\left (f+g\,x\right )}^{5/2}}{5\,e^2\,g}-\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}\,\sqrt {d\,g-e\,f}\,\left (7\,c\,g\,d^2-4\,c\,f\,d\,e+3\,a\,g\,e^2\right )}{7\,c\,d^3\,g^2-11\,c\,d^2\,e\,f\,g+4\,c\,d\,e^2\,f^2+3\,a\,d\,e^2\,g^2-3\,a\,e^3\,f\,g}\right )\,\sqrt {d\,g-e\,f}\,\left (7\,c\,g\,d^2-4\,c\,f\,d\,e+3\,a\,g\,e^2\right )}{e^{9/2}} \] Input:

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

Output:

(f + g*x)^(1/2)*((2*a*g^2 + 2*c*f^2)/(e^2*g) + (2*((4*c*(d*g - e*f))/(e^3* 
g) + (4*c*f)/(e^2*g))*(d*g - e*f))/e - (2*c*(d*g - e*f)^2)/(e^4*g)) - (f + 
 g*x)^(3/2)*((4*c*(d*g - e*f))/(3*e^3*g) + (4*c*f)/(3*e^2*g)) + ((f + g*x) 
^(1/2)*(c*d^3*g^2 - a*e^3*f*g + a*d*e^2*g^2 - c*d^2*e*f*g))/(e^5*(f + g*x) 
 - e^5*f + d*e^4*g) + (2*c*(f + g*x)^(5/2))/(5*e^2*g) - (atan((e^(1/2)*(f 
+ g*x)^(1/2)*(d*g - e*f)^(1/2)*(3*a*e^2*g + 7*c*d^2*g - 4*c*d*e*f))/(7*c*d 
^3*g^2 - 3*a*e^3*f*g + 3*a*d*e^2*g^2 + 4*c*d*e^2*f^2 - 11*c*d^2*e*f*g))*(d 
*g - e*f)^(1/2)*(3*a*e^2*g + 7*c*d^2*g - 4*c*d*e*f))/e^(9/2)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 485, normalized size of antiderivative = 2.80 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^2} \, dx=\frac {-45 \sqrt {e}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) a d \,e^{2} g^{2}-45 \sqrt {e}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) a \,e^{3} g^{2} x -105 \sqrt {e}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) c \,d^{3} g^{2}+60 \sqrt {e}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) c \,d^{2} e f g -105 \sqrt {e}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) c \,d^{2} e \,g^{2} x +60 \sqrt {e}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) c d \,e^{2} f g x +45 \sqrt {g x +f}\, a d \,e^{3} g^{2}-15 \sqrt {g x +f}\, a \,e^{4} f g +30 \sqrt {g x +f}\, a \,e^{4} g^{2} x +105 \sqrt {g x +f}\, c \,d^{3} e \,g^{2}-95 \sqrt {g x +f}\, c \,d^{2} e^{2} f g +70 \sqrt {g x +f}\, c \,d^{2} e^{2} g^{2} x +6 \sqrt {g x +f}\, c d \,e^{3} f^{2}-68 \sqrt {g x +f}\, c d \,e^{3} f g x -14 \sqrt {g x +f}\, c d \,e^{3} g^{2} x^{2}+6 \sqrt {g x +f}\, c \,e^{4} f^{2} x +12 \sqrt {g x +f}\, c \,e^{4} f g \,x^{2}+6 \sqrt {g x +f}\, c \,e^{4} g^{2} x^{3}}{15 e^{5} g \left (e x +d \right )} \] Input:

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

Output:

( - 45*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - 
e*f)))*a*d*e**2*g**2 - 45*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/( 
sqrt(e)*sqrt(d*g - e*f)))*a*e**3*g**2*x - 105*sqrt(e)*sqrt(d*g - e*f)*atan 
((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*c*d**3*g**2 + 60*sqrt(e)*sqr 
t(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*c*d**2*e*f* 
g - 105*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - 
 e*f)))*c*d**2*e*g**2*x + 60*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e 
)/(sqrt(e)*sqrt(d*g - e*f)))*c*d*e**2*f*g*x + 45*sqrt(f + g*x)*a*d*e**3*g* 
*2 - 15*sqrt(f + g*x)*a*e**4*f*g + 30*sqrt(f + g*x)*a*e**4*g**2*x + 105*sq 
rt(f + g*x)*c*d**3*e*g**2 - 95*sqrt(f + g*x)*c*d**2*e**2*f*g + 70*sqrt(f + 
 g*x)*c*d**2*e**2*g**2*x + 6*sqrt(f + g*x)*c*d*e**3*f**2 - 68*sqrt(f + g*x 
)*c*d*e**3*f*g*x - 14*sqrt(f + g*x)*c*d*e**3*g**2*x**2 + 6*sqrt(f + g*x)*c 
*e**4*f**2*x + 12*sqrt(f + g*x)*c*e**4*f*g*x**2 + 6*sqrt(f + g*x)*c*e**4*g 
**2*x**3)/(15*e**5*g*(d + e*x))