Integrand size = 24, antiderivative size = 203 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^3} \, dx=\frac {2 c (e f-3 d g) \sqrt {f+g x}}{e^4}-\frac {\left (3 a e^2 g-c d (8 e f-11 d g)\right ) \sqrt {f+g x}}{4 e^4 (d+e x)}+\frac {2 c (f+g x)^{3/2}}{3 e^3}-\frac {\left (c d^2+a e^2\right ) (f+g x)^{3/2}}{2 e^3 (d+e x)^2}-\frac {\left (3 a e^2 g^2+c \left (8 e^2 f^2-40 d e f g+35 d^2 g^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{9/2} \sqrt {e f-d g}} \] Output:
2*c*(-3*d*g+e*f)*(g*x+f)^(1/2)/e^4-1/4*(3*a*e^2*g-c*d*(-11*d*g+8*e*f))*(g* x+f)^(1/2)/e^4/(e*x+d)+2/3*c*(g*x+f)^(3/2)/e^3-1/2*(a*e^2+c*d^2)*(g*x+f)^( 3/2)/e^3/(e*x+d)^2-1/4*(3*a*e^2*g^2+c*(35*d^2*g^2-40*d*e*f*g+8*e^2*f^2))*a rctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/e^(9/2)/(-d*g+e*f)^(1/2)
Time = 1.00 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.90 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^3} \, dx=\frac {\sqrt {f+g x} \left (-3 a e^2 (2 e f+3 d g+5 e g x)+c \left (-105 d^3 g+25 d^2 e (2 f-7 g x)+8 d e^2 x (11 f-7 g x)+8 e^3 x^2 (4 f+g x)\right )\right )}{12 e^4 (d+e x)^2}+\frac {\left (3 a e^2 g^2+c \left (8 e^2 f^2-40 d e f g+35 d^2 g^2\right )\right ) \arctan \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{4 e^{9/2} \sqrt {-e f+d g}} \] Input:
Integrate[((f + g*x)^(3/2)*(a + c*x^2))/(d + e*x)^3,x]
Output:
(Sqrt[f + g*x]*(-3*a*e^2*(2*e*f + 3*d*g + 5*e*g*x) + c*(-105*d^3*g + 25*d^ 2*e*(2*f - 7*g*x) + 8*d*e^2*x*(11*f - 7*g*x) + 8*e^3*x^2*(4*f + g*x))))/(1 2*e^4*(d + e*x)^2) + ((3*a*e^2*g^2 + c*(8*e^2*f^2 - 40*d*e*f*g + 35*d^2*g^ 2))*ArcTan[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g]])/(4*e^(9/2)*Sqrt[-( e*f) + d*g])
Time = 0.69 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.32, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {649, 25, 1580, 2345, 1467, 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.
| \(\displaystyle \int \frac {\left (a+c x^2\right ) (f+g x)^{3/2}}{(d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 649 |
| \(\displaystyle 2 \int -\frac {(f+g x)^2 \left (c f^2-2 c (f+g x) f+a g^2+c (f+g x)^2\right )}{(e f-d g-e (f+g x))^3}d\sqrt {f+g x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int \frac {(f+g x)^2 \left (c f^2-2 c (f+g x) f+a g^2+c (f+g x)^2\right )}{(e f-d g-e (f+g x))^3}d\sqrt {f+g x}\) |
| \(\Big \downarrow \) 1580 |
| \(\displaystyle 2 \left (\frac {\int \frac {4 c e^3 (f+g x)^3-4 c e^2 (e f+d g) (f+g x)^2+4 e \left (c d^2+a e^2\right ) g^2 (f+g x)+\left (c d^2+a e^2\right ) g^2 (e f-d g)}{(e f-d g-e (f+g x))^2}d\sqrt {f+g x}}{4 e^4}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right ) (e f-d g)}{4 e^4 (-d g-e (f+g x)+e f)^2}\right )\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle 2 \left (\frac {\frac {g \sqrt {f+g x} \left (5 a e^2 g-c d (8 e f-13 d g)\right )}{2 (-d g-e (f+g x)+e f)}-\frac {\int \frac {8 c e^2 (e f-d g) (f+g x)^2-16 c d e g (e f-d g) (f+g x)+g (e f-d g) \left (3 a e^2 g-c d (8 e f-11 d g)\right )}{e f-d g-e (f+g x)}d\sqrt {f+g x}}{2 (e f-d g)}}{4 e^4}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right ) (e f-d g)}{4 e^4 (-d g-e (f+g x)+e f)^2}\right )\) |
| \(\Big \downarrow \) 1467 |
| \(\displaystyle 2 \left (\frac {\frac {g \sqrt {f+g x} \left (5 a e^2 g-c d (8 e f-13 d g)\right )}{2 (-d g-e (f+g x)+e f)}-\frac {\int \left (-8 c (e f-3 d g) (e f-d g)-8 c e (f+g x) (e f-d g)+\frac {8 c f^3 e^3+3 a f g^2 e^3-3 a d g^3 e^2-48 c d f^2 g e^2+75 c d^2 f g^2 e-35 c d^3 g^3}{e f-d g-e (f+g x)}\right )d\sqrt {f+g x}}{2 (e f-d g)}}{4 e^4}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right ) (e f-d g)}{4 e^4 (-d g-e (f+g x)+e f)^2}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (\frac {\frac {g \sqrt {f+g x} \left (5 a e^2 g-c d (8 e f-13 d g)\right )}{2 (-d g-e (f+g x)+e f)}-\frac {\frac {\sqrt {e f-d g} \left (3 a e^2 g^2+c \left (35 d^2 g^2-40 d e f g+8 e^2 f^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e}}-\frac {8}{3} c e (f+g x)^{3/2} (e f-d g)-8 c \sqrt {f+g x} (e f-3 d g) (e f-d g)}{2 (e f-d g)}}{4 e^4}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right ) (e f-d g)}{4 e^4 (-d g-e (f+g x)+e f)^2}\right )\) |
Input:
Int[((f + g*x)^(3/2)*(a + c*x^2))/(d + e*x)^3,x]
Output:
2*(-1/4*((c*d^2 + a*e^2)*g^2*(e*f - d*g)*Sqrt[f + g*x])/(e^4*(e*f - d*g - e*(f + g*x))^2) + ((g*(5*a*e^2*g - c*d*(8*e*f - 13*d*g))*Sqrt[f + g*x])/(2 *(e*f - d*g - e*(f + g*x))) - (-8*c*(e*f - 3*d*g)*(e*f - d*g)*Sqrt[f + g*x ] - (8*c*e*(e*f - d*g)*(f + g*x)^(3/2))/3 + (Sqrt[e*f - d*g]*(3*a*e^2*g^2 + c*(8*e^2*f^2 - 40*d*e*f*g + 35*d^2*g^2))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x]) /Sqrt[e*f - d*g]])/Sqrt[e])/(2*(e*f - d*g)))/(4*e^4))
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]
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
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]
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]
Time = 0.97 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.87
| method | result | size |
| pseudoelliptic | \(-\frac {3 \left (-\left (\left (a \,g^{2}+\frac {8 c \,f^{2}}{3}\right ) e^{2}-\frac {40 c d e f g}{3}+\frac {35 c \,d^{2} g^{2}}{3}\right ) \left (e x +d \right )^{2} \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )+\sqrt {g x +f}\, \left (\left (-\frac {32 x^{2} \left (\frac {g x}{4}+f \right ) c}{9}+\frac {2 \left (\frac {5 g x}{2}+f \right ) a}{3}\right ) e^{3}+\left (-\frac {88 x \left (-\frac {7 g x}{11}+f \right ) c}{9}+a g \right ) d \,e^{2}-\frac {50 \left (-\frac {7 g x}{2}+f \right ) c \,d^{2} e}{9}+\frac {35 c \,d^{3} g}{3}\right ) \sqrt {\left (d g -e f \right ) e}\right )}{4 \sqrt {\left (d g -e f \right ) e}\, e^{4} \left (e x +d \right )^{2}}\) | \(176\) |
| risch | \(-\frac {2 c \left (-e g x +9 d g -4 e f \right ) \sqrt {g x +f}}{3 e^{4}}+\frac {\frac {2 \left (-\frac {5}{8} a \,e^{3} g^{2}-\frac {13}{8} c \,d^{2} e \,g^{2}+c d \,e^{2} f g \right ) \left (g x +f \right )^{\frac {3}{2}}-\frac {g \left (3 a d \,e^{2} g^{2}-3 a \,e^{3} f g +11 c \,d^{3} g^{2}-19 c \,d^{2} e f g +8 c d \,e^{2} f^{2}\right ) \sqrt {g x +f}}{4}}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (3 a \,e^{2} g^{2}+35 c \,d^{2} g^{2}-40 c d e f g +8 c \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \sqrt {\left (d g -e f \right ) e}}}{e^{4}}\) | \(218\) |
| derivativedivides | \(-\frac {2 c \left (-\frac {e \left (g x +f \right )^{\frac {3}{2}}}{3}+3 d g \sqrt {g x +f}-e f \sqrt {g x +f}\right )}{e^{4}}+\frac {\frac {2 \left (\left (-\frac {5}{8} a \,e^{3} g^{2}-\frac {13}{8} c \,d^{2} e \,g^{2}+c d \,e^{2} f g \right ) \left (g x +f \right )^{\frac {3}{2}}-\frac {g \left (3 a d \,e^{2} g^{2}-3 a \,e^{3} f g +11 c \,d^{3} g^{2}-19 c \,d^{2} e f g +8 c d \,e^{2} f^{2}\right ) \sqrt {g x +f}}{8}\right )}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (3 a \,e^{2} g^{2}+35 c \,d^{2} g^{2}-40 c d e f g +8 c \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \sqrt {\left (d g -e f \right ) e}}}{e^{4}}\) | \(230\) |
| default | \(-\frac {2 c \left (-\frac {e \left (g x +f \right )^{\frac {3}{2}}}{3}+3 d g \sqrt {g x +f}-e f \sqrt {g x +f}\right )}{e^{4}}+\frac {\frac {2 \left (\left (-\frac {5}{8} a \,e^{3} g^{2}-\frac {13}{8} c \,d^{2} e \,g^{2}+c d \,e^{2} f g \right ) \left (g x +f \right )^{\frac {3}{2}}-\frac {g \left (3 a d \,e^{2} g^{2}-3 a \,e^{3} f g +11 c \,d^{3} g^{2}-19 c \,d^{2} e f g +8 c d \,e^{2} f^{2}\right ) \sqrt {g x +f}}{8}\right )}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (3 a \,e^{2} g^{2}+35 c \,d^{2} g^{2}-40 c d e f g +8 c \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \sqrt {\left (d g -e f \right ) e}}}{e^{4}}\) | \(230\) |
Input:
int((g*x+f)^(3/2)*(c*x^2+a)/(e*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-3/4/((d*g-e*f)*e)^(1/2)*(-((a*g^2+8/3*c*f^2)*e^2-40/3*c*d*e*f*g+35/3*c*d^ 2*g^2)*(e*x+d)^2*arctan(e*(g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2))+(g*x+f)^(1/2) *((-32/9*x^2*(1/4*g*x+f)*c+2/3*(5/2*g*x+f)*a)*e^3+(-88/9*x*(-7/11*g*x+f)*c +a*g)*d*e^2-50/9*(-7/2*g*x+f)*c*d^2*e+35/3*c*d^3*g)*((d*g-e*f)*e)^(1/2))/e ^4/(e*x+d)^2
Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (177) = 354\).
Time = 0.13 (sec) , antiderivative size = 858, normalized size of antiderivative = 4.23 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^3} \, dx =\text {Too large to display} \] Input:
integrate((g*x+f)^(3/2)*(c*x^2+a)/(e*x+d)^3,x, algorithm="fricas")
Output:
[1/24*(3*(8*c*d^2*e^2*f^2 - 40*c*d^3*e*f*g + (35*c*d^4 + 3*a*d^2*e^2)*g^2 + (8*c*e^4*f^2 - 40*c*d*e^3*f*g + (35*c*d^2*e^2 + 3*a*e^4)*g^2)*x^2 + 2*(8 *c*d*e^3*f^2 - 40*c*d^2*e^2*f*g + (35*c*d^3*e + 3*a*d*e^3)*g^2)*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*(8*(c*e^5*f*g - c*d*e^4*g^2)*x^3 + 2*(25*c*d^2*e^3 - 3*a *e^5)*f^2 - (155*c*d^3*e^2 + 3*a*d*e^4)*f*g + 3*(35*c*d^4*e + 3*a*d^2*e^3) *g^2 + 8*(4*c*e^5*f^2 - 11*c*d*e^4*f*g + 7*c*d^2*e^3*g^2)*x^2 + (88*c*d*e^ 4*f^2 - (263*c*d^2*e^3 + 15*a*e^5)*f*g + 5*(35*c*d^3*e^2 + 3*a*d*e^4)*g^2) *x)*sqrt(g*x + f))/(d^2*e^6*f - d^3*e^5*g + (e^8*f - d*e^7*g)*x^2 + 2*(d*e ^7*f - d^2*e^6*g)*x), 1/12*(3*(8*c*d^2*e^2*f^2 - 40*c*d^3*e*f*g + (35*c*d^ 4 + 3*a*d^2*e^2)*g^2 + (8*c*e^4*f^2 - 40*c*d*e^3*f*g + (35*c*d^2*e^2 + 3*a *e^4)*g^2)*x^2 + 2*(8*c*d*e^3*f^2 - 40*c*d^2*e^2*f*g + (35*c*d^3*e + 3*a*d *e^3)*g^2)*x)*sqrt(-e^2*f + d*e*g)*arctan(sqrt(-e^2*f + d*e*g)*sqrt(g*x + f)/(e*g*x + e*f)) + (8*(c*e^5*f*g - c*d*e^4*g^2)*x^3 + 2*(25*c*d^2*e^3 - 3 *a*e^5)*f^2 - (155*c*d^3*e^2 + 3*a*d*e^4)*f*g + 3*(35*c*d^4*e + 3*a*d^2*e^ 3)*g^2 + 8*(4*c*e^5*f^2 - 11*c*d*e^4*f*g + 7*c*d^2*e^3*g^2)*x^2 + (88*c*d* e^4*f^2 - (263*c*d^2*e^3 + 15*a*e^5)*f*g + 5*(35*c*d^3*e^2 + 3*a*d*e^4)*g^ 2)*x)*sqrt(g*x + f))/(d^2*e^6*f - d^3*e^5*g + (e^8*f - d*e^7*g)*x^2 + 2*(d *e^7*f - d^2*e^6*g)*x)]
Timed out. \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^3} \, dx=\text {Timed out} \] Input:
integrate((g*x+f)**(3/2)*(c*x**2+a)/(e*x+d)**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((g*x+f)^(3/2)*(c*x^2+a)/(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
Time = 0.12 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.39 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^3} \, dx=\frac {{\left (8 \, c e^{2} f^{2} - 40 \, c d e f g + 35 \, c d^{2} g^{2} + 3 \, a e^{2} g^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {-e^{2} f + d e g}}\right )}{4 \, \sqrt {-e^{2} f + d e g} e^{4}} + \frac {8 \, {\left (g x + f\right )}^{\frac {3}{2}} c d e^{2} f g - 8 \, \sqrt {g x + f} c d e^{2} f^{2} g - 13 \, {\left (g x + f\right )}^{\frac {3}{2}} c d^{2} e g^{2} - 5 \, {\left (g x + f\right )}^{\frac {3}{2}} a e^{3} g^{2} + 19 \, \sqrt {g x + f} c d^{2} e f g^{2} + 3 \, \sqrt {g x + f} a e^{3} f g^{2} - 11 \, \sqrt {g x + f} c d^{3} g^{3} - 3 \, \sqrt {g x + f} a d e^{2} g^{3}}{4 \, {\left ({\left (g x + f\right )} e - e f + d g\right )}^{2} e^{4}} + \frac {2 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} c e^{6} + 3 \, \sqrt {g x + f} c e^{6} f - 9 \, \sqrt {g x + f} c d e^{5} g\right )}}{3 \, e^{9}} \] Input:
integrate((g*x+f)^(3/2)*(c*x^2+a)/(e*x+d)^3,x, algorithm="giac")
Output:
1/4*(8*c*e^2*f^2 - 40*c*d*e*f*g + 35*c*d^2*g^2 + 3*a*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) + 1/4*(8*(g*x + f)^(3/2)*c*d*e^2*f*g - 8*sqrt(g*x + f)*c*d*e^2*f^2*g - 13*(g*x + f)^(3/2 )*c*d^2*e*g^2 - 5*(g*x + f)^(3/2)*a*e^3*g^2 + 19*sqrt(g*x + f)*c*d^2*e*f*g ^2 + 3*sqrt(g*x + f)*a*e^3*f*g^2 - 11*sqrt(g*x + f)*c*d^3*g^3 - 3*sqrt(g*x + f)*a*d*e^2*g^3)/(((g*x + f)*e - e*f + d*g)^2*e^4) + 2/3*((g*x + f)^(3/2 )*c*e^6 + 3*sqrt(g*x + f)*c*e^6*f - 9*sqrt(g*x + f)*c*d*e^5*g)/e^9
Time = 5.92 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.39 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^3} \, dx=\sqrt {f+g\,x}\,\left (\frac {2\,c\,\left (3\,e^3\,f-3\,d\,e^2\,g\right )}{e^6}-\frac {4\,c\,f}{e^3}\right )-\frac {{\left (f+g\,x\right )}^{3/2}\,\left (\frac {13\,c\,d^2\,e\,g^2}{4}-2\,c\,f\,d\,e^2\,g+\frac {5\,a\,e^3\,g^2}{4}\right )+\sqrt {f+g\,x}\,\left (\frac {11\,c\,d^3\,g^3}{4}-\frac {19\,c\,d^2\,e\,f\,g^2}{4}+2\,c\,d\,e^2\,f^2\,g+\frac {3\,a\,d\,e^2\,g^3}{4}-\frac {3\,a\,e^3\,f\,g^2}{4}\right )}{e^6\,{\left (f+g\,x\right )}^2-\left (f+g\,x\right )\,\left (2\,e^6\,f-2\,d\,e^5\,g\right )+e^6\,f^2+d^2\,e^4\,g^2-2\,d\,e^5\,f\,g}+\frac {2\,c\,{\left (f+g\,x\right )}^{3/2}}{3\,e^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (35\,c\,d^2\,g^2-40\,c\,d\,e\,f\,g+8\,c\,e^2\,f^2+3\,a\,e^2\,g^2\right )}{4\,e^{9/2}\,\sqrt {d\,g-e\,f}} \] Input:
int(((f + g*x)^(3/2)*(a + c*x^2))/(d + e*x)^3,x)
Output:
(f + g*x)^(1/2)*((2*c*(3*e^3*f - 3*d*e^2*g))/e^6 - (4*c*f)/e^3) - ((f + g* x)^(3/2)*((5*a*e^3*g^2)/4 + (13*c*d^2*e*g^2)/4 - 2*c*d*e^2*f*g) + (f + g*x )^(1/2)*((11*c*d^3*g^3)/4 + (3*a*d*e^2*g^3)/4 - (3*a*e^3*f*g^2)/4 + 2*c*d* e^2*f^2*g - (19*c*d^2*e*f*g^2)/4))/(e^6*(f + g*x)^2 - (f + g*x)*(2*e^6*f - 2*d*e^5*g) + e^6*f^2 + d^2*e^4*g^2 - 2*d*e^5*f*g) + (2*c*(f + g*x)^(3/2)) /(3*e^3) + (atan((e^(1/2)*(f + g*x)^(1/2))/(d*g - e*f)^(1/2))*(3*a*e^2*g^2 + 35*c*d^2*g^2 + 8*c*e^2*f^2 - 40*c*d*e*f*g))/(4*e^(9/2)*(d*g - e*f)^(1/2 ))
Time = 0.24 (sec) , antiderivative size = 891, normalized size of antiderivative = 4.39 \[ \int \frac {(f+g x)^{3/2} \left (a+c x^2\right )}{(d+e x)^3} \, dx =\text {Too large to display} \] Input:
int((g*x+f)^(3/2)*(c*x^2+a)/(e*x+d)^3,x)
Output:
(9*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f) ))*a*d**2*e**2*g**2 + 18*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(s qrt(e)*sqrt(d*g - e*f)))*a*d*e**3*g**2*x + 9*sqrt(e)*sqrt(d*g - e*f)*atan( (sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*a*e**4*g**2*x**2 + 105*sqrt(e )*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*c*d**4 *g**2 - 120*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d *g - e*f)))*c*d**3*e*f*g + 210*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x) *e)/(sqrt(e)*sqrt(d*g - e*f)))*c*d**3*e*g**2*x + 24*sqrt(e)*sqrt(d*g - e*f )*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*c*d**2*e**2*f**2 - 240 *sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f))) *c*d**2*e**2*f*g*x + 105*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(s qrt(e)*sqrt(d*g - e*f)))*c*d**2*e**2*g**2*x**2 + 48*sqrt(e)*sqrt(d*g - e*f )*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*c*d*e**3*f**2*x - 120* sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))* c*d*e**3*f*g*x**2 + 24*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqr t(e)*sqrt(d*g - e*f)))*c*e**4*f**2*x**2 - 9*sqrt(f + g*x)*a*d**2*e**3*g**2 + 3*sqrt(f + g*x)*a*d*e**4*f*g - 15*sqrt(f + g*x)*a*d*e**4*g**2*x + 6*sqr t(f + g*x)*a*e**5*f**2 + 15*sqrt(f + g*x)*a*e**5*f*g*x - 105*sqrt(f + g*x) *c*d**4*e*g**2 + 155*sqrt(f + g*x)*c*d**3*e**2*f*g - 175*sqrt(f + g*x)*c*d **3*e**2*g**2*x - 50*sqrt(f + g*x)*c*d**2*e**3*f**2 + 263*sqrt(f + g*x)...