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

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

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 365, normalized size of antiderivative = 1.00, 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 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)])

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 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 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]

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*}

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Mathematica [C]  time = 3.91, size = 549, normalized size = 1.50 \[ \frac {\sqrt {d+e x} \left (\frac {2 \left (a+c x^2\right ) (5 A e-4 B d+3 B e x)}{e^2}-\frac {4 \left (e^2 \left (a+c x^2\right ) \left (-\sqrt {-d-\frac {i \sqrt {a} e}{\sqrt {c}}}\right ) \left (3 a B e^2-5 A c d e+4 B c d^2\right )+\sqrt {c} (d+e x)^{3/2} \left (\sqrt {a} e-i \sqrt {c} d\right ) \sqrt {\frac {e \left (x+\frac {i \sqrt {a}}{\sqrt {c}}\right )}{d+e x}} \sqrt {-\frac {-e x+\frac {i \sqrt {a} e}{\sqrt {c}}}{d+e x}} \left (-3 a B e^2+5 A c d e-4 B c d^2\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {-d-\frac {i \sqrt {a} e}{\sqrt {c}}}}{\sqrt {d+e x}}\right )|\frac {\sqrt {c} d-i \sqrt {a} e}{\sqrt {c} d+i \sqrt {a} e}\right )+\sqrt {a} \sqrt {c} e (d+e x)^{3/2} \left (\sqrt {c} d+i \sqrt {a} e\right ) \sqrt {\frac {e \left (x+\frac {i \sqrt {a}}{\sqrt {c}}\right )}{d+e x}} \sqrt {-\frac {-e x+\frac {i \sqrt {a} e}{\sqrt {c}}}{d+e x}} \left (-3 i \sqrt {a} B e-5 A \sqrt {c} e+4 B \sqrt {c} d\right ) F\left (i \sinh ^{-1}\left (\frac {\sqrt {-d-\frac {i \sqrt {a} e}{\sqrt {c}}}}{\sqrt {d+e x}}\right )|\frac {\sqrt {c} d-i \sqrt {a} e}{\sqrt {c} d+i \sqrt {a} e}\right )\right )}{c e^4 (d+e x) \sqrt {-d-\frac {i \sqrt {a} e}{\sqrt {c}}}}\right )}{15 \sqrt {a+c x^2}} \]

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])

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fricas [F]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + a} {\left (B x + A\right )}}{\sqrt {e x + d}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{2} + a} {\left (B x + A\right )}}{\sqrt {e x + d}}\,{d x} \]

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)

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maple [B]  time = 0.11, size = 1826, normalized size = 5.00 \[ \text {result too large to display} \]

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*(-a*c)^(1/2)*(-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2)*((-c*x+(-a*c
)^(1/2))/(c*d+(-a*c)^(1/2)*e)*e)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*d+(-a*c)^(1/2)*e)*e)^(1/2)*EllipticF((-(e*x+d)/
(-c*d+(-a*c)^(1/2)*e)*c)^(1/2),(-(-c*d+(-a*c)^(1/2)*e)/(c*d+(-a*c)^(1/2)*e))^(1/2))*a*e^4+10*A*(-a*c)^(1/2)*(-
(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*d+(-a*c)^(1/2)*e)*e)^(1/2)*((c*x+(-a*c)^(1/2))/
(-c*d+(-a*c)^(1/2)*e)*e)^(1/2)*EllipticF((-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2),(-(-c*d+(-a*c)^(1/2)*e)/(c*d
+(-a*c)^(1/2)*e))^(1/2))*c*d^2*e^2-10*A*a*c*(-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*d
+(-a*c)^(1/2)*e)*e)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*d+(-a*c)^(1/2)*e)*e)^(1/2)*EllipticE((-(e*x+d)/(-c*d+(-a*c)^
(1/2)*e)*c)^(1/2),(-(-c*d+(-a*c)^(1/2)*e)/(c*d+(-a*c)^(1/2)*e))^(1/2))*d*e^3-10*A*(-(e*x+d)/(-c*d+(-a*c)^(1/2)
*e)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*d+(-a*c)^(1/2)*e)*e)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*d+(-a*c)^(1/2)*e)*e)^(
1/2)*EllipticE((-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2),(-(-c*d+(-a*c)^(1/2)*e)/(c*d+(-a*c)^(1/2)*e))^(1/2))*c
^2*d^3*e-8*B*(-a*c)^(1/2)*(-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*d+(-a*c)^(1/2)*e)*e
)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*d+(-a*c)^(1/2)*e)*e)^(1/2)*EllipticF((-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2),
(-(-c*d+(-a*c)^(1/2)*e)/(c*d+(-a*c)^(1/2)*e))^(1/2))*a*d*e^3-8*B*(-a*c)^(1/2)*(-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*
c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*d+(-a*c)^(1/2)*e)*e)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*d+(-a*c)^(1/2)*e)*e)^(1/2)
*EllipticF((-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2),(-(-c*d+(-a*c)^(1/2)*e)/(c*d+(-a*c)^(1/2)*e))^(1/2))*c*d^3
*e-6*B*a^2*(-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*d+(-a*c)^(1/2)*e)*e)^(1/2)*((c*x+(
-a*c)^(1/2))/(-c*d+(-a*c)^(1/2)*e)*e)^(1/2)*EllipticF((-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2),(-(-c*d+(-a*c)^
(1/2)*e)/(c*d+(-a*c)^(1/2)*e))^(1/2))*e^4-6*B*(-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c
*d+(-a*c)^(1/2)*e)*e)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*d+(-a*c)^(1/2)*e)*e)^(1/2)*EllipticF((-(e*x+d)/(-c*d+(-a*c
)^(1/2)*e)*c)^(1/2),(-(-c*d+(-a*c)^(1/2)*e)/(c*d+(-a*c)^(1/2)*e))^(1/2))*a*c*d^2*e^2+6*B*a^2*(-(e*x+d)/(-c*d+(
-a*c)^(1/2)*e)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*d+(-a*c)^(1/2)*e)*e)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*d+(-a*c)^(1
/2)*e)*e)^(1/2)*EllipticE((-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2),(-(-c*d+(-a*c)^(1/2)*e)/(c*d+(-a*c)^(1/2)*e
))^(1/2))*e^4+14*B*(-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*d+(-a*c)^(1/2)*e)*e)^(1/2)
*((c*x+(-a*c)^(1/2))/(-c*d+(-a*c)^(1/2)*e)*e)^(1/2)*EllipticE((-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2),(-(-c*d
+(-a*c)^(1/2)*e)/(c*d+(-a*c)^(1/2)*e))^(1/2))*a*c*d^2*e^2+8*B*(-(e*x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2)*((-c*x+
(-a*c)^(1/2))/(c*d+(-a*c)^(1/2)*e)*e)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*d+(-a*c)^(1/2)*e)*e)^(1/2)*EllipticE((-(e*
x+d)/(-c*d+(-a*c)^(1/2)*e)*c)^(1/2),(-(-c*d+(-a*c)^(1/2)*e)/(c*d+(-a*c)^(1/2)*e))^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{2} + a} {\left (B x + A\right )}}{\sqrt {e x + d}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {c\,x^2+a}\,\left (A+B\,x\right )}{\sqrt {d+e\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \sqrt {a + c x^{2}}}{\sqrt {d + e x}}\, dx \]

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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