3.1473 \(\int \frac{(A+B x) \sqrt{a+c x^2}}{\sqrt{d+e x}} \, dx\)

Optimal. Leaf size=365 \[ \frac{4 \sqrt{-a} \sqrt{\frac{c x^2}{a}+1} \left (a e^2+c d^2\right ) (4 B d-5 A e) \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right ),-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{15 \sqrt{c} e^3 \sqrt{a+c x^2} \sqrt{d+e x}}-\frac{4 \sqrt{-a} \sqrt{\frac{c x^2}{a}+1} \sqrt{d+e x} \left (3 a B e^2-5 A c d e+4 B c d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{15 \sqrt{c} e^3 \sqrt{a+c x^2} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}}}-\frac{2 \sqrt{a+c x^2} \sqrt{d+e x} (-5 A e+4 B d-3 B e x)}{15 e^2} \]

[Out]

(-2*Sqrt[d + e*x]*(4*B*d - 5*A*e - 3*B*e*x)*Sqrt[a + c*x^2])/(15*e^2) - (4*Sqrt[-a]*(4*B*c*d^2 - 5*A*c*d*e + 3
*a*B*e^2)*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)])/(15*Sqrt[c]*e^3*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]*Sqrt[a + c*x^
2]) + (4*Sqrt[-a]*(4*B*d - 5*A*e)*(c*d^2 + a*e^2)*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]*Sqrt[1 +
(c*x^2)/a]*EllipticF[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*e)/(Sqrt[-a]*Sqrt[c]*d - a*e)])/(15
*Sqrt[c]*e^3*Sqrt[d + e*x]*Sqrt[a + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.336735, antiderivative size = 365, 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 = {815, 844, 719, 424, 419} \[ \frac{4 \sqrt{-a} \sqrt{\frac{c x^2}{a}+1} \left (a e^2+c d^2\right ) (4 B d-5 A e) \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}} F\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{15 \sqrt{c} e^3 \sqrt{a+c x^2} \sqrt{d+e x}}-\frac{4 \sqrt{-a} \sqrt{\frac{c x^2}{a}+1} \sqrt{d+e x} \left (3 a B e^2-5 A c d e+4 B c d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{15 \sqrt{c} e^3 \sqrt{a+c x^2} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}}}-\frac{2 \sqrt{a+c x^2} \sqrt{d+e x} (-5 A e+4 B d-3 B e x)}{15 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x]*(4*B*d - 5*A*e - 3*B*e*x)*Sqrt[a + c*x^2])/(15*e^2) - (4*Sqrt[-a]*(4*B*c*d^2 - 5*A*c*d*e + 3
*a*B*e^2)*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)])/(15*Sqrt[c]*e^3*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]*Sqrt[a + c*x^
2]) + (4*Sqrt[-a]*(4*B*d - 5*A*e)*(c*d^2 + a*e^2)*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)]*Sqrt[1 +
(c*x^2)/a]*EllipticF[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*e)/(Sqrt[-a]*Sqrt[c]*d - a*e)])/(15
*Sqrt[c]*e^3*Sqrt[d + e*x]*Sqrt[a + c*x^2])

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 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}}{\sqrt{d+e x}} \, dx &=-\frac{2 \sqrt{d+e x} (4 B d-5 A e-3 B e x) \sqrt{a+c x^2}}{15 e^2}+\frac{4 \int \frac{-\frac{1}{2} a c e (B d-5 A e)+\frac{1}{2} c \left (4 B c d^2-5 A c d e+3 a B e^2\right ) x}{\sqrt{d+e x} \sqrt{a+c x^2}} \, dx}{15 c e^2}\\ &=-\frac{2 \sqrt{d+e x} (4 B d-5 A e-3 B e x) \sqrt{a+c x^2}}{15 e^2}-\frac{\left (2 (4 B d-5 A e) \left (c d^2+a e^2\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{a+c x^2}} \, dx}{15 e^3}+\frac{\left (2 \left (4 B c d^2-5 A c d e+3 a B e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{a+c x^2}} \, dx}{15 e^3}\\ &=-\frac{2 \sqrt{d+e x} (4 B d-5 A e-3 B e x) \sqrt{a+c x^2}}{15 e^2}+\frac{\left (4 a \left (4 B c d^2-5 A c d e+3 a B e^2\right ) \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 )}{15 \sqrt{-a} \sqrt{c} 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 (4 B d-5 A e) \left (c d^2+a 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 )}{15 \sqrt{-a} \sqrt{c} e^3 \sqrt{d+e x} \sqrt{a+c x^2}}\\ &=-\frac{2 \sqrt{d+e x} (4 B d-5 A e-3 B e x) \sqrt{a+c x^2}}{15 e^2}-\frac{4 \sqrt{-a} \left (4 B c d^2-5 A c d e+3 a B e^2\right ) \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 )}{15 \sqrt{c} e^3 \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}} \sqrt{a+c x^2}}+\frac{4 \sqrt{-a} (4 B d-5 A e) \left (c d^2+a 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 )}{15 \sqrt{c} e^3 \sqrt{d+e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 4.05271, size = 549, normalized size = 1.5 \[ \frac{\sqrt{d+e x} \left (\frac{2 \left (a+c x^2\right ) (5 A e-4 B d+3 B e x)}{e^2}-\frac{4 \left (\sqrt{a} \sqrt{c} e (d+e x)^{3/2} \left (\sqrt{c} d+i \sqrt{a} e\right ) \sqrt{\frac{e \left (x+\frac{i \sqrt{a}}{\sqrt{c}}\right )}{d+e x}} \sqrt{-\frac{-e x+\frac{i \sqrt{a} e}{\sqrt{c}}}{d+e x}} \left (-3 i \sqrt{a} B e-5 A \sqrt{c} e+4 B \sqrt{c} d\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}{\sqrt{d+e x}}\right ),\frac{\sqrt{c} d-i \sqrt{a} e}{\sqrt{c} d+i \sqrt{a} e}\right )+e^2 \left (a+c x^2\right ) \left (-\sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}\right ) \left (3 a B e^2-5 A c d e+4 B c d^2\right )+\sqrt{c} (d+e x)^{3/2} \left (\sqrt{a} e-i \sqrt{c} d\right ) \sqrt{\frac{e \left (x+\frac{i \sqrt{a}}{\sqrt{c}}\right )}{d+e x}} \sqrt{-\frac{-e x+\frac{i \sqrt{a} e}{\sqrt{c}}}{d+e x}} \left (-3 a B e^2+5 A c d e-4 B c d^2\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}{\sqrt{d+e x}}\right )|\frac{\sqrt{c} d-i \sqrt{a} e}{\sqrt{c} d+i \sqrt{a} e}\right )\right )}{c e^4 (d+e x) \sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}\right )}{15 \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*((2*(-4*B*d + 5*A*e + 3*B*e*x)*(a + c*x^2))/e^2 - (4*(-(e^2*Sqrt[-d - (I*Sqrt[a]*e)/Sqrt[c]]*(4
*B*c*d^2 - 5*A*c*d*e + 3*a*B*e^2)*(a + c*x^2)) + Sqrt[c]*((-I)*Sqrt[c]*d + Sqrt[a]*e)*(-4*B*c*d^2 + 5*A*c*d*e
- 3*a*B*e^2)*Sqrt[(e*((I*Sqrt[a])/Sqrt[c] + x))/(d + e*x)]*Sqrt[-(((I*Sqrt[a]*e)/Sqrt[c] - e*x)/(d + e*x))]*(d
+ e*x)^(3/2)*EllipticE[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[a]*Sqrt[c]*e*(Sqrt[c]*d + I*Sqrt[a]*e)*(4*B*Sqrt[c]*d - (3*I)*Sqrt[a]*B*e - 5
*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))]*(
d + e*x)^(3/2)*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)]))/(c*e^4*Sqrt[-d - (I*Sqrt[a]*e)/Sqrt[c]]*(d + e*x))))/(15*Sqrt[a + c*x^2])

________________________________________________________________________________________

Maple [B]  time = 0.034, size = 1826, normalized size = 5. \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)^(1/2),x)

[Out]

-2/15*(c*x^2+a)^(1/2)*(e*x+d)^(1/2)/c*(10*A*(-(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)*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)^(1/2)*a*e^4+10*A*(-(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)*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)^(1/2)*c*d^2*e^2-10*A*(-(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)*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*d*e^3-10*A*(-(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)*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-8*B*(-(
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)*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)^(1/2)*a*d*e^3-8*B*(-(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)*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)^(1/2)*c*d^3*e-6*B*(-(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)*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^2*e^4-6*B*(-(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)*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^2*e^2+6*B*(-(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)*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^2*e^4+14*B*(-(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)*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*d^2*e^2+8*B*(-(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)*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^4-3*B*x^4*c^2*e^4-5*A*x^3*c^2*e^4+B*x^3*c^2*d*e^3-5*A*x^2*c
^2*d*e^3-3*B*x^2*a*c*e^4+4*B*x^2*c^2*d^2*e^2-5*A*x*a*c*e^4+B*x*a*c*d*e^3-5*A*a*c*d*e^3+4*B*a*c*d^2*e^2)/(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 )}}{\sqrt{e x + d}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)/sqrt(e*x + d), 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}}}{\sqrt{d + e x}}\, 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)**(1/2),x)

[Out]

Integral((A + B*x)*sqrt(a + c*x**2)/sqrt(d + e*x), 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 )}}{\sqrt{e x + d}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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