3.5.74 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{5/2} (f+g x)^3} \, dx\)

Optimal. Leaf size=246 \[ -\frac {15 c^2 d^2 \sqrt {c d f-a e g} \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{4 g^{7/2}}+\frac {15 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 g^3 \sqrt {d+e x}}-\frac {5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{4 g^2 (d+e x)^{3/2} (f+g x)}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 g (d+e x)^{5/2} (f+g x)^2} \]

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Rubi [A]  time = 0.34, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {862, 864, 874, 205} \begin {gather*} -\frac {15 c^2 d^2 \sqrt {c d f-a e g} \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{4 g^{7/2}}+\frac {15 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 g^3 \sqrt {d+e x}}-\frac {5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{4 g^2 (d+e x)^{3/2} (f+g x)}-\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{2 g (d+e x)^{5/2} (f+g x)^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)^3),x]

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 864

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*(m - n - 1)), x] - Dist[(m*(c*e*f + c*d*g - b*e*g
))/(e^2*g*(m - n - 1)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c,
d, e, f, g, 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] && EqQ[m + p, 0] && GtQ[p, 0] && NeQ[m - n - 1, 0] &&  !IGtQ[n, 0] &&  !(IntegerQ[n + p] && LtQ[n + p + 2,
0]) && RationalQ[n]

Rule 874

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[
2*e^2, Subst[Int[1/(c*(e*f + d*g) - b*e*g + e^2*g*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; Fre
eQ[{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]

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)^3} \, dx &=-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 g (d+e x)^{5/2} (f+g x)^2}+\frac {(5 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)^2} \, dx}{4 g}\\ &=-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 g^2 (d+e x)^{3/2} (f+g x)}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 g (d+e x)^{5/2} (f+g x)^2}+\frac {\left (15 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)} \, dx}{8 g^2}\\ &=\frac {15 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 g^3 \sqrt {d+e x}}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 g^2 (d+e x)^{3/2} (f+g x)}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 g (d+e x)^{5/2} (f+g x)^2}-\frac {\left (15 c^2 d^2 (c d f-a e g)\right ) \int \frac {\sqrt {d+e x}}{(f+g x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 g^3}\\ &=\frac {15 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 g^3 \sqrt {d+e x}}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 g^2 (d+e x)^{3/2} (f+g x)}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 g (d+e x)^{5/2} (f+g x)^2}-\frac {\left (15 c^2 d^2 e^2 (c d f-a e g)\right ) \operatorname {Subst}\left (\int \frac {1}{-e \left (c d^2+a e^2\right ) g+c d e (e f+d g)+e^2 g x^2} \, dx,x,\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}}\right )}{4 g^3}\\ &=\frac {15 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 g^3 \sqrt {d+e x}}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 g^2 (d+e x)^{3/2} (f+g x)}-\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{2 g (d+e x)^{5/2} (f+g x)^2}-\frac {15 c^2 d^2 \sqrt {c d f-a e g} \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d f-a e g} \sqrt {d+e x}}\right )}{4 g^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 79, normalized size = 0.32 \begin {gather*} \frac {2 c^2 d^2 ((d+e x) (a e+c d x))^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};\frac {g (a e+c d x)}{a e g-c d f}\right )}{7 (d+e x)^{7/2} (c d f-a e g)^3} \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)^3),x]

[Out]

(2*c^2*d^2*((a*e + c*d*x)*(d + e*x))^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, (g*(a*e + c*d*x))/(-(c*d*f) + a*e*g)
])/(7*(c*d*f - a*e*g)^3*(d + e*x)^(7/2))

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IntegrateAlgebraic [F]  time = 180.04, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[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)^3),x]

[Out]

$Aborted

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fricas [A]  time = 0.62, size = 683, normalized size = 2.78 \begin {gather*} \left [\frac {15 \, {\left (c^{2} d^{2} e g^{2} x^{3} + c^{2} d^{3} f^{2} + {\left (2 \, c^{2} d^{2} e f g + c^{2} d^{3} g^{2}\right )} x^{2} + {\left (c^{2} d^{2} e f^{2} + 2 \, c^{2} d^{3} f g\right )} x\right )} \sqrt {-\frac {c d f - a e g}{g}} \log \left (-\frac {c d e g x^{2} - c d^{2} f + 2 \, a d e g - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} g \sqrt {-\frac {c d f - a e g}{g}} - {\left (c d e f - {\left (c d^{2} + 2 \, a e^{2}\right )} g\right )} x}{e g x^{2} + d f + {\left (e f + d g\right )} x}\right ) + 2 \, {\left (8 \, c^{2} d^{2} g^{2} x^{2} + 15 \, c^{2} d^{2} f^{2} - 5 \, a c d e f g - 2 \, a^{2} e^{2} g^{2} + {\left (25 \, c^{2} d^{2} f g - 9 \, 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}}{8 \, {\left (e g^{5} x^{3} + d f^{2} g^{3} + {\left (2 \, e f g^{4} + d g^{5}\right )} x^{2} + {\left (e f^{2} g^{3} + 2 \, d f g^{4}\right )} x\right )}}, \frac {15 \, {\left (c^{2} d^{2} e g^{2} x^{3} + c^{2} d^{3} f^{2} + {\left (2 \, c^{2} d^{2} e f g + c^{2} d^{3} g^{2}\right )} x^{2} + {\left (c^{2} d^{2} e f^{2} + 2 \, c^{2} d^{3} f g\right )} x\right )} \sqrt {\frac {c d f - a e g}{g}} \arctan \left (\frac {\sqrt {e x + d} \sqrt {\frac {c d f - a e g}{g}}}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}\right ) + {\left (8 \, c^{2} d^{2} g^{2} x^{2} + 15 \, c^{2} d^{2} f^{2} - 5 \, a c d e f g - 2 \, a^{2} e^{2} g^{2} + {\left (25 \, c^{2} d^{2} f g - 9 \, 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}}{4 \, {\left (e g^{5} x^{3} + d f^{2} g^{3} + {\left (2 \, e f g^{4} + d g^{5}\right )} x^{2} + {\left (e f^{2} g^{3} + 2 \, d f 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)^3,x, algorithm="fricas")

[Out]

[1/8*(15*(c^2*d^2*e*g^2*x^3 + c^2*d^3*f^2 + (2*c^2*d^2*e*f*g + c^2*d^3*g^2)*x^2 + (c^2*d^2*e*f^2 + 2*c^2*d^3*f
*g)*x)*sqrt(-(c*d*f - a*e*g)/g)*log(-(c*d*e*g*x^2 - c*d^2*f + 2*a*d*e*g - 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 +
a*e^2)*x)*sqrt(e*x + d)*g*sqrt(-(c*d*f - a*e*g)/g) - (c*d*e*f - (c*d^2 + 2*a*e^2)*g)*x)/(e*g*x^2 + d*f + (e*f
+ d*g)*x)) + 2*(8*c^2*d^2*g^2*x^2 + 15*c^2*d^2*f^2 - 5*a*c*d*e*f*g - 2*a^2*e^2*g^2 + (25*c^2*d^2*f*g - 9*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))/(e*g^5*x^3 + d*f^2*g^3 + (2*e*f*g^4 + d*
g^5)*x^2 + (e*f^2*g^3 + 2*d*f*g^4)*x), 1/4*(15*(c^2*d^2*e*g^2*x^3 + c^2*d^3*f^2 + (2*c^2*d^2*e*f*g + c^2*d^3*g
^2)*x^2 + (c^2*d^2*e*f^2 + 2*c^2*d^3*f*g)*x)*sqrt((c*d*f - a*e*g)/g)*arctan(sqrt(e*x + d)*sqrt((c*d*f - a*e*g)
/g)/sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)) + (8*c^2*d^2*g^2*x^2 + 15*c^2*d^2*f^2 - 5*a*c*d*e*f*g - 2*a^2
*e^2*g^2 + (25*c^2*d^2*f*g - 9*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))/(e*g
^5*x^3 + d*f^2*g^3 + (2*e*f*g^4 + d*g^5)*x^2 + (e*f^2*g^3 + 2*d*f*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)^3,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.03, size = 526, normalized size = 2.14 \begin {gather*} -\frac {\sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (15 a \,c^{2} d^{2} e \,g^{3} x^{2} \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )-15 c^{3} d^{3} f \,g^{2} x^{2} \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )+30 a \,c^{2} d^{2} e f \,g^{2} x \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )-30 c^{3} d^{3} f^{2} g x \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )+15 a \,c^{2} d^{2} e \,f^{2} g \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )-15 c^{3} d^{3} f^{3} \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )-8 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, c^{2} d^{2} g^{2} x^{2}+9 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, a c d e \,g^{2} x -25 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, c^{2} d^{2} f g x +2 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, a^{2} e^{2} g^{2}+5 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, a c d e f g -15 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, c^{2} d^{2} f^{2}\right )}{4 \sqrt {e x +d}\, \sqrt {c d x +a e}\, \left (g x +f \right )^{2} \sqrt {\left (a e g -c d f \right ) g}\, 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)^3,x)

[Out]

-1/4*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(15*arctanh((c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*x^2*a*c^
2*d^2*e*g^3-15*arctanh((c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*x^2*c^3*d^3*f*g^2+30*arctanh((c*d*x+a*e)^(
1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*x*a*c^2*d^2*e*f*g^2-30*arctanh((c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*x*
c^3*d^3*f^2*g+15*arctanh((c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*a*c^2*d^2*e*f^2*g-15*arctanh((c*d*x+a*e)
^(1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*c^3*d^3*f^3-8*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/2)*c^2*d^2*g^2*x^2+9*((
a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/2)*a*c*d*e*g^2*x-25*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/2)*c^2*d^2*f*g
*x+2*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/2)*a^2*e^2*g^2+5*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/2)*a*c*d*e
*f*g-15*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/2)*c^2*d^2*f^2)/(e*x+d)^(1/2)/(c*d*x+a*e)^(1/2)/g^3/(g*x+f)^2/(
(a*e*g-c*d*f)*g)^(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 )}^{3}}\,{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)^3,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)^3), 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 )}^3\,{\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)^3*(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)^3*(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)**3,x)

[Out]

Timed out

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