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

Optimal. Leaf size=236 \[ -\frac {2 (c d f-a e g)^{5/2} \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{g^{7/2}}+\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)^2}{g^3 \sqrt {d+e x}}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g)}{3 g^2 (d+e x)^{3/2}}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {864, 874, 205} \[ \frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)^2}{g^3 \sqrt {d+e x}}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2} (c d f-a e g)}{3 g^2 (d+e x)^{3/2}}-\frac {2 (c d f-a e g)^{5/2} \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{g^{7/2}}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 g (d+e x)^{5/2}} \]

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)),x]

[Out]

(2*(c*d*f - a*e*g)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(g^3*Sqrt[d + e*x]) - (2*(c*d*f - a*e*g)*(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)) + (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)) - (2*(c*d*f - a*e*g)^(5/2)*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])])/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 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)} \, 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}}-\frac {\left (c d e^2 f+c d^2 e g-e \left (c d^2+a e^2\right ) g\right ) \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)} \, dx}{e^2 g}\\ &=-\frac {2 (c d f-a e g) \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}}+\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}}+\frac {(c d f-a e g)^2 \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}{g^2}\\ &=\frac {2 (c d f-a e g)^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}}-\frac {2 (c d f-a e g) \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}}+\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}}-\frac {(c d f-a e g)^3 \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}{g^3}\\ &=\frac {2 (c d f-a e g)^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}}-\frac {2 (c d f-a e g) \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}}+\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}}-\frac {\left (2 e^2 (c d f-a e g)^3\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 )}{g^3}\\ &=\frac {2 (c d f-a e g)^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}}-\frac {2 (c d f-a e g) \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}}+\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}}-\frac {2 (c d f-a e g)^{5/2} \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 )}{g^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 145, normalized size = 0.61 \[ \frac {((d+e x) (a e+c d x))^{5/2} \left (-\frac {10 (c d f-a e g)^{5/2} \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c d f-a e g}}\right )}{g^{5/2} (a e+c d x)^{5/2}}+\frac {10 (a e g-c d f) (4 a e g+c d (g x-3 f))}{3 g^2 (a e+c d x)^2}+2\right )}{5 g (d+e x)^{5/2}} \]

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)),x]

[Out]

(((a*e + c*d*x)*(d + e*x))^(5/2)*(2 + (10*(-(c*d*f) + a*e*g)*(4*a*e*g + c*d*(-3*f + g*x)))/(3*g^2*(a*e + c*d*x
)^2) - (10*(c*d*f - a*e*g)^(5/2)*ArcTan[(Sqrt[g]*Sqrt[a*e + c*d*x])/Sqrt[c*d*f - a*e*g]])/(g^(5/2)*(a*e + c*d*
x)^(5/2))))/(5*g*(d + e*x)^(5/2))

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fricas [A]  time = 1.02, size = 587, normalized size = 2.49 \[ \left [\frac {15 \, {\left (c^{2} d^{3} f^{2} - 2 \, a c d^{2} e f g + a^{2} d e^{2} g^{2} + {\left (c^{2} d^{2} e f^{2} - 2 \, a c d e^{2} f g + a^{2} e^{3} g^{2}\right )} x\right )} \sqrt {-\frac {c d f - a e g}{g}} \log \left (-\frac {c d e g x^{2} - c d^{2} f + 2 \, a d e g - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} g \sqrt {-\frac {c d f - a e g}{g}} - {\left (c d e f - {\left (c d^{2} + 2 \, a e^{2}\right )} g\right )} x}{e g x^{2} + d f + {\left (e f + d g\right )} x}\right ) + 2 \, {\left (3 \, c^{2} d^{2} g^{2} x^{2} + 15 \, c^{2} d^{2} f^{2} - 35 \, a c d e f g + 23 \, a^{2} e^{2} g^{2} - {\left (5 \, c^{2} d^{2} f g - 11 \, a c d e g^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{15 \, {\left (e g^{3} x + d g^{3}\right )}}, \frac {2 \, {\left (15 \, {\left (c^{2} d^{3} f^{2} - 2 \, a c d^{2} e f g + a^{2} d e^{2} g^{2} + {\left (c^{2} d^{2} e f^{2} - 2 \, a c d e^{2} f g + a^{2} e^{3} g^{2}\right )} x\right )} \sqrt {\frac {c d f - a e g}{g}} \arctan \left (\frac {\sqrt {e x + d} \sqrt {\frac {c d f - a e g}{g}}}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}\right ) + {\left (3 \, c^{2} d^{2} g^{2} x^{2} + 15 \, c^{2} d^{2} f^{2} - 35 \, a c d e f g + 23 \, a^{2} e^{2} g^{2} - {\left (5 \, c^{2} d^{2} f g - 11 \, a c d e g^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}\right )}}{15 \, {\left (e g^{3} x + d g^{3}\right )}}\right ] \]

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),x, algorithm="fricas")

[Out]

[1/15*(15*(c^2*d^3*f^2 - 2*a*c*d^2*e*f*g + a^2*d*e^2*g^2 + (c^2*d^2*e*f^2 - 2*a*c*d*e^2*f*g + a^2*e^3*g^2)*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*(3*c^2*d^2*g^2*x^2 + 15*c^2*d^2*f^2 - 35*a*c*d*e*f*g + 23*a^2*e^2*g^2 - (5*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))/(e*g^3*x + d*g^3), 2/15*(15*(c^2*d^3*f^2 - 2*
a*c*d^2*e*f*g + a^2*d*e^2*g^2 + (c^2*d^2*e*f^2 - 2*a*c*d*e^2*f*g + a^2*e^3*g^2)*x)*sqrt((c*d*f - a*e*g)/g)*arc
tan(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)) + (3*c^2*d^2*g^2*x^2 +
15*c^2*d^2*f^2 - 35*a*c*d*e*f*g + 23*a^2*e^2*g^2 - (5*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))/(e*g^3*x + d*g^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.02, size = 431, normalized size = 1.83 \[ -\frac {2 \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (15 a^{3} e^{3} g^{3} \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )-45 a^{2} c d \,e^{2} f \,g^{2} \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )+45 a \,c^{2} d^{2} e \,f^{2} g \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )-15 c^{3} d^{3} f^{3} \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )-3 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, c^{2} d^{2} g^{2} x^{2}-11 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, a c d e \,g^{2} x +5 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, c^{2} d^{2} f g x -23 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, a^{2} e^{2} g^{2}+35 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, a c d e f g -15 \sqrt {\left (a e g -c d f \right ) g}\, \sqrt {c d x +a e}\, c^{2} d^{2} f^{2}\right )}{15 \sqrt {e x +d}\, \sqrt {c d x +a e}\, \sqrt {\left (a e g -c d f \right ) g}\, g^{3}} \]

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),x)

[Out]

-2/15*(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)*a^3*e^3
*g^3-45*arctanh((c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*a^2*c*d*e^2*f*g^2+45*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
-3*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/2)*c^2*d^2*g^2*x^2-11*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/2)*a*c*
d*e*g^2*x+5*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/2)*c^2*d^2*f*g*x-23*((a*e*g-c*d*f)*g)^(1/2)*(c*d*x+a*e)^(1/
2)*a^2*e^2*g^2+35*((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/((a*e*g-c*d*f)*g)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}}{{\left (e x + d\right )}^{\frac {5}{2}} {\left (g x + f\right )}}\,{d x} \]

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),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)), x)

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

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)*(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)*(d + e*x)^(5/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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),x)

[Out]

Timed out

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