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

Optimal. Leaf size=219 \[ \frac {2 g^{3/2} \sqrt {d+e x} \sqrt {a e+c d x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{c^{5/2} d^{5/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 g \sqrt {d+e x} \sqrt {f+g x}}{c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)^{3/2} (f+g x)^{3/2}}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {866, 891, 63, 217, 206} \[ \frac {2 g^{3/2} \sqrt {d+e x} \sqrt {a e+c d x} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{c^{5/2} d^{5/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 g \sqrt {d+e x} \sqrt {f+g x}}{c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)^{3/2} (f+g x)^{3/2}}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^(5/2)*(f + g*x)^(3/2))/(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)^(3/2))/(3*c*d*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (2*g*Sqrt[d + e*x
]*Sqrt[f + g*x])/(c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (2*g^(3/2)*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])])/(c^(5/2)*d^(5/2)*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 866

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 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 {(d+e x)^{5/2} (f+g x)^{3/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)^{3/2} (f+g x)^{3/2}}{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 {g \int \frac {(d+e x)^{3/2} \sqrt {f+g x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{c d}\\ &=-\frac {2 (d+e x)^{3/2} (f+g x)^{3/2}}{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 {2 g \sqrt {d+e x} \sqrt {f+g x}}{c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {g^2 \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)^{3/2}}{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 {2 g \sqrt {d+e x} \sqrt {f+g x}}{c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (g^2 \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}{c^2 d^2 \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)^{3/2}}{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 {2 g \sqrt {d+e x} \sqrt {f+g x}}{c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (2 g^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 )}{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)^{3/2}}{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 {2 g \sqrt {d+e x} \sqrt {f+g x}}{c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (2 g^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 )}{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)^{3/2}}{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 {2 g \sqrt {d+e x} \sqrt {f+g x}}{c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {2 g^{3/2} \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 )}{c^{5/2} d^{5/2} \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 [C]  time = 0.11, size = 102, normalized size = 0.47 \[ -\frac {2 (d+e x)^{3/2} (f+g x)^{3/2} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {g (a e+c d x)}{a e g-c d f}\right )}{3 c d ((d+e x) (a e+c d x))^{3/2} \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^(5/2)*(f + g*x)^(3/2))/(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)^(3/2)*Hypergeometric2F1[-3/2, -3/2, -1/2, (g*(a*e + c*d*x))/(-(c*d*f) + a*e*g)])
/(3*c*d*((a*e + c*d*x)*(d + e*x))^(3/2)*((c*d*(f + g*x))/(c*d*f - a*e*g))^(3/2))

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fricas [A]  time = 2.22, size = 755, normalized size = 3.45 \[ \left [-\frac {4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (4 \, c d g x + c d f + 3 \, a e g\right )} \sqrt {e x + d} \sqrt {g x + f} - 3 \, {\left (c^{2} d^{2} e g x^{3} + a^{2} d e^{2} g + {\left (c^{2} d^{3} + 2 \, a c d e^{2}\right )} g x^{2} + {\left (2 \, a c d^{2} e + a^{2} e^{3}\right )} g x\right )} \sqrt {\frac {g}{c d}} \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} + 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} + 4 \, {\left (2 \, c^{2} d^{2} g x + c^{2} d^{2} f + a c d e g\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 {g}{c d}} + {\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 )}{6 \, {\left (c^{4} d^{4} e x^{3} + a^{2} c^{2} d^{3} e^{2} + {\left (c^{4} d^{5} + 2 \, a c^{3} d^{3} e^{2}\right )} x^{2} + {\left (2 \, a c^{3} d^{4} e + a^{2} c^{2} d^{2} e^{3}\right )} x\right )}}, -\frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (4 \, c d g x + c d f + 3 \, a e g\right )} \sqrt {e x + d} \sqrt {g x + f} + 3 \, {\left (c^{2} d^{2} e g x^{3} + a^{2} d e^{2} g + {\left (c^{2} d^{3} + 2 \, a c d e^{2}\right )} g x^{2} + {\left (2 \, a c d^{2} e + a^{2} e^{3}\right )} g x\right )} \sqrt {-\frac {g}{c d}} \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} c d \sqrt {-\frac {g}{c d}}}{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 )}{3 \, {\left (c^{4} d^{4} e x^{3} + a^{2} c^{2} d^{3} e^{2} + {\left (c^{4} d^{5} + 2 \, a c^{3} d^{3} e^{2}\right )} x^{2} + {\left (2 \, a c^{3} d^{4} e + a^{2} c^{2} d^{2} e^{3}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)*(g*x+f)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(4*c*d*g*x + c*d*f + 3*a*e*g)*sqrt(e*x + d)*sqrt(g*x + f)
- 3*(c^2*d^2*e*g*x^3 + a^2*d*e^2*g + (c^2*d^3 + 2*a*c*d*e^2)*g*x^2 + (2*a*c*d^2*e + a^2*e^3)*g*x)*sqrt(g/(c*d
))*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 + 8*(c^2*d^2*e*f*g + (c^2*d^3 + a
*c*d*e^2)*g^2)*x^2 + 4*(2*c^2*d^2*g*x + c^2*d^2*f + a*c*d*e*g)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqr
t(e*x + d)*sqrt(g*x + f)*sqrt(g/(c*d)) + (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)))/(c^4*d^4*e*x^3 + a^2*c^2*d^3*e^2 + (c^4*d^5 + 2*a*c^3*d^3*e^2)*x^2 + (2*a*c^3*d^4*e
+ a^2*c^2*d^2*e^3)*x), -1/3*(2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(4*c*d*g*x + c*d*f + 3*a*e*g)*sqrt(
e*x + d)*sqrt(g*x + f) + 3*(c^2*d^2*e*g*x^3 + a^2*d*e^2*g + (c^2*d^3 + 2*a*c*d*e^2)*g*x^2 + (2*a*c*d^2*e + a^2
*e^3)*g*x)*sqrt(-g/(c*d))*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)*c*d
*sqrt(-g/(c*d))/(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)))/(c^4*d^4*e*x^3 + a^2
*c^2*d^3*e^2 + (c^4*d^5 + 2*a*c^3*d^3*e^2)*x^2 + (2*a*c^3*d^4*e + a^2*c^2*d^2*e^3)*x)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)*(g*x+f)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 5.17Unable to transpose Error: Bad Argument Value

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maple [A]  time = 0.03, size = 343, normalized size = 1.57 \[ \frac {\sqrt {g x +f}\, \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (3 c^{2} d^{2} g^{2} x^{2} \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right )+6 a c d e \,g^{2} x \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right )+3 a^{2} e^{2} g^{2} \ln \left (\frac {2 c d g x +a e g +c d f +2 \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}}{2 \sqrt {c d g}}\right )-8 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, c d g x -6 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, a e g -2 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, c d f \right )}{3 \sqrt {c d g}\, \left (c d x +a e \right )^{2} \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {e x +d}\, c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)*(g*x+f)^(3/2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(5/2),x)

[Out]

1/3*(g*x+f)^(1/2)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(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))*x^2*c^2*d^2*g^2+6*a*c*d*e*g^2*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))+3*a^2*e^2*g^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))-8*(c*d*g)^(1/2)*((g*x+f)*(c*d*x+a*e))^(1/2)*c*d*g*x-6*(c*d*g)^(
1/2)*((g*x+f)*(c*d*x+a*e))^(1/2)*a*e*g-2*(c*d*g)^(1/2)*((g*x+f)*(c*d*x+a*e))^(1/2)*c*d*f)/(c*d*g)^(1/2)/(c*d*x
+a*e)^2/((g*x+f)*(c*d*x+a*e))^(1/2)/d^2/c^2/(e*x+d)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{\frac {5}{2}} {\left (g x + f\right )}^{\frac {3}{2}}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)*(g*x+f)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(5/2)*(g*x + f)^(3/2)/(c*d*e*x^2 + 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 \[ \int \frac {{\left (f+g\,x\right )}^{3/2}\,{\left (d+e\,x\right )}^{5/2}}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)^(3/2)*(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)^(3/2)*(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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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((e*x+d)**(5/2)*(g*x+f)**(3/2)/(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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