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

Optimal. Leaf size=244 \[ \frac {3 \sqrt {d+e x} \sqrt {a e+c d x} (c d f-a e g)^2 \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{4 c^{5/2} d^{5/2} \sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {3 \sqrt {f+g x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)}{4 c^2 d^2 \sqrt {d+e x}}+\frac {(f+g x)^{3/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 c d \sqrt {d+e x}} \]

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Rubi [A]  time = 0.37, antiderivative size = 244, 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 = {870, 891, 63, 217, 206} \begin {gather*} \frac {3 \sqrt {f+g x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)}{4 c^2 d^2 \sqrt {d+e x}}+\frac {3 \sqrt {d+e x} \sqrt {a e+c d x} (c d f-a e g)^2 \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{4 c^{5/2} d^{5/2} \sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {(f+g x)^{3/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 c d \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(c*d*f - a*e*g)*Sqrt[f + g*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*c^2*d^2*Sqrt[d + e*x]) + ((f
+ g*x)^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(2*c*d*Sqrt[d + e*x]) + (3*(c*d*f - a*e*g)^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])])/(4*c^(5/2)*d^(5
/2)*Sqrt[g]*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 870

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

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 {\sqrt {d+e x} (f+g x)^{3/2}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac {(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c d \sqrt {d+e x}}+\frac {\left (3 \left (c d e^2 f+c d^2 e g-e \left (c d^2+a e^2\right ) g\right )\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}{4 c d e^2}\\ &=\frac {3 (c d f-a e g) \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^2 d^2 \sqrt {d+e x}}+\frac {(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c d \sqrt {d+e x}}+\frac {\left (3 (c d f-a e g)^2\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}{8 c^2 d^2}\\ &=\frac {3 (c d f-a e g) \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^2 d^2 \sqrt {d+e x}}+\frac {(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c d \sqrt {d+e x}}+\frac {\left (3 (c d f-a e 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}{8 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac {3 (c d f-a e g) \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^2 d^2 \sqrt {d+e x}}+\frac {(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c d \sqrt {d+e x}}+\frac {\left (3 (c d f-a e 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 )}{4 c^3 d^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac {3 (c d f-a e g) \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^2 d^2 \sqrt {d+e x}}+\frac {(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c d \sqrt {d+e x}}+\frac {\left (3 (c d f-a e 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 )}{4 c^3 d^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac {3 (c d f-a e g) \sqrt {f+g x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^2 d^2 \sqrt {d+e x}}+\frac {(f+g x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c d \sqrt {d+e x}}+\frac {3 (c d f-a e g)^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 )}{4 c^{5/2} d^{5/2} \sqrt {g} \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 = 0.47, size = 234, normalized size = 0.96 \begin {gather*} \frac {\sqrt {d+e x} \sqrt {f+g x} \sqrt {a e+c d x} \left (\sqrt {c} \sqrt {d} \sqrt {g} \sqrt {a e+c d x} \sqrt {\frac {c d (f+g x)}{c d f-a e g}} (c d (5 f+2 g x)-3 a e g)+3 \sqrt {c d} (c d f-a e g)^{3/2} \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 )\right )}{4 c^{5/2} d^{5/2} \sqrt {g} \sqrt {(d+e x) (a e+c d x)} \sqrt {\frac {c d (f+g x)}{c d f-a e g}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[d + e*x]*(f + g*x)^(3/2))/Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 7.75, size = 206, normalized size = 0.84 \begin {gather*} \frac {\sqrt {d+e x} \sqrt {a e+c d x} \left (\frac {3 (c d f-a e g)^2 \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {f+g x}}\right )}{4 c^{5/2} d^{5/2} \sqrt {g}}+\frac {(c d f-a e g)^2 \left (\frac {5 c d \sqrt {a e+c d x}}{\sqrt {f+g x}}-\frac {3 g (a e+c d x)^{3/2}}{(f+g x)^{3/2}}\right )}{4 c^2 d^2 \left (c d-\frac {g (a e+c d x)}{f+g x}\right )^2}\right )}{\sqrt {(d+e x) (a e+c d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[d + e*x]*(f + g*x)^(3/2))/Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2],x]

[Out]

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

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fricas [A]  time = 1.24, size = 655, normalized size = 2.68 \begin {gather*} \left [\frac {4 \, {\left (2 \, c^{2} d^{2} g^{2} x + 5 \, c^{2} d^{2} f g - 3 \, a c d 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} + 3 \, {\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 {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 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d g x + c d f + a e g\right )} \sqrt {c d g} \sqrt {e x + d} \sqrt {g x + f} + 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 )}{16 \, {\left (c^{3} d^{3} e g x + c^{3} d^{4} g\right )}}, \frac {2 \, {\left (2 \, c^{2} d^{2} g^{2} x + 5 \, c^{2} d^{2} f g - 3 \, a c d 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} - 3 \, {\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 {-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 {-c d g} \sqrt {e x + d} \sqrt {g x + f}}{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 )}{8 \, {\left (c^{3} d^{3} e g x + c^{3} d^{4} g\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^(3/2)*(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(4*(2*c^2*d^2*g^2*x + 5*c^2*d^2*f*g - 3*a*c*d*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) + 3*(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*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*sqrt(
c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*g*x + c*d*f + a*e*g)*sqrt(c*d*g)*sqrt(e*x + d)*sqrt(g*x + f) + 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)))/(c^3*d^3*e*g*x + c^3*d^4*g), 1/8*(2*(2*c^2*d^2*g^2*x + 5*c^2*d^2*f*g - 3*a
*c*d*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) - 3*(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*g)*arctan(2*sqrt(c*d*e
*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(-c*d*g)*sqrt(e*x + d)*sqrt(g*x + f)/(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^3*d^3*e*g*x + c^3*d^4*g)]

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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((g*x+f)^(3/2)*(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.03, size = 328, normalized size = 1.34 \begin {gather*} \frac {\sqrt {g x +f}\, \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (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 )-6 a c d e f g \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 c^{2} d^{2} f^{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 )+4 \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 +10 \sqrt {c d g}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, c d f \right )}{8 \sqrt {e x +d}\, \sqrt {\left (g x +f \right ) \left (c d x +a e \right )}\, \sqrt {c d g}\, c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(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))*a^2*e^2*g^2-6*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))*a*c*d*e*f*g+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))*c^2*d^2*f^2+4*(c*d*g)^(1/2)*((g*x+f)*(c*d*x+a*e))^(1/2)*x*c*d*g-6*(c*d*g)^(1/2)*(
(g*x+f)*(c*d*x+a*e))^(1/2)*a*e*g+10*(c*d*g)^(1/2)*((g*x+f)*(c*d*x+a*e))^(1/2)*c*d*f)/(e*x+d)^(1/2)/((g*x+f)*(c
*d*x+a*e))^(1/2)/c^2/d^2/(c*d*g)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^(3/2)*(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (f+g\,x\right )}^{3/2}\,\sqrt {d+e\,x}}{\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d + e x} \left (f + g x\right )^{\frac {3}{2}}}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**(3/2)*(e*x+d)**(1/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

Integral(sqrt(d + e*x)*(f + g*x)**(3/2)/sqrt((d + e*x)*(a*e + c*d*x)), x)

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