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

Optimal. Leaf size=364 \[ \frac {4 \sqrt {-a} \sqrt {\frac {c x^2}{a}+1} \left (a g^2+c f^2\right ) (4 e f-5 d g) \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {-a} g+\sqrt {c} f}} F\left (\sin ^{-1}\left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )|-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 \sqrt {c} g^3 \sqrt {a+c x^2} \sqrt {f+g x}}-\frac {2 \sqrt {a+c x^2} \sqrt {f+g x} (-5 d g+4 e f-3 e g x)}{15 g^2}-\frac {4 \sqrt {-a} \sqrt {\frac {c x^2}{a}+1} \sqrt {f+g x} \left (3 a e g^2+c f (4 e f-5 d g)\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )|-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 \sqrt {c} g^3 \sqrt {a+c x^2} \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {-a} g+\sqrt {c} f}}} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 364, 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} \[ \frac {4 \sqrt {-a} \sqrt {\frac {c x^2}{a}+1} \left (a g^2+c f^2\right ) (4 e f-5 d g) \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {-a} g+\sqrt {c} f}} F\left (\sin ^{-1}\left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )|-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 \sqrt {c} g^3 \sqrt {a+c x^2} \sqrt {f+g x}}-\frac {2 \sqrt {a+c x^2} \sqrt {f+g x} (-5 d g+4 e f-3 e g x)}{15 g^2}-\frac {4 \sqrt {-a} \sqrt {\frac {c x^2}{a}+1} \sqrt {f+g x} \left (3 a e g^2+c f (4 e f-5 d g)\right ) E\left (\sin ^{-1}\left (\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )|-\frac {2 a g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 \sqrt {c} g^3 \sqrt {a+c x^2} \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {-a} g+\sqrt {c} f}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[f + g*x]*(4*e*f - 5*d*g - 3*e*g*x)*Sqrt[a + c*x^2])/(15*g^2) - (4*Sqrt[-a]*(3*a*e*g^2 + c*f*(4*e*f -
5*d*g))*Sqrt[f + g*x]*Sqrt[1 + (c*x^2)/a]*EllipticE[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*g)/(
Sqrt[-a]*Sqrt[c]*f - a*g)])/(15*Sqrt[c]*g^3*Sqrt[(Sqrt[c]*(f + g*x))/(Sqrt[c]*f + Sqrt[-a]*g)]*Sqrt[a + c*x^2]
) + (4*Sqrt[-a]*(4*e*f - 5*d*g)*(c*f^2 + a*g^2)*Sqrt[(Sqrt[c]*(f + g*x))/(Sqrt[c]*f + Sqrt[-a]*g)]*Sqrt[1 + (c
*x^2)/a]*EllipticF[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*g)/(Sqrt[-a]*Sqrt[c]*f - a*g)])/(15*S
qrt[c]*g^3*Sqrt[f + g*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 {(d+e x) \sqrt {a+c x^2}}{\sqrt {f+g x}} \, dx &=-\frac {2 \sqrt {f+g x} (4 e f-5 d g-3 e g x) \sqrt {a+c x^2}}{15 g^2}+\frac {4 \int \frac {-\frac {1}{2} a c g (e f-5 d g)+\frac {1}{2} c \left (3 a e g^2+c f (4 e f-5 d g)\right ) x}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx}{15 c g^2}\\ &=-\frac {2 \sqrt {f+g x} (4 e f-5 d g-3 e g x) \sqrt {a+c x^2}}{15 g^2}-\frac {\left (2 (4 e f-5 d g) \left (c f^2+a g^2\right )\right ) \int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx}{15 g^3}+\frac {\left (2 \left (3 a e g^2+c f (4 e f-5 d g)\right )\right ) \int \frac {\sqrt {f+g x}}{\sqrt {a+c x^2}} \, dx}{15 g^3}\\ &=-\frac {2 \sqrt {f+g x} (4 e f-5 d g-3 e g x) \sqrt {a+c x^2}}{15 g^2}+\frac {\left (4 a \left (3 a e g^2+c f (4 e f-5 d g)\right ) \sqrt {f+g x} \sqrt {1+\frac {c x^2}{a}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {2 a \sqrt {c} g x^2}{\sqrt {-a} \left (c f-\frac {a \sqrt {c} g}{\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} g^3 \sqrt {\frac {c (f+g x)}{c f-\frac {a \sqrt {c} g}{\sqrt {-a}}}} \sqrt {a+c x^2}}-\frac {\left (4 a (4 e f-5 d g) \left (c f^2+a g^2\right ) \sqrt {\frac {c (f+g x)}{c f-\frac {a \sqrt {c} g}{\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} g x^2}{\sqrt {-a} \left (c f-\frac {a \sqrt {c} g}{\sqrt {-a}}\right )}}} \, dx,x,\frac {\sqrt {1-\frac {\sqrt {c} x}{\sqrt {-a}}}}{\sqrt {2}}\right )}{15 \sqrt {-a} \sqrt {c} g^3 \sqrt {f+g x} \sqrt {a+c x^2}}\\ &=-\frac {2 \sqrt {f+g x} (4 e f-5 d g-3 e g x) \sqrt {a+c x^2}}{15 g^2}-\frac {4 \sqrt {-a} \left (3 a e g^2+c f (4 e f-5 d g)\right ) \sqrt {f+g 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 g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 \sqrt {c} g^3 \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {c} f+\sqrt {-a} g}} \sqrt {a+c x^2}}+\frac {4 \sqrt {-a} (4 e f-5 d g) \left (c f^2+a g^2\right ) \sqrt {\frac {\sqrt {c} (f+g x)}{\sqrt {c} f+\sqrt {-a} g}} \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 g}{\sqrt {-a} \sqrt {c} f-a g}\right )}{15 \sqrt {c} g^3 \sqrt {f+g x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 4.83, size = 545, normalized size = 1.50 \[ \frac {\sqrt {f+g x} \left (\frac {2 \left (a+c x^2\right ) (5 d g-4 e f+3 e g x)}{g^2}+\frac {4 \left (g^2 \left (a+c x^2\right ) \sqrt {-f-\frac {i \sqrt {a} g}{\sqrt {c}}} \left (3 a e g^2+c f (4 e f-5 d g)\right )-\sqrt {c} (f+g x)^{3/2} \left (-\sqrt {a} g+i \sqrt {c} f\right ) \sqrt {\frac {g \left (x+\frac {i \sqrt {a}}{\sqrt {c}}\right )}{f+g x}} \sqrt {-\frac {-g x+\frac {i \sqrt {a} g}{\sqrt {c}}}{f+g x}} \left (3 a e g^2+c f (4 e f-5 d g)\right ) E\left (i \sinh ^{-1}\left (\frac {\sqrt {-f-\frac {i \sqrt {a} g}{\sqrt {c}}}}{\sqrt {f+g x}}\right )|\frac {\sqrt {c} f-i \sqrt {a} g}{\sqrt {c} f+i \sqrt {a} g}\right )+\sqrt {a} \sqrt {c} g (f+g x)^{3/2} \left (\sqrt {c} f+i \sqrt {a} g\right ) \sqrt {\frac {g \left (x+\frac {i \sqrt {a}}{\sqrt {c}}\right )}{f+g x}} \sqrt {-\frac {-g x+\frac {i \sqrt {a} g}{\sqrt {c}}}{f+g x}} \left (\sqrt {c} (5 d g-4 e f)+3 i \sqrt {a} e g\right ) F\left (i \sinh ^{-1}\left (\frac {\sqrt {-f-\frac {i \sqrt {a} g}{\sqrt {c}}}}{\sqrt {f+g x}}\right )|\frac {\sqrt {c} f-i \sqrt {a} g}{\sqrt {c} f+i \sqrt {a} g}\right )\right )}{c g^4 (f+g x) \sqrt {-f-\frac {i \sqrt {a} g}{\sqrt {c}}}}\right )}{15 \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[f + g*x]*((2*(-4*e*f + 5*d*g + 3*e*g*x)*(a + c*x^2))/g^2 + (4*(g^2*Sqrt[-f - (I*Sqrt[a]*g)/Sqrt[c]]*(3*a
*e*g^2 + c*f*(4*e*f - 5*d*g))*(a + c*x^2) - Sqrt[c]*(I*Sqrt[c]*f - Sqrt[a]*g)*(3*a*e*g^2 + c*f*(4*e*f - 5*d*g)
)*Sqrt[(g*((I*Sqrt[a])/Sqrt[c] + x))/(f + g*x)]*Sqrt[-(((I*Sqrt[a]*g)/Sqrt[c] - g*x)/(f + g*x))]*(f + g*x)^(3/
2)*EllipticE[I*ArcSinh[Sqrt[-f - (I*Sqrt[a]*g)/Sqrt[c]]/Sqrt[f + g*x]], (Sqrt[c]*f - I*Sqrt[a]*g)/(Sqrt[c]*f +
I*Sqrt[a]*g)] + Sqrt[a]*Sqrt[c]*g*(Sqrt[c]*f + I*Sqrt[a]*g)*((3*I)*Sqrt[a]*e*g + Sqrt[c]*(-4*e*f + 5*d*g))*Sq
rt[(g*((I*Sqrt[a])/Sqrt[c] + x))/(f + g*x)]*Sqrt[-(((I*Sqrt[a]*g)/Sqrt[c] - g*x)/(f + g*x))]*(f + g*x)^(3/2)*E
llipticF[I*ArcSinh[Sqrt[-f - (I*Sqrt[a]*g)/Sqrt[c]]/Sqrt[f + g*x]], (Sqrt[c]*f - I*Sqrt[a]*g)/(Sqrt[c]*f + I*S
qrt[a]*g)]))/(c*g^4*Sqrt[-f - (I*Sqrt[a]*g)/Sqrt[c]]*(f + g*x))))/(15*Sqrt[a + c*x^2])

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + a} {\left (e x + d\right )}}{\sqrt {g x + f}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 1828, normalized size = 5.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(c*x^2+a)^(1/2)*(g*x+f)^(1/2)*(6*(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*f+(-a*c
)^(1/2)*g)*g)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*EllipticF((-(g*x+f)/(-c*f+(-a*c)^(1/2)*
g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*a^2*e*g^4+6*(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c
)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*
EllipticF((-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*a*c*e*
f^2*g^2-10*(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((c*x+(
-a*c)^(1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*EllipticF((-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^
(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*(-a*c)^(1/2)*a*d*g^4+8*(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2)*((-c*x+(
-a*c)^(1/2))/(c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*EllipticF((-(g*x
+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*(-a*c)^(1/2)*a*e*f*g^3
-10*(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((c*x+(-a*c)^(
1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*EllipticF((-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g
)/(c*f+(-a*c)^(1/2)*g))^(1/2))*(-a*c)^(1/2)*c*d*f^2*g^2+8*(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2)*((-c*x+(-a*
c)^(1/2))/(c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*EllipticF((-(g*x+f)
/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*(-a*c)^(1/2)*c*e*f^3*g-6*
(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((c*x+(-a*c)^(1/2)
)/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*EllipticE((-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c
*f+(-a*c)^(1/2)*g))^(1/2))*a^2*e*g^4+10*(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*f+(-a
*c)^(1/2)*g)*g)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*EllipticE((-(g*x+f)/(-c*f+(-a*c)^(1/2
)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*a*c*d*f*g^3-14*(-(g*x+f)/(-c*f+(-a*c)^(1/2)
*g)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(
1/2)*EllipticE((-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*a
*c*e*f^2*g^2+10*(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2)*((-c*x+(-a*c)^(1/2))/(c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((
c*x+(-a*c)^(1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*EllipticE((-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-
a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*c^2*d*f^3*g-8*(-(g*x+f)/(-c*f+(-a*c)^(1/2)*g)*c)^(1/2)*((-c*x+(-a*c
)^(1/2))/(c*f+(-a*c)^(1/2)*g)*g)^(1/2)*((c*x+(-a*c)^(1/2))/(-c*f+(-a*c)^(1/2)*g)*g)^(1/2)*EllipticE((-(g*x+f)/
(-c*f+(-a*c)^(1/2)*g)*c)^(1/2),(-(-c*f+(-a*c)^(1/2)*g)/(c*f+(-a*c)^(1/2)*g))^(1/2))*c^2*e*f^4+3*x^4*c^2*e*g^4+
5*x^3*c^2*d*g^4-x^3*c^2*e*f*g^3+3*x^2*a*c*e*g^4+5*x^2*c^2*d*f*g^3-4*x^2*c^2*e*f^2*g^2+5*x*a*c*d*g^4-x*a*c*e*f*
g^3+5*a*c*d*f*g^3-4*a*c*e*f^2*g^2)/c/(c*g*x^3+c*f*x^2+a*g*x+a*f)/g^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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