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

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

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Rubi [A]  time = 0.37, 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 = {862, 891, 63, 217, 206} \begin {gather*} -\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^{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 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 63

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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 862

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 891

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*x)/e)^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 ) \operatorname {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 ) \operatorname {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 = 1.37, size = 224, normalized size = 0.82 \begin {gather*} \frac {2 \sqrt {(d+e x) (a e+c d x)} \left (\frac {15 \sqrt {c} \sqrt {d} (c d f-a e g)^{5/2} \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{5/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {g} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d f-a e g}}\right )}{\sqrt {c d} \sqrt {a e+c d x}}-\sqrt {g} \left (3 a^2 e^2 g^2+a c d e g (5 f+11 g x)+c^2 d^2 \left (15 f^2+35 f g x+23 g^2 x^2\right )\right )\right )}{15 g^{7/2} \sqrt {d+e x} (f+g 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*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-(Sqrt[g]*(3*a^2*e^2*g^2 + a*c*d*e*g*(5*f + 11*g*x) + c^2*d^2*(15*f^2 + 35*f
*g*x + 23*g^2*x^2))) + (15*Sqrt[c]*Sqrt[d]*(c*d*f - a*e*g)^(5/2)*((c*d*(f + g*x))/(c*d*f - a*e*g))^(5/2)*ArcSi
nh[(Sqrt[c]*Sqrt[d]*Sqrt[g]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sqrt[c*d*f - a*e*g])])/(Sqrt[c*d]*Sqrt[a*e + c*d*x])
))/(15*g^(7/2)*Sqrt[d + e*x]*(f + g*x)^(5/2))

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IntegrateAlgebraic [A]  time = 2.43, size = 230, normalized size = 0.84 \begin {gather*} \frac {g^{5/2} \left (\frac {(d g+e g x) (a e g+c d g x)}{g^2}\right )^{5/2} \left (-\frac {2 \sqrt {a e g+c d (f+g x)-c d f} \left (3 a^2 e^2 g^2+11 a c d e g (f+g x)-6 a c d e f g+3 c^2 d^2 f^2+23 c^2 d^2 (f+g x)^2-11 c^2 d^2 f (f+g x)\right )}{15 g^{7/2} (f+g x)^{5/2}}-\frac {2 c^2 d^2 \sqrt {c d} \log \left (\sqrt {a e g+c d (f+g x)-c d f}-\sqrt {c d} \sqrt {f+g x}\right )}{g^{7/2}}\right )}{(d+e x)^{5/2} (a e g+c d g x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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]

(g^(5/2)*(((a*e*g + c*d*g*x)*(d*g + e*g*x))/g^2)^(5/2)*((-2*Sqrt[-(c*d*f) + a*e*g + c*d*(f + g*x)]*(3*c^2*d^2*
f^2 - 6*a*c*d*e*f*g + 3*a^2*e^2*g^2 - 11*c^2*d^2*f*(f + g*x) + 11*a*c*d*e*g*(f + g*x) + 23*c^2*d^2*(f + g*x)^2
))/(15*g^(7/2)*(f + g*x)^(5/2)) - (2*c^2*d^2*Sqrt[c*d]*Log[-(Sqrt[c*d]*Sqrt[f + g*x]) + Sqrt[-(c*d*f) + a*e*g
+ c*d*(f + g*x)]])/g^(7/2)))/((d + e*x)^(5/2)*(a*e*g + c*d*g*x)^(5/2))

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fricas [A]  time = 1.10, size = 933, normalized size = 3.41 \begin {gather*} \left [-\frac {4 \, {\left (23 \, c^{2} d^{2} g^{2} x^{2} + 15 \, c^{2} d^{2} f^{2} + 5 \, a c d e f g + 3 \, a^{2} e^{2} g^{2} + {\left (35 \, c^{2} d^{2} f g + 11 \, a c d e g^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} \sqrt {g x + f} - 15 \, {\left (c^{2} d^{2} e g^{3} x^{4} + c^{2} d^{3} f^{3} + {\left (3 \, c^{2} d^{2} e f g^{2} + c^{2} d^{3} g^{3}\right )} x^{3} + 3 \, {\left (c^{2} d^{2} e f^{2} g + c^{2} d^{3} f g^{2}\right )} x^{2} + {\left (c^{2} d^{2} e f^{3} + 3 \, c^{2} d^{3} f^{2} g\right )} x\right )} \sqrt {\frac {c d}{g}} \log \left (-\frac {8 \, c^{2} d^{2} e g^{2} x^{3} + c^{2} d^{3} f^{2} + 6 \, a c d^{2} e f g + a^{2} d e^{2} g^{2} + 4 \, {\left (2 \, c d g^{2} x + c d f g + a e g^{2}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} \sqrt {g x + f} \sqrt {\frac {c d}{g}} + 8 \, {\left (c^{2} d^{2} e f g + {\left (c^{2} d^{3} + a c d e^{2}\right )} g^{2}\right )} x^{2} + {\left (c^{2} d^{2} e f^{2} + 2 \, {\left (4 \, c^{2} d^{3} + 3 \, a c d e^{2}\right )} f g + {\left (8 \, a c d^{2} e + a^{2} e^{3}\right )} g^{2}\right )} x}{e x + d}\right )}{30 \, {\left (e g^{6} x^{4} + d f^{3} g^{3} + {\left (3 \, e f g^{5} + d g^{6}\right )} x^{3} + 3 \, {\left (e f^{2} g^{4} + d f g^{5}\right )} x^{2} + {\left (e f^{3} g^{3} + 3 \, d f^{2} g^{4}\right )} x\right )}}, -\frac {2 \, {\left (23 \, c^{2} d^{2} g^{2} x^{2} + 15 \, c^{2} d^{2} f^{2} + 5 \, a c d e f g + 3 \, a^{2} e^{2} g^{2} + {\left (35 \, c^{2} d^{2} f g + 11 \, a c d e g^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} \sqrt {g x + f} + 15 \, {\left (c^{2} d^{2} e g^{3} x^{4} + c^{2} d^{3} f^{3} + {\left (3 \, c^{2} d^{2} e f g^{2} + c^{2} d^{3} g^{3}\right )} x^{3} + 3 \, {\left (c^{2} d^{2} e f^{2} g + c^{2} d^{3} f g^{2}\right )} x^{2} + {\left (c^{2} d^{2} e f^{3} + 3 \, c^{2} d^{3} f^{2} g\right )} x\right )} \sqrt {-\frac {c d}{g}} \arctan \left (\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} \sqrt {g x + f} \sqrt {-\frac {c d}{g}} g}{2 \, c d e g x^{2} + c d^{2} f + a d e g + {\left (c d e f + {\left (2 \, c d^{2} + a e^{2}\right )} g\right )} x}\right )}{15 \, {\left (e g^{6} x^{4} + d f^{3} g^{3} + {\left (3 \, e f g^{5} + d g^{6}\right )} x^{3} + 3 \, {\left (e f^{2} g^{4} + d f g^{5}\right )} x^{2} + {\left (e f^{3} g^{3} + 3 \, d f^{2} g^{4}\right )} x\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 + 15*c^2*d^2*f^2 + 5*a*c*d*e*f*g + 3*a^2*e^2*g^2 + (35*c^2*d^2*f*g + 11*a*c*d*e*
g^2)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*sqrt(g*x + f) - 15*(c^2*d^2*e*g^3*x^4 + c^2*
d^3*f^3 + (3*c^2*d^2*e*f*g^2 + c^2*d^3*g^3)*x^3 + 3*(c^2*d^2*e*f^2*g + c^2*d^3*f*g^2)*x^2 + (c^2*d^2*e*f^3 + 3
*c^2*d^3*f^2*g)*x)*sqrt(c*d/g)*log(-(8*c^2*d^2*e*g^2*x^3 + c^2*d^3*f^2 + 6*a*c*d^2*e*f*g + a^2*d*e^2*g^2 + 4*(
2*c*d*g^2*x + c*d*f*g + a*e*g^2)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*sqrt(g*x + f)*sqrt(
c*d/g) + 8*(c^2*d^2*e*f*g + (c^2*d^3 + a*c*d*e^2)*g^2)*x^2 + (c^2*d^2*e*f^2 + 2*(4*c^2*d^3 + 3*a*c*d*e^2)*f*g
+ (8*a*c*d^2*e + a^2*e^3)*g^2)*x)/(e*x + d)))/(e*g^6*x^4 + d*f^3*g^3 + (3*e*f*g^5 + d*g^6)*x^3 + 3*(e*f^2*g^4
+ d*f*g^5)*x^2 + (e*f^3*g^3 + 3*d*f^2*g^4)*x), -1/15*(2*(23*c^2*d^2*g^2*x^2 + 15*c^2*d^2*f^2 + 5*a*c*d*e*f*g +
3*a^2*e^2*g^2 + (35*c^2*d^2*f*g + 11*a*c*d*e*g^2)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d
)*sqrt(g*x + f) + 15*(c^2*d^2*e*g^3*x^4 + c^2*d^3*f^3 + (3*c^2*d^2*e*f*g^2 + c^2*d^3*g^3)*x^3 + 3*(c^2*d^2*e*f
^2*g + c^2*d^3*f*g^2)*x^2 + (c^2*d^2*e*f^3 + 3*c^2*d^3*f^2*g)*x)*sqrt(-c*d/g)*arctan(2*sqrt(c*d*e*x^2 + a*d*e
+ (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*sqrt(g*x + f)*sqrt(-c*d/g)*g/(2*c*d*e*g*x^2 + c*d^2*f + a*d*e*g + (c*d*e*f
+ (2*c*d^2 + a*e^2)*g)*x)))/(e*g^6*x^4 + d*f^3*g^3 + (3*e*f*g^5 + d*g^6)*x^3 + 3*(e*f^2*g^4 + d*f*g^5)*x^2 + (
e*f^3*g^3 + 3*d*f^2*g^4)*x)]

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giac [F(-1)]  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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maple [B]  time = 0.04, size = 511, normalized size = 1.86 \begin {gather*} \frac {\sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (15 c^{3} d^{3} g^{3} x^{3} \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 {c d g}}{2 \sqrt {c d g}}\right )+45 c^{3} d^{3} f \,g^{2} x^{2} \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 {c d g}}{2 \sqrt {c d g}}\right )+45 c^{3} d^{3} f^{2} g x \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 {c d g}}{2 \sqrt {c d g}}\right )+15 c^{3} d^{3} f^{3} \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 {c d g}}{2 \sqrt {c d g}}\right )-46 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}\, c^{2} d^{2} g^{2} x^{2}-22 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, a c d e \,g^{2} x -70 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, c^{2} d^{2} f g x -6 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}\, a^{2} e^{2} g^{2}-10 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}\, a c d e f g -30 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}\, c^{2} d^{2} f^{2}\right )}{15 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}\, \left (g x +f \right )^{\frac {5}{2}} \sqrt {e x +d}\, g^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(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)*(
c*d*g)^(1/2))/(c*d*g)^(1/2))*x^3*c^3*d^3*g^3+45*c^3*d^3*f*g^2*x^2*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)*(c*d*g)^(1/2))/(c*d*g)^(1/2))+45*c^3*d^3*f^2*g*x*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)*(c*d*g)^(1/2))/(c*d*g)^(1/2))+15*c^3*d^3*f^3*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)*(c*d*g)^(1/2))/(c*d*g)^(1/2))-46*((g*x+f)*(c*d*x+a*e))^(1/2)*(c*d*g)^(1/2)*c^2*d^2*g^2*x^2-22*(c*d*g
)^(1/2)*((g*x+f)*(c*d*x+a*e))^(1/2)*a*c*d*e*g^2*x-70*(c*d*g)^(1/2)*((g*x+f)*(c*d*x+a*e))^(1/2)*c^2*d^2*f*g*x-6
*((g*x+f)*(c*d*x+a*e))^(1/2)*(c*d*g)^(1/2)*a^2*e^2*g^2-10*((g*x+f)*(c*d*x+a*e))^(1/2)*(c*d*g)^(1/2)*a*c*d*e*f*
g-30*((g*x+f)*(c*d*x+a*e))^(1/2)*(c*d*g)^(1/2)*c^2*d^2*f^2)/((g*x+f)*(c*d*x+a*e))^(1/2)/(c*d*g)^(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*} \int \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{\frac {5}{2}} {\left (g x + f\right )}^{\frac {7}{2}}}\,{d x} \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*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/((e*x + d)^(5/2)*(g*x + f)^(7/2)), x)

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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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sympy [F(-1)]  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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