3.8.27 \(\int \frac {(d+e x)^{5/2} (f+g x)^{5/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\) [727]

Optimal. Leaf size=289 \[ -\frac {2 (d+e x)^{3/2} (f+g x)^{5/2}}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 g \sqrt {d+e x} (f+g x)^{3/2}}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 g^2 \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3 \sqrt {d+e x}}+\frac {5 g^{3/2} (c d f-a e g) \sqrt {a e+c d x} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{c^{7/2} d^{7/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \]

[Out]

-2/3*(e*x+d)^(3/2)*(g*x+f)^(5/2)/c/d/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-10/3*g*(g*x+f)^(3/2)*(e*x+d)^(1/2
)/c^2/d^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+5*g^(3/2)*(-a*e*g+c*d*f)*arctanh(g^(1/2)*(c*d*x+a*e)^(1/2)/c
^(1/2)/d^(1/2)/(g*x+f)^(1/2))*(c*d*x+a*e)^(1/2)*(e*x+d)^(1/2)/c^(7/2)/d^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2
)^(1/2)+5*g^2*(g*x+f)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^3/d^3/(e*x+d)^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {880, 884, 905, 65, 223, 212} \begin {gather*} \frac {5 g^{3/2} \sqrt {d+e x} \sqrt {a e+c d x} (c d f-a e g) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{c^{7/2} d^{7/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {5 g^2 \sqrt {f+g x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c^3 d^3 \sqrt {d+e x}}-\frac {10 g \sqrt {d+e x} (f+g x)^{3/2}}{3 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)^{3/2} (f+g x)^{5/2}}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^(5/2)*(f + g*x)^(5/2))/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]

[Out]

(-2*(d + e*x)^(3/2)*(f + g*x)^(5/2))/(3*c*d*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (10*g*Sqrt[d + e*
x]*(f + g*x)^(3/2))/(3*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (5*g^2*Sqrt[f + g*x]*Sqrt[a*d*e
+ (c*d^2 + a*e^2)*x + c*d*e*x^2])/(c^3*d^3*Sqrt[d + e*x]) + (5*g^(3/2)*(c*d*f - a*e*g)*Sqrt[a*e + c*d*x]*Sqrt[
d + e*x]*ArcTanh[(Sqrt[g]*Sqrt[a*e + c*d*x])/(Sqrt[c]*Sqrt[d]*Sqrt[f + g*x])])/(c^(7/2)*d^(7/2)*Sqrt[a*d*e + (
c*d^2 + a*e^2)*x + c*d*e*x^2])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 880

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Simp[e*(d + e*x)^(m - 1)*(f + g*x)^n*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] - Dist[e*g*(n/(c*(p + 1))), I
nt[(d + e*x)^(m - 1)*(f + g*x)^(n - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &&
NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0] &
& LtQ[p, -1] && GtQ[n, 0]

Rule 884

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Simp[(-e)*(d + e*x)^(m - 1)*(f + g*x)^n*((a + b*x + c*x^2)^(p + 1)/(c*(m - n - 1))), x] - Dist[n*((c*e*f + c*d
*g - b*e*g)/(c*e*(m - n - 1))), Int[(d + e*x)^m*(f + g*x)^(n - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b,
c, d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !Int
egerQ[p] && EqQ[m + p, 0] && GtQ[n, 0] && NeQ[m - n - 1, 0] && (IntegerQ[2*p] || IntegerQ[n])

Rule 905

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Dist[(a + b*x + c*x^2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*
(f + g*x)^n*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2
- 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(d+e x)^{5/2} (f+g x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)^{3/2} (f+g x)^{5/2}}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {(5 g) \int \frac {(d+e x)^{3/2} (f+g x)^{3/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 c d}\\ &=-\frac {2 (d+e x)^{3/2} (f+g x)^{5/2}}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 g \sqrt {d+e x} (f+g x)^{3/2}}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (5 g^2\right ) \int \frac {\sqrt {d+e x} \sqrt {f+g x}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c^2 d^2}\\ &=-\frac {2 (d+e x)^{3/2} (f+g x)^{5/2}}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 g \sqrt {d+e x} (f+g x)^{3/2}}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 g^2 \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3 \sqrt {d+e x}}+\frac {\left (5 g^2 (c d f-a e g)\right ) \int \frac {\sqrt {d+e x}}{\sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 c^3 d^3}\\ &=-\frac {2 (d+e x)^{3/2} (f+g x)^{5/2}}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 g \sqrt {d+e x} (f+g x)^{3/2}}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 g^2 \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3 \sqrt {d+e x}}+\frac {\left (5 g^2 (c d f-a e g) \sqrt {a e+c d x} \sqrt {d+e x}\right ) \int \frac {1}{\sqrt {a e+c d x} \sqrt {f+g x}} \, dx}{2 c^3 d^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac {2 (d+e x)^{3/2} (f+g x)^{5/2}}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 g \sqrt {d+e x} (f+g x)^{3/2}}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 g^2 \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3 \sqrt {d+e x}}+\frac {\left (5 g^2 (c d f-a e g) \sqrt {a e+c d x} \sqrt {d+e x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {f-\frac {a e g}{c d}+\frac {g x^2}{c d}}} \, dx,x,\sqrt {a e+c d x}\right )}{c^4 d^4 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac {2 (d+e x)^{3/2} (f+g x)^{5/2}}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 g \sqrt {d+e x} (f+g x)^{3/2}}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 g^2 \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3 \sqrt {d+e x}}+\frac {\left (5 g^2 (c d f-a e g) \sqrt {a e+c d x} \sqrt {d+e x}\right ) \text {Subst}\left (\int \frac {1}{1-\frac {g x^2}{c d}} \, dx,x,\frac {\sqrt {a e+c d x}}{\sqrt {f+g x}}\right )}{c^4 d^4 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac {2 (d+e x)^{3/2} (f+g x)^{5/2}}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 g \sqrt {d+e x} (f+g x)^{3/2}}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 g^2 \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3 \sqrt {d+e x}}+\frac {5 g^{3/2} (c d f-a e g) \sqrt {a e+c d x} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{c^{7/2} d^{7/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end {align*}

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Mathematica [A]
time = 1.08, size = 179, normalized size = 0.62 \begin {gather*} \frac {(d+e x)^{3/2} \left (\sqrt {\frac {c d}{g}} \sqrt {f+g x} \left (15 a^2 e^2 g^2-10 a c d e g (f-2 g x)+c^2 d^2 \left (-2 f^2-14 f g x+3 g^2 x^2\right )\right )-15 g (c d f-a e g) (a e+c d x)^{3/2} \log \left (\sqrt {a e+c d x}-\sqrt {\frac {c d}{g}} \sqrt {f+g x}\right )\right )}{3 c^3 d^3 \sqrt {\frac {c d}{g}} ((a e+c d x) (d+e x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^(5/2)*(f + g*x)^(5/2))/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]

[Out]

((d + e*x)^(3/2)*(Sqrt[(c*d)/g]*Sqrt[f + g*x]*(15*a^2*e^2*g^2 - 10*a*c*d*e*g*(f - 2*g*x) + c^2*d^2*(-2*f^2 - 1
4*f*g*x + 3*g^2*x^2)) - 15*g*(c*d*f - a*e*g)*(a*e + c*d*x)^(3/2)*Log[Sqrt[a*e + c*d*x] - Sqrt[(c*d)/g]*Sqrt[f
+ g*x]]))/(3*c^3*d^3*Sqrt[(c*d)/g]*((a*e + c*d*x)*(d + e*x))^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(641\) vs. \(2(245)=490\).
time = 0.15, size = 642, normalized size = 2.22

method	result	size
default	\(-\frac {\left (15 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) a \,c^{2} d^{2} e \,g^{3} x^{2}-15 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) c^{3} d^{3} f \,g^{2} x^{2}+30 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) a^{2} c d \,e^{2} g^{3} x -30 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) a \,c^{2} d^{2} e f \,g^{2} x +15 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) a^{3} e^{3} g^{3}-15 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) a^{2} c d \,e^{2} f \,g^{2}-6 c^{2} d^{2} g^{2} x^{2} \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}-40 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, a c d e \,g^{2} x +28 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, c^{2} d^{2} f g x -30 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, a^{2} e^{2} g^{2}+20 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, a c d e f g +4 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, c^{2} d^{2} f^{2}\right ) \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \sqrt {g x +f}}{6 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \left (c d x +a e \right )^{2} \sqrt {d g c}\, c^{3} d^{3} \sqrt {e x +d}}\)	\(642\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)*(g*x+f)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/6*(15*ln(1/2*(2*c*d*g*x+a*e*g+c*d*f+2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2))/(d*g*c)^(1/2))*a*c^2*d^2*e
*g^3*x^2-15*ln(1/2*(2*c*d*g*x+a*e*g+c*d*f+2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2))/(d*g*c)^(1/2))*c^3*d^3*
f*g^2*x^2+30*ln(1/2*(2*c*d*g*x+a*e*g+c*d*f+2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2))/(d*g*c)^(1/2))*a^2*c*d
*e^2*g^3*x-30*ln(1/2*(2*c*d*g*x+a*e*g+c*d*f+2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2))/(d*g*c)^(1/2))*a*c^2*
d^2*e*f*g^2*x+15*ln(1/2*(2*c*d*g*x+a*e*g+c*d*f+2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2))/(d*g*c)^(1/2))*a^3
*e^3*g^3-15*ln(1/2*(2*c*d*g*x+a*e*g+c*d*f+2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2))/(d*g*c)^(1/2))*a^2*c*d*
e^2*f*g^2-6*c^2*d^2*g^2*x^2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2)-40*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(
1/2)*a*c*d*e*g^2*x+28*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2)*c^2*d^2*f*g*x-30*((g*x+f)*(c*d*x+a*e))^(1/2)*(
d*g*c)^(1/2)*a^2*e^2*g^2+20*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2)*a*c*d*e*f*g+4*((g*x+f)*(c*d*x+a*e))^(1/2
)*(d*g*c)^(1/2)*c^2*d^2*f^2)*((c*d*x+a*e)*(e*x+d))^(1/2)*(g*x+f)^(1/2)/((g*x+f)*(c*d*x+a*e))^(1/2)/(c*d*x+a*e)
^2/(d*g*c)^(1/2)/c^3/d^3/(e*x+d)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)*(g*x+f)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((g*x + f)^(5/2)*(x*e + d)^(5/2)/(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)^(5/2), x)

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Fricas [A]
time = 3.95, size = 1061, normalized size = 3.67 \begin {gather*} \left [\frac {4 \, {\left (3 \, c^{2} d^{2} g^{2} x^{2} - 14 \, c^{2} d^{2} f g x - 2 \, c^{2} d^{2} f^{2} + 15 \, a^{2} g^{2} e^{2} + 10 \, {\left (2 \, a c d g^{2} x - a c d f g\right )} e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {g x + f} \sqrt {x e + d} - 15 \, {\left (c^{3} d^{4} f g x^{2} - a^{3} g^{2} x e^{4} - {\left (2 \, a^{2} c d g^{2} x^{2} - a^{2} c d f g x + a^{3} d g^{2}\right )} e^{3} - {\left (a c^{2} d^{2} g^{2} x^{3} - 2 \, a c^{2} d^{2} f g x^{2} + 2 \, a^{2} c d^{2} g^{2} x - a^{2} c d^{2} f g\right )} e^{2} + {\left (c^{3} d^{3} f g x^{3} - a c^{2} d^{3} g^{2} x^{2} + 2 \, a c^{2} d^{3} f g x\right )} e\right )} \sqrt {\frac {g}{c d}} \log \left (-\frac {8 \, c^{2} d^{3} g^{2} x^{2} + 8 \, c^{2} d^{3} f g x + c^{2} d^{3} f^{2} + a^{2} g^{2} x e^{3} - 4 \, {\left (2 \, c^{2} d^{2} g x + c^{2} d^{2} f + a c d g e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {g x + f} \sqrt {x e + d} \sqrt {\frac {g}{c d}} + {\left (8 \, a c d g^{2} x^{2} + 6 \, a c d f g x + a^{2} d g^{2}\right )} e^{2} + {\left (8 \, c^{2} d^{2} g^{2} x^{3} + 8 \, c^{2} d^{2} f g x^{2} + 6 \, a c d^{2} f g + {\left (c^{2} d^{2} f^{2} + 8 \, a c d^{2} g^{2}\right )} x\right )} e}{x e + d}\right )}{12 \, {\left (c^{5} d^{6} x^{2} + a^{2} c^{3} d^{3} x e^{3} + {\left (2 \, a c^{4} d^{4} x^{2} + a^{2} c^{3} d^{4}\right )} e^{2} + {\left (c^{5} d^{5} x^{3} + 2 \, a c^{4} d^{5} x\right )} e\right )}}, \frac {2 \, {\left (3 \, c^{2} d^{2} g^{2} x^{2} - 14 \, c^{2} d^{2} f g x - 2 \, c^{2} d^{2} f^{2} + 15 \, a^{2} g^{2} e^{2} + 10 \, {\left (2 \, a c d g^{2} x - a c d f g\right )} e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {g x + f} \sqrt {x e + d} - 15 \, {\left (c^{3} d^{4} f g x^{2} - a^{3} g^{2} x e^{4} - {\left (2 \, a^{2} c d g^{2} x^{2} - a^{2} c d f g x + a^{3} d g^{2}\right )} e^{3} - {\left (a c^{2} d^{2} g^{2} x^{3} - 2 \, a c^{2} d^{2} f g x^{2} + 2 \, a^{2} c d^{2} g^{2} x - a^{2} c d^{2} f g\right )} e^{2} + {\left (c^{3} d^{3} f g x^{3} - a c^{2} d^{3} g^{2} x^{2} + 2 \, a c^{2} d^{3} f g x\right )} e\right )} \sqrt {-\frac {g}{c d}} \arctan \left (\frac {2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {g x + f} \sqrt {x e + d} c d \sqrt {-\frac {g}{c d}}}{2 \, c d^{2} g x + c d^{2} f + a g x e^{2} + {\left (2 \, c d g x^{2} + c d f x + a d g\right )} e}\right )}{6 \, {\left (c^{5} d^{6} x^{2} + a^{2} c^{3} d^{3} x e^{3} + {\left (2 \, a c^{4} d^{4} x^{2} + a^{2} c^{3} d^{4}\right )} e^{2} + {\left (c^{5} d^{5} x^{3} + 2 \, a c^{4} d^{5} x\right )} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)*(g*x+f)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(4*(3*c^2*d^2*g^2*x^2 - 14*c^2*d^2*f*g*x - 2*c^2*d^2*f^2 + 15*a^2*g^2*e^2 + 10*(2*a*c*d*g^2*x - a*c*d*f*
g)*e)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(g*x + f)*sqrt(x*e + d) - 15*(c^3*d^4*f*g*x^2 - a^3*g^2*
x*e^4 - (2*a^2*c*d*g^2*x^2 - a^2*c*d*f*g*x + a^3*d*g^2)*e^3 - (a*c^2*d^2*g^2*x^3 - 2*a*c^2*d^2*f*g*x^2 + 2*a^2
*c*d^2*g^2*x - a^2*c*d^2*f*g)*e^2 + (c^3*d^3*f*g*x^3 - a*c^2*d^3*g^2*x^2 + 2*a*c^2*d^3*f*g*x)*e)*sqrt(g/(c*d))
*log(-(8*c^2*d^3*g^2*x^2 + 8*c^2*d^3*f*g*x + c^2*d^3*f^2 + a^2*g^2*x*e^3 - 4*(2*c^2*d^2*g*x + c^2*d^2*f + a*c*
d*g*e)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(g*x + f)*sqrt(x*e + d)*sqrt(g/(c*d)) + (8*a*c*d*g^2*x^
2 + 6*a*c*d*f*g*x + a^2*d*g^2)*e^2 + (8*c^2*d^2*g^2*x^3 + 8*c^2*d^2*f*g*x^2 + 6*a*c*d^2*f*g + (c^2*d^2*f^2 + 8
*a*c*d^2*g^2)*x)*e)/(x*e + d)))/(c^5*d^6*x^2 + a^2*c^3*d^3*x*e^3 + (2*a*c^4*d^4*x^2 + a^2*c^3*d^4)*e^2 + (c^5*
d^5*x^3 + 2*a*c^4*d^5*x)*e), 1/6*(2*(3*c^2*d^2*g^2*x^2 - 14*c^2*d^2*f*g*x - 2*c^2*d^2*f^2 + 15*a^2*g^2*e^2 + 1
0*(2*a*c*d*g^2*x - a*c*d*f*g)*e)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(g*x + f)*sqrt(x*e + d) - 15*
(c^3*d^4*f*g*x^2 - a^3*g^2*x*e^4 - (2*a^2*c*d*g^2*x^2 - a^2*c*d*f*g*x + a^3*d*g^2)*e^3 - (a*c^2*d^2*g^2*x^3 -
2*a*c^2*d^2*f*g*x^2 + 2*a^2*c*d^2*g^2*x - a^2*c*d^2*f*g)*e^2 + (c^3*d^3*f*g*x^3 - a*c^2*d^3*g^2*x^2 + 2*a*c^2*
d^3*f*g*x)*e)*sqrt(-g/(c*d))*arctan(2*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(g*x + f)*sqrt(x*e + d)*
c*d*sqrt(-g/(c*d))/(2*c*d^2*g*x + c*d^2*f + a*g*x*e^2 + (2*c*d*g*x^2 + c*d*f*x + a*d*g)*e)))/(c^5*d^6*x^2 + a^
2*c^3*d^3*x*e^3 + (2*a*c^4*d^4*x^2 + a^2*c^3*d^4)*e^2 + (c^5*d^5*x^3 + 2*a*c^4*d^5*x)*e)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(5/2)*(g*x+f)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)*(g*x+f)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="giac")

[Out]

integrate((g*x + f)^(5/2)*(x*e + d)^(5/2)/(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (f+g\,x\right )}^{5/2}\,{\left (d+e\,x\right )}^{5/2}}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)^(5/2)*(d + e*x)^(5/2))/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2),x)

[Out]

int(((f + g*x)^(5/2)*(d + e*x)^(5/2))/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2), x)

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