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

Optimal. Leaf size=139 \[ \frac {2 e \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d g \sqrt {d+e x}}-\frac {2 (e f-d g) \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d f-a e g}}\right )}{g^{3/2} \sqrt {c d f-a e g}} \]

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Rubi [A]  time = 0.19, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {880, 874, 205} \begin {gather*} \frac {2 e \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d g \sqrt {d+e x}}-\frac {2 (e f-d 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 )}{g^{3/2} \sqrt {c d f-a e g}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rule 880

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Simp[(e^2*(d + e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + b*x + c*x^2)^(p + 1))/(c*g*(n + p + 2)), x] - Dist[(b*e*g*(
n + 1) + c*e*f*(p + 1) - c*d*g*(2*n + p + 3))/(c*g*(n + p + 2)), Int[(d + e*x)^(m - 1)*(f + g*x)^n*(a + b*x +
c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && Eq
Q[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p - 1, 0] &&  !LtQ[n, -1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {(d+e x)^{3/2}}{(f+g x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac {2 e \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d g \sqrt {d+e x}}-\frac {\left (2 \left (\frac {1}{2} c d e^2 f-\frac {1}{2} c d^2 e g\right )\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}{c d e g}\\ &=\frac {2 e \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d g \sqrt {d+e x}}-\frac {\left (2 e^2 (e f-d 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 )}{g}\\ &=\frac {2 e \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d g \sqrt {d+e x}}-\frac {2 (e f-d 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 )}{g^{3/2} \sqrt {c d f-a e g}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 140, normalized size = 1.01 \begin {gather*} \frac {2 \sqrt {d+e x} \left (e \sqrt {g} (a e+c d x) \sqrt {c d f-a e g}+c d (d g-e f) \sqrt {a e+c d x} \tan ^{-1}\left (\frac {\sqrt {g} \sqrt {a e+c d x}}{\sqrt {c d f-a e g}}\right )\right )}{c d g^{3/2} \sqrt {(d+e x) (a e+c d x)} \sqrt {c d f-a e g}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [C]  time = 20.82, size = 2866, normalized size = 20.62 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(e*Sqrt[d + e*x]*(c*d^2 - 2*a*e^2 - 2*c*d*(d + e*x))) - 2*e*Sqrt[c*d*e]*Sqrt[d + e*x]*Sqrt[-((c*d^2*(d + e*x
))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e])/(-(c*d*Sqrt[c*d*e]*g*(d + e*x)) + c*d*e*g*Sqrt[-((c*d^2*(d + e*x
))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e]) - ((2*I)*Sqrt[c]*Sqrt[d]*(-(e*f) + d*g)^(3/2)*ArcTanh[(Sqrt[e]*S
qrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]*Sqr
t[d + e*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(
d + e*x)^2)/e])])/(g^(3/2)*Sqrt[c*d*f - a*e*g]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqr
t[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]) + ((2*I)*Sqrt[c]*Sqrt[d]*(-(e*f) + d*g)^(3/2)*ArcTanh[(Sqrt[e]*S
qrt[-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]*Sqr
t[d + e*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(
d + e*x)^2)/e])])/(g^(3/2)*Sqrt[c*d*f - a*e*g]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqr
t[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]) + (-((d*e*Sqrt[d + e*x])/g) - (2*e^(3/2)*Sqrt[c*d*e]*f*(d + e*x)
*ArcTanh[(Sqrt[e]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[
c*d*f - a*e*g]]*Sqrt[d + e*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e
*(d + e*x) + (c*d*(d + e*x)^2)/e])])/(g^(3/2)*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt
[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]) + (2*d*Sqrt[e]*Sqrt[c*d*e]*(d + e*x)*ArcTanh[(Sqrt[e]*Sqrt[-2*c*d
*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]*Sqrt[d + e*x]
)/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2
)/e])])/(Sqrt[g]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c
*d*f - a*e*g]]) + (2*e^(5/2)*f*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e]*ArcTanh[(Sqr
t[e]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g
]]*Sqrt[d + e*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) +
(c*d*(d + e*x)^2)/e])])/(g^(3/2)*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*
f) + d*g]*Sqrt[c*d*f - a*e*g]]) - (2*d*e^(3/2)*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)
/e]*ArcTanh[(Sqrt[e]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sq
rt[c*d*f - a*e*g]]*Sqrt[d + e*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) +
a*e*(d + e*x) + (c*d*(d + e*x)^2)/e])])/(Sqrt[g]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g - (2*I)*Sqrt[c]*Sqrt[d]*S
qrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]) - (2*e^(3/2)*Sqrt[c*d*e]*f*(d + e*x)*ArcTanh[(Sqrt[e]*Sqrt[-2*
c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]*Sqrt[d + e
*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x
)^2)/e])])/(g^(3/2)*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqr
t[c*d*f - a*e*g]]) + (2*d*Sqrt[e]*Sqrt[c*d*e]*(d + e*x)*ArcTanh[(Sqrt[e]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g +
(2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]*Sqrt[d + e*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(
d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e])])/(Sqrt[g]*Sqrt[-2*c
*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]) + (2*e^(5/
2)*f*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e]*ArcTanh[(Sqrt[e]*Sqrt[-2*c*d*e*f + c*d
^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]*Sqrt[d + e*x])/(-(Sqrt[
c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e])])/(g
^(3/2)*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e
*g]]) - (2*d*e^(3/2)*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e*x)^2)/e]*ArcTanh[(Sqrt[e]*Sqrt[
-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*Sqrt[c*d*f - a*e*g]]*Sqrt[d
+ e*x])/(-(Sqrt[c*d*e]*Sqrt[g]*(d + e*x)) + e*Sqrt[g]*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d +
e*x)^2)/e])])/(Sqrt[g]*Sqrt[-2*c*d*e*f + c*d^2*g + a*e^2*g + (2*I)*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[-(e*f) + d*g]*
Sqrt[c*d*f - a*e*g]]))/(-(Sqrt[c*d*e]*(d + e*x)) + e*Sqrt[-((c*d^2*(d + e*x))/e) + a*e*(d + e*x) + (c*d*(d + e
*x)^2)/e])

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fricas [A]  time = 0.44, size = 511, normalized size = 3.68 \begin {gather*} \left [\frac {{\left (c d^{2} e f - c d^{3} g + {\left (c d e^{2} f - c d^{2} e g\right )} x\right )} \sqrt {-c d f g + a e g^{2}} \log \left (-\frac {c d e g x^{2} - c d^{2} f + 2 \, a d e g - {\left (c d e f - {\left (c d^{2} + 2 \, a e^{2}\right )} g\right )} x - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {-c d f g + a e g^{2}} \sqrt {e x + d}}{e g x^{2} + d f + {\left (e f + d g\right )} x}\right ) + 2 \, {\left (c d e f g - a e^{2} 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}}{c^{2} d^{3} f g^{2} - a c d^{2} e g^{3} + {\left (c^{2} d^{2} e f g^{2} - a c d e^{2} g^{3}\right )} x}, \frac {2 \, {\left ({\left (c d^{2} e f - c d^{3} g + {\left (c d e^{2} f - c d^{2} e g\right )} x\right )} \sqrt {c d f g - a e g^{2}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {c d f g - a e g^{2}} \sqrt {e x + d}}{c d e g x^{2} + a d e g + {\left (c d^{2} + a e^{2}\right )} g x}\right ) + {\left (c d e f g - a e^{2} 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}\right )}}{c^{2} d^{3} f g^{2} - a c d^{2} e g^{3} + {\left (c^{2} d^{2} e f g^{2} - a c d e^{2} g^{3}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((c*d^2*e*f - c*d^3*g + (c*d*e^2*f - c*d^2*e*g)*x)*sqrt(-c*d*f*g + a*e*g^2)*log(-(c*d*e*g*x^2 - c*d^2*f + 2*a
*d*e*g - (c*d*e*f - (c*d^2 + 2*a*e^2)*g)*x - 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(-c*d*f*g + a*e
*g^2)*sqrt(e*x + d))/(e*g*x^2 + d*f + (e*f + d*g)*x)) + 2*(c*d*e*f*g - a*e^2*g^2)*sqrt(c*d*e*x^2 + a*d*e + (c*
d^2 + a*e^2)*x)*sqrt(e*x + d))/(c^2*d^3*f*g^2 - a*c*d^2*e*g^3 + (c^2*d^2*e*f*g^2 - a*c*d*e^2*g^3)*x), 2*((c*d^
2*e*f - c*d^3*g + (c*d*e^2*f - c*d^2*e*g)*x)*sqrt(c*d*f*g - a*e*g^2)*arctan(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 +
a*e^2)*x)*sqrt(c*d*f*g - a*e*g^2)*sqrt(e*x + d)/(c*d*e*g*x^2 + a*d*e*g + (c*d^2 + a*e^2)*g*x)) + (c*d*e*f*g -
a*e^2*g^2)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(c^2*d^3*f*g^2 - a*c*d^2*e*g^3 + (c^2*d^
2*e*f*g^2 - a*c*d*e^2*g^3)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 163, normalized size = 1.17 \begin {gather*} -\frac {2 \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (c \,d^{2} g \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )-c d e f \arctanh \left (\frac {\sqrt {c d x +a e}\, g}{\sqrt {\left (a e g -c d f \right ) g}}\right )-\sqrt {c d x +a e}\, \sqrt {\left (a e g -c d f \right ) g}\, e \right )}{\sqrt {e x +d}\, \sqrt {c d x +a e}\, \sqrt {\left (a e g -c d f \right ) g}\, c d g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(arctanh((c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*c*d^2*g-arcta
nh((c*d*x+a*e)^(1/2)/((a*e*g-c*d*f)*g)^(1/2)*g)*c*d*e*f-(c*d*x+a*e)^(1/2)*e*((a*e*g-c*d*f)*g)^(1/2))/(e*x+d)^(
1/2)/(c*d*x+a*e)^(1/2)/c/d/g/((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 (e x + d\right )}^{\frac {3}{2}}}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (g x + f\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^{3/2}}{\left (f+g\,x\right )\,\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((d + e*x)^(3/2)/((f + g*x)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)),x)

[Out]

int((d + e*x)^(3/2)/((f + g*x)*(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 {\left (d + e x\right )^{\frac {3}{2}}}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )} \left (f + g x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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