3.1474 \(\int \frac{(A+B x) \sqrt{a+c x^2}}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=352 \[ -\frac{4 \sqrt{-a} \sqrt{\frac{c x^2}{a}+1} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}} \left (a B e^2-3 A c d e+4 B c d^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right ),-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 \sqrt{c} e^3 \sqrt{a+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{a+c x^2} (-3 A e+4 B d+B e x)}{3 e^2 \sqrt{d+e x}}+\frac{4 \sqrt{-a} \sqrt{c} \sqrt{\frac{c x^2}{a}+1} \sqrt{d+e x} (4 B d-3 A e) E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 e^3 \sqrt{a+c x^2} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}}} \]

[Out]

(2*(4*B*d - 3*A*e + B*e*x)*Sqrt[a + c*x^2])/(3*e^2*Sqrt[d + e*x]) + (4*Sqrt[-a]*Sqrt[c]*(4*B*d - 3*A*e)*Sqrt[d
+ e*x]*Sqrt[1 + (c*x^2)/a]*EllipticE[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*e)/(Sqrt[-a]*Sqrt[
c]*d - a*e)])/(3*e^3*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]*Sqrt[a + c*x^2]) - (4*Sqrt[-a]*(4*B*c*
d^2 - 3*A*c*d*e + a*B*e^2)*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]*Sqrt[1 + (c*x^2)/a]*EllipticF[Ar
cSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*e)/(Sqrt[-a]*Sqrt[c]*d - a*e)])/(3*Sqrt[c]*e^3*Sqrt[d + e*
x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.252054, antiderivative size = 352, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {813, 844, 719, 424, 419} \[ -\frac{4 \sqrt{-a} \sqrt{\frac{c x^2}{a}+1} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}} \left (a B e^2-3 A c d e+4 B c d^2\right ) F\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 \sqrt{c} e^3 \sqrt{a+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{a+c x^2} (-3 A e+4 B d+B e x)}{3 e^2 \sqrt{d+e x}}+\frac{4 \sqrt{-a} \sqrt{c} \sqrt{\frac{c x^2}{a}+1} \sqrt{d+e x} (4 B d-3 A e) E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 e^3 \sqrt{a+c x^2} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(4*B*d - 3*A*e + B*e*x)*Sqrt[a + c*x^2])/(3*e^2*Sqrt[d + e*x]) + (4*Sqrt[-a]*Sqrt[c]*(4*B*d - 3*A*e)*Sqrt[d
+ e*x]*Sqrt[1 + (c*x^2)/a]*EllipticE[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*e)/(Sqrt[-a]*Sqrt[
c]*d - a*e)])/(3*e^3*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]*Sqrt[a + c*x^2]) - (4*Sqrt[-a]*(4*B*c*
d^2 - 3*A*c*d*e + a*B*e^2)*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]*Sqrt[1 + (c*x^2)/a]*EllipticF[Ar
cSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*e)/(Sqrt[-a]*Sqrt[c]*d - a*e)])/(3*Sqrt[c]*e^3*Sqrt[d + e*
x]*Sqrt[a + c*x^2])

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 719

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*a*Rt[-(c/a), 2]*(d + e*x)^m*Sqrt[
1 + (c*x^2)/a])/(c*Sqrt[a + c*x^2]*((c*(d + e*x))/(c*d - a*e*Rt[-(c/a), 2]))^m), Subst[Int[(1 + (2*a*e*Rt[-(c/
a), 2]*x^2)/(c*d - a*e*Rt[-(c/a), 2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(1 - Rt[-(c/a), 2]*x)/2]], x] /; FreeQ[{a,
c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m^2, 1/4]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rubi steps

\begin{align*} \int \frac{(A+B x) \sqrt{a+c x^2}}{(d+e x)^{3/2}} \, dx &=\frac{2 (4 B d-3 A e+B e x) \sqrt{a+c x^2}}{3 e^2 \sqrt{d+e x}}-\frac{2 \int \frac{-a B e+c (4 B d-3 A e) x}{\sqrt{d+e x} \sqrt{a+c x^2}} \, dx}{3 e^2}\\ &=\frac{2 (4 B d-3 A e+B e x) \sqrt{a+c x^2}}{3 e^2 \sqrt{d+e x}}-\frac{(2 c (4 B d-3 A e)) \int \frac{\sqrt{d+e x}}{\sqrt{a+c x^2}} \, dx}{3 e^3}+\frac{\left (2 \left (4 B c d^2-3 A c d e+a B e^2\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{a+c x^2}} \, dx}{3 e^3}\\ &=\frac{2 (4 B d-3 A e+B e x) \sqrt{a+c x^2}}{3 e^2 \sqrt{d+e x}}-\frac{\left (4 a \sqrt{c} (4 B d-3 A e) \sqrt{d+e x} \sqrt{1+\frac{c x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{2 a \sqrt{c} e x^2}{\sqrt{-a} \left (c d-\frac{a \sqrt{c} e}{\sqrt{-a}}\right )}}}{\sqrt{1-x^2}} \, dx,x,\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )}{3 \sqrt{-a} e^3 \sqrt{\frac{c (d+e x)}{c d-\frac{a \sqrt{c} e}{\sqrt{-a}}}} \sqrt{a+c x^2}}+\frac{\left (4 a \left (4 B c d^2-3 A c d e+a B e^2\right ) \sqrt{\frac{c (d+e x)}{c d-\frac{a \sqrt{c} e}{\sqrt{-a}}}} \sqrt{1+\frac{c x^2}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+\frac{2 a \sqrt{c} e x^2}{\sqrt{-a} \left (c d-\frac{a \sqrt{c} e}{\sqrt{-a}}\right )}}} \, dx,x,\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )}{3 \sqrt{-a} \sqrt{c} e^3 \sqrt{d+e x} \sqrt{a+c x^2}}\\ &=\frac{2 (4 B d-3 A e+B e x) \sqrt{a+c x^2}}{3 e^2 \sqrt{d+e x}}+\frac{4 \sqrt{-a} \sqrt{c} (4 B d-3 A e) \sqrt{d+e x} \sqrt{1+\frac{c x^2}{a}} E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 e^3 \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}} \sqrt{a+c x^2}}-\frac{4 \sqrt{-a} \left (4 B c d^2-3 A c d e+a B e^2\right ) \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}} \sqrt{1+\frac{c x^2}{a}} F\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{3 \sqrt{c} e^3 \sqrt{d+e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 3.46145, size = 512, normalized size = 1.45 \[ \frac{\sqrt{d+e x} \left (\frac{2 \left (a+c x^2\right ) (-3 A e+4 B d+B e x)}{e^2 (d+e x)}+\frac{2 (d+e x) \left (-\frac{2 \sqrt{a} e \sqrt{\frac{e \left (x+\frac{i \sqrt{a}}{\sqrt{c}}\right )}{d+e x}} \sqrt{-\frac{-e x+\frac{i \sqrt{a} e}{\sqrt{c}}}{d+e x}} \left (-i \sqrt{a} B e+3 A \sqrt{c} e-4 B \sqrt{c} d\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}{\sqrt{d+e x}}\right ),\frac{\sqrt{c} d-i \sqrt{a} e}{\sqrt{c} d+i \sqrt{a} e}\right )}{\sqrt{d+e x}}+\frac{2 e^2 \left (a+c x^2\right ) \sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}} (3 A e-4 B d)}{(d+e x)^2}+\frac{2 \sqrt{c} \left (\sqrt{a} e-i \sqrt{c} d\right ) (3 A e-4 B d) \sqrt{\frac{e \left (x+\frac{i \sqrt{a}}{\sqrt{c}}\right )}{d+e x}} \sqrt{-\frac{-e x+\frac{i \sqrt{a} e}{\sqrt{c}}}{d+e x}} E\left (i \sinh ^{-1}\left (\frac{\sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}{\sqrt{d+e x}}\right )|\frac{\sqrt{c} d-i \sqrt{a} e}{\sqrt{c} d+i \sqrt{a} e}\right )}{\sqrt{d+e x}}\right )}{e^4 \sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}\right )}{3 \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*((2*(4*B*d - 3*A*e + B*e*x)*(a + c*x^2))/(e^2*(d + e*x)) + (2*(d + e*x)*((2*e^2*(-4*B*d + 3*A*e
)*Sqrt[-d - (I*Sqrt[a]*e)/Sqrt[c]]*(a + c*x^2))/(d + e*x)^2 + (2*Sqrt[c]*((-I)*Sqrt[c]*d + Sqrt[a]*e)*(-4*B*d
+ 3*A*e)*Sqrt[(e*((I*Sqrt[a])/Sqrt[c] + x))/(d + e*x)]*Sqrt[-(((I*Sqrt[a]*e)/Sqrt[c] - e*x)/(d + e*x))]*Ellipt
icE[I*ArcSinh[Sqrt[-d - (I*Sqrt[a]*e)/Sqrt[c]]/Sqrt[d + e*x]], (Sqrt[c]*d - I*Sqrt[a]*e)/(Sqrt[c]*d + I*Sqrt[a
]*e)])/Sqrt[d + e*x] - (2*Sqrt[a]*e*(-4*B*Sqrt[c]*d - I*Sqrt[a]*B*e + 3*A*Sqrt[c]*e)*Sqrt[(e*((I*Sqrt[a])/Sqrt
[c] + x))/(d + e*x)]*Sqrt[-(((I*Sqrt[a]*e)/Sqrt[c] - e*x)/(d + e*x))]*EllipticF[I*ArcSinh[Sqrt[-d - (I*Sqrt[a]
*e)/Sqrt[c]]/Sqrt[d + e*x]], (Sqrt[c]*d - I*Sqrt[a]*e)/(Sqrt[c]*d + I*Sqrt[a]*e)])/Sqrt[d + e*x]))/(e^4*Sqrt[-
d - (I*Sqrt[a]*e)/Sqrt[c]])))/(3*Sqrt[a + c*x^2])

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Maple [B]  time = 0.049, size = 1463, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(c*x^2+a)^(1/2)*(e*x+d)^(1/2)*(6*A*EllipticF((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d
)/((-a*c)^(1/2)*e+c*d))^(1/2))*a*c*e^3*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^
(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)+6*A*EllipticF((-(e*x+d)*c/((-a*c)^(1/2)*
e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*c*d*e^2*(-a*c)^(1/2)*(-(e*x+d)*c/((-a*c)^(1/
2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d)
)^(1/2)-6*A*EllipticE((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/
2))*a*c*e^3*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(
-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)-6*A*EllipticE((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/
2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*c^2*d^2*e*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*
e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)-6*B*EllipticF((-(e*x+d)*c/((-a
*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*a*c*d*e^2*(-(e*x+d)*c/((-a*c)^(1/2
)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))
^(1/2)-2*B*EllipticF((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2
))*a*e^3*(-a*c)^(1/2)*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/
2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)-8*B*EllipticF((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-(
(-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*c*d^2*e*(-a*c)^(1/2)*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*
((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)+8*B*Ellip
ticE((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*c)^(1/2)*e+c*d))^(1/2))*a*c*d*e^2*(-(
e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/
((-a*c)^(1/2)*e-c*d))^(1/2)+8*B*EllipticE((-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2),(-((-a*c)^(1/2)*e-c*d)/((-a*
c)^(1/2)*e+c*d))^(1/2))*c^2*d^3*(-(e*x+d)*c/((-a*c)^(1/2)*e-c*d))^(1/2)*((-c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e
+c*d))^(1/2)*((c*x+(-a*c)^(1/2))*e/((-a*c)^(1/2)*e-c*d))^(1/2)+B*x^3*c^2*e^3-3*A*x^2*c^2*e^3+4*B*x^2*c^2*d*e^2
+B*x*a*c*e^3-3*a*A*e^3*c+4*a*B*d*e^2*c)/c/(c*e*x^3+c*d*x^2+a*e*x+a*d)/e^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + a}{\left (B x + A\right )}}{{\left (e x + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + a}{\left (B x + A\right )} \sqrt{e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)*sqrt(e*x + d)/(e^2*x^2 + 2*d*e*x + d^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \sqrt{a + c x^{2}}}{\left (d + e x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(1/2)/(e*x+d)**(3/2),x)

[Out]

Integral((A + B*x)*sqrt(a + c*x**2)/(d + e*x)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + a}{\left (B x + A\right )}}{{\left (e x + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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