3.8.56 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{5/2} (f+g x)^{7/2}} \, dx\) [756]

Optimal. Leaf size=274 \[ -\frac {2 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt {d+e x} \sqrt {f+g x}}-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} (f+g x)^{3/2}}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2} (f+g x)^{5/2}}+\frac {2 c^{5/2} d^{5/2} \sqrt {a e+c d x} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{g^{7/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {876, 905, 65, 223, 212} \begin {gather*} \frac {2 c^{5/2} d^{5/2} \sqrt {d+e x} \sqrt {a e+c d x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{g^{7/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g^3 \sqrt {d+e x} \sqrt {f+g x}}-\frac {2 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} (f+g x)^{3/2}}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2} (f+g x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/((d + e*x)^(5/2)*(f + g*x)^(7/2)),x]

[Out]

(-2*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(g^3*Sqrt[d + e*x]*Sqrt[f + g*x]) - (2*c*d*(a*d*e + (
c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(3*g^2*(d + e*x)^(3/2)*(f + g*x)^(3/2)) - (2*(a*d*e + (c*d^2 + a*e^2)*x +
c*d*e*x^2)^(5/2))/(5*g*(d + e*x)^(5/2)*(f + g*x)^(5/2)) + (2*c^(5/2)*d^(5/2)*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])])/(g^(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 876

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

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 {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^{7/2}} \, dx &=-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2} (f+g x)^{5/2}}+\frac {(c d) \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2} (f+g x)^{5/2}} \, dx}{g}\\ &=-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} (f+g x)^{3/2}}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2} (f+g x)^{5/2}}+\frac {\left (c^2 d^2\right ) \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x} (f+g x)^{3/2}} \, dx}{g^2}\\ &=-\frac {2 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt {d+e x} \sqrt {f+g x}}-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} (f+g x)^{3/2}}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2} (f+g x)^{5/2}}+\frac {\left (c^3 d^3\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}{g^3}\\ &=-\frac {2 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt {d+e x} \sqrt {f+g x}}-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} (f+g x)^{3/2}}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2} (f+g x)^{5/2}}+\frac {\left (c^3 d^3 \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}{g^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac {2 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt {d+e x} \sqrt {f+g x}}-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} (f+g x)^{3/2}}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2} (f+g x)^{5/2}}+\frac {\left (2 c^2 d^2 \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 )}{g^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac {2 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt {d+e x} \sqrt {f+g x}}-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} (f+g x)^{3/2}}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2} (f+g x)^{5/2}}+\frac {\left (2 c^2 d^2 \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 )}{g^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=-\frac {2 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{g^3 \sqrt {d+e x} \sqrt {f+g x}}-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 g^2 (d+e x)^{3/2} (f+g x)^{3/2}}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2} (f+g x)^{5/2}}+\frac {2 c^{5/2} d^{5/2} \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 )}{g^{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 = 0.27, size = 166, normalized size = 0.61 \begin {gather*} \frac {2 ((a e+c d x) (d+e x))^{5/2} \left (-\frac {\sqrt {g} \left (3 g^2+\frac {5 c d g (f+g x)}{a e+c d x}+\frac {15 c^2 d^2 (f+g x)^2}{(a e+c d x)^2}\right )}{(f+g x)^{5/2}}+\frac {15 c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {f+g x}}{\sqrt {g} \sqrt {a e+c d x}}\right )}{(a e+c d x)^{5/2}}\right )}{15 g^{7/2} (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/((d + e*x)^(5/2)*(f + g*x)^(7/2)),x]

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(5/2)*(-((Sqrt[g]*(3*g^2 + (5*c*d*g*(f + g*x))/(a*e + c*d*x) + (15*c^2*d^2*(f + g
*x)^2)/(a*e + c*d*x)^2))/(f + g*x)^(5/2)) + (15*c^(5/2)*d^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[f + g*x])/(Sqrt[
g]*Sqrt[a*e + c*d*x])])/(a*e + c*d*x)^(5/2)))/(15*g^(7/2)*(d + e*x)^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(500\) vs. \(2(230)=460\).
time = 0.14, size = 501, normalized size = 1.83

method	result	size
default	\(\frac {\sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (15 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) c^{3} d^{3} g^{3} x^{3}+45 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) c^{3} d^{3} f \,g^{2} x^{2}+45 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) c^{3} d^{3} f^{2} g x +15 \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}}{2 \sqrt {d g c}}\right ) c^{3} d^{3} f^{3}-46 c^{2} d^{2} g^{2} x^{2} \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}-22 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, a c d e \,g^{2} x -70 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, c^{2} d^{2} f g x -6 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, a^{2} e^{2} g^{2}-10 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, a c d e f g -30 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, c^{2} d^{2} f^{2}\right )}{15 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {d g c}\, g^{3} \left (g x +f \right )^{\frac {5}{2}} \sqrt {e x +d}}\)	\(501\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/15*((c*d*x+a*e)*(e*x+d))^(1/2)*(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*g^3*x^3+45*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+45*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^2*g*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))*c^3*d^3*f^3-46*c^2*d^2*g^2*x^2*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2)-22*((g*x+f)*(c*d*x+a*
e))^(1/2)*(d*g*c)^(1/2)*a*c*d*e*g^2*x-70*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2)*c^2*d^2*f*g*x-6*((g*x+f)*(c
*d*x+a*e))^(1/2)*(d*g*c)^(1/2)*a^2*e^2*g^2-10*((g*x+f)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2)*a*c*d*e*f*g-30*((g*x+f
)*(c*d*x+a*e))^(1/2)*(d*g*c)^(1/2)*c^2*d^2*f^2)/((g*x+f)*(c*d*x+a*e))^(1/2)/(d*g*c)^(1/2)/g^3/(g*x+f)^(5/2)/(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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(7/2),x, algorithm="maxima")

[Out]

integrate((c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/((g*x + f)^(7/2)*(x*e + d)^(5/2)), x)

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Fricas [A]
time = 1.40, size = 949, normalized size = 3.46 \begin {gather*} \left [-\frac {4 \, {\left (23 \, c^{2} d^{2} g^{2} x^{2} + 35 \, c^{2} d^{2} f g x + 15 \, c^{2} d^{2} f^{2} + 3 \, a^{2} g^{2} e^{2} + {\left (11 \, a c d g^{2} x + 5 \, 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^{2} d^{3} g^{3} x^{3} + 3 \, c^{2} d^{3} f g^{2} x^{2} + 3 \, c^{2} d^{3} f^{2} g x + c^{2} d^{3} f^{3} + {\left (c^{2} d^{2} g^{3} x^{4} + 3 \, c^{2} d^{2} f g^{2} x^{3} + 3 \, c^{2} d^{2} f^{2} g x^{2} + c^{2} d^{2} f^{3} x\right )} e\right )} \sqrt {\frac {c d}{g}} \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 d g^{2} x + c d f g + a g^{2} 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 {c d}{g}} + {\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 )}{30 \, {\left (d g^{6} x^{3} + 3 \, d f g^{5} x^{2} + 3 \, d f^{2} g^{4} x + d f^{3} g^{3} + {\left (g^{6} x^{4} + 3 \, f g^{5} x^{3} + 3 \, f^{2} g^{4} x^{2} + f^{3} g^{3} x\right )} e\right )}}, -\frac {2 \, {\left (23 \, c^{2} d^{2} g^{2} x^{2} + 35 \, c^{2} d^{2} f g x + 15 \, c^{2} d^{2} f^{2} + 3 \, a^{2} g^{2} e^{2} + {\left (11 \, a c d g^{2} x + 5 \, 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^{2} d^{3} g^{3} x^{3} + 3 \, c^{2} d^{3} f g^{2} x^{2} + 3 \, c^{2} d^{3} f^{2} g x + c^{2} d^{3} f^{3} + {\left (c^{2} d^{2} g^{3} x^{4} + 3 \, c^{2} d^{2} f g^{2} x^{3} + 3 \, c^{2} d^{2} f^{2} g x^{2} + c^{2} d^{2} f^{3} x\right )} e\right )} \sqrt {-\frac {c d}{g}} \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} \sqrt {-\frac {c d}{g}} g}{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 )}{15 \, {\left (d g^{6} x^{3} + 3 \, d f g^{5} x^{2} + 3 \, d f^{2} g^{4} x + d f^{3} g^{3} + {\left (g^{6} x^{4} + 3 \, f g^{5} x^{3} + 3 \, f^{2} g^{4} x^{2} + f^{3} g^{3} x\right )} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(7/2),x, algorithm="fricas")

[Out]

[-1/30*(4*(23*c^2*d^2*g^2*x^2 + 35*c^2*d^2*f*g*x + 15*c^2*d^2*f^2 + 3*a^2*g^2*e^2 + (11*a*c*d*g^2*x + 5*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^2*d^3*g^3*x^3 + 3*c^2*
d^3*f*g^2*x^2 + 3*c^2*d^3*f^2*g*x + c^2*d^3*f^3 + (c^2*d^2*g^3*x^4 + 3*c^2*d^2*f*g^2*x^3 + 3*c^2*d^2*f^2*g*x^2
+ c^2*d^2*f^3*x)*e)*sqrt(c*d/g)*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*d*g^2*x + c*d*f*g + a*g^2*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(
c*d/g) + (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)))/(d*g^6*x^3 + 3*d*f*g^5*x^2 + 3*d*f^2*g^4*x + d*f^3*g^3
+ (g^6*x^4 + 3*f*g^5*x^3 + 3*f^2*g^4*x^2 + f^3*g^3*x)*e), -1/15*(2*(23*c^2*d^2*g^2*x^2 + 35*c^2*d^2*f*g*x + 1
5*c^2*d^2*f^2 + 3*a^2*g^2*e^2 + (11*a*c*d*g^2*x + 5*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^2*d^3*g^3*x^3 + 3*c^2*d^3*f*g^2*x^2 + 3*c^2*d^3*f^2*g*x + c^2*d^3*f^3 + (c
^2*d^2*g^3*x^4 + 3*c^2*d^2*f*g^2*x^3 + 3*c^2*d^2*f^2*g*x^2 + c^2*d^2*f^3*x)*e)*sqrt(-c*d/g)*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)*sqrt(-c*d/g)*g/(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)))/(d*g^6*x^3 + 3*d*f*g^5*x^2 + 3*d*f^2*g^4*x + d*f^3*g^3 + (g^6*x^4 +
3*f*g^5*x^3 + 3*f^2*g^4*x^2 + f^3*g^3*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((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**(5/2)/(g*x+f)**(7/2),x)

[Out]

Timed out

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Giac [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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(5/2)/(g*x+f)^(7/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}}{{\left (f+g\,x\right )}^{7/2}\,{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/((f + g*x)^(7/2)*(d + e*x)^(5/2)),x)

[Out]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/((f + g*x)^(7/2)*(d + e*x)^(5/2)), x)

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