3.625 \(\int \sqrt{f+g x} \sqrt{a+c x^2} \, dx\)

Optimal. Leaf size=362 \[ -\frac{4 \sqrt{-a} f \sqrt{\frac{c x^2}{a}+1} \left (a g^2+c f^2\right ) \sqrt{\frac{\sqrt{c} (f+g x)}{\sqrt{-a} g+\sqrt{c} f}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right ),-\frac{2 a g}{\sqrt{-a} \sqrt{c} f-a g}\right )}{15 \sqrt{c} g^2 \sqrt{a+c x^2} \sqrt{f+g x}}+\frac{4 \sqrt{-a} \sqrt{\frac{c x^2}{a}+1} \sqrt{f+g x} \left (c f^2-3 a g^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a g}{\sqrt{-a} \sqrt{c} f-a g}\right )}{15 \sqrt{c} g^2 \sqrt{a+c x^2} \sqrt{\frac{\sqrt{c} (f+g x)}{\sqrt{-a} g+\sqrt{c} f}}}+\frac{2 \sqrt{a+c x^2} (f+g x)^{3/2}}{5 g}-\frac{4 f \sqrt{a+c x^2} \sqrt{f+g x}}{15 g} \]

[Out]

(-4*f*Sqrt[f + g*x]*Sqrt[a + c*x^2])/(15*g) + (2*(f + g*x)^(3/2)*Sqrt[a + c*x^2])/(5*g) + (4*Sqrt[-a]*(c*f^2 -
3*a*g^2)*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^2*Sqrt[(Sqrt[c]*(f + g*x))/(Sqrt[c]*f + Sqrt[-a]*g)]*Sqrt[a + c*x^
2]) - (4*Sqrt[-a]*f*(c*f^2 + a*g^2)*Sqrt[(Sqrt[c]*(f + g*x))/(Sqrt[c]*f + Sqrt[-a]*g)]*Sqrt[1 + (c*x^2)/a]*Ell
ipticF[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*g)/(Sqrt[-a]*Sqrt[c]*f - a*g)])/(15*Sqrt[c]*g^2*S
qrt[f + g*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.317375, antiderivative size = 362, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {735, 833, 844, 719, 424, 419} \[ -\frac{4 \sqrt{-a} f \sqrt{\frac{c x^2}{a}+1} \left (a g^2+c f^2\right ) \sqrt{\frac{\sqrt{c} (f+g x)}{\sqrt{-a} g+\sqrt{c} f}} F\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a g}{\sqrt{-a} \sqrt{c} f-a g}\right )}{15 \sqrt{c} g^2 \sqrt{a+c x^2} \sqrt{f+g x}}+\frac{4 \sqrt{-a} \sqrt{\frac{c x^2}{a}+1} \sqrt{f+g x} \left (c f^2-3 a g^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a g}{\sqrt{-a} \sqrt{c} f-a g}\right )}{15 \sqrt{c} g^2 \sqrt{a+c x^2} \sqrt{\frac{\sqrt{c} (f+g x)}{\sqrt{-a} g+\sqrt{c} f}}}+\frac{2 \sqrt{a+c x^2} (f+g x)^{3/2}}{5 g}-\frac{4 f \sqrt{a+c x^2} \sqrt{f+g x}}{15 g} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[f + g*x]*Sqrt[a + c*x^2],x]

[Out]

(-4*f*Sqrt[f + g*x]*Sqrt[a + c*x^2])/(15*g) + (2*(f + g*x)^(3/2)*Sqrt[a + c*x^2])/(5*g) + (4*Sqrt[-a]*(c*f^2 -
3*a*g^2)*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^2*Sqrt[(Sqrt[c]*(f + g*x))/(Sqrt[c]*f + Sqrt[-a]*g)]*Sqrt[a + c*x^
2]) - (4*Sqrt[-a]*f*(c*f^2 + a*g^2)*Sqrt[(Sqrt[c]*(f + g*x))/(Sqrt[c]*f + Sqrt[-a]*g)]*Sqrt[1 + (c*x^2)/a]*Ell
ipticF[ArcSin[Sqrt[1 - (Sqrt[c]*x)/Sqrt[-a]]/Sqrt[2]], (-2*a*g)/(Sqrt[-a]*Sqrt[c]*f - a*g)])/(15*Sqrt[c]*g^2*S
qrt[f + g*x]*Sqrt[a + c*x^2])

Rule 735

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] + Dist[(2*p)/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(a + c*x^2)^(p - 1),
x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 833

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

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 \sqrt{f+g x} \sqrt{a+c x^2} \, dx &=\frac{2 (f+g x)^{3/2} \sqrt{a+c x^2}}{5 g}+\frac{2 \int \frac{(a g-c f x) \sqrt{f+g x}}{\sqrt{a+c x^2}} \, dx}{5 g}\\ &=-\frac{4 f \sqrt{f+g x} \sqrt{a+c x^2}}{15 g}+\frac{2 (f+g x)^{3/2} \sqrt{a+c x^2}}{5 g}+\frac{4 \int \frac{2 a c f g-\frac{1}{2} c \left (c f^2-3 a g^2\right ) x}{\sqrt{f+g x} \sqrt{a+c x^2}} \, dx}{15 c g}\\ &=-\frac{4 f \sqrt{f+g x} \sqrt{a+c x^2}}{15 g}+\frac{2 (f+g x)^{3/2} \sqrt{a+c x^2}}{5 g}+\frac{1}{15} \left (2 \left (3 a-\frac{c f^2}{g^2}\right )\right ) \int \frac{\sqrt{f+g x}}{\sqrt{a+c x^2}} \, dx+\frac{1}{15} \left (2 f \left (a+\frac{c f^2}{g^2}\right )\right ) \int \frac{1}{\sqrt{f+g x} \sqrt{a+c x^2}} \, dx\\ &=-\frac{4 f \sqrt{f+g x} \sqrt{a+c x^2}}{15 g}+\frac{2 (f+g x)^{3/2} \sqrt{a+c x^2}}{5 g}+\frac{\left (4 a \left (3 a-\frac{c f^2}{g^2}\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} \sqrt{\frac{c (f+g x)}{c f-\frac{a \sqrt{c} g}{\sqrt{-a}}}} \sqrt{a+c x^2}}+\frac{\left (4 a f \left (a+\frac{c f^2}{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} \sqrt{f+g x} \sqrt{a+c x^2}}\\ &=-\frac{4 f \sqrt{f+g x} \sqrt{a+c x^2}}{15 g}+\frac{2 (f+g x)^{3/2} \sqrt{a+c x^2}}{5 g}-\frac{4 \sqrt{-a} \left (3 a-\frac{c f^2}{g^2}\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} \sqrt{\frac{\sqrt{c} (f+g x)}{\sqrt{c} f+\sqrt{-a} g}} \sqrt{a+c x^2}}-\frac{4 \sqrt{-a} f \left (a+\frac{c f^2}{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} \sqrt{f+g x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 2.96802, size = 536, normalized size = 1.48 \[ \frac{\sqrt{f+g x} \left (\frac{2 \left (a+c x^2\right ) (f+3 g x)}{g}-\frac{4 \left (-\sqrt{a} \sqrt{c} g (f+g x)^{3/2} \left (4 i \sqrt{a} \sqrt{c} f g-3 a g^2+c f^2\right ) \sqrt{\frac{g \left (x+\frac{i \sqrt{a}}{\sqrt{c}}\right )}{f+g x}} \sqrt{-\frac{-g x+\frac{i \sqrt{a} g}{\sqrt{c}}}{f+g x}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-f-\frac{i \sqrt{a} g}{\sqrt{c}}}}{\sqrt{f+g x}}\right ),\frac{\sqrt{c} f-i \sqrt{a} g}{\sqrt{c} f+i \sqrt{a} g}\right )+g^2 \sqrt{-f-\frac{i \sqrt{a} g}{\sqrt{c}}} \left (-3 a^2 g^2+a c \left (f^2-3 g^2 x^2\right )+c^2 f^2 x^2\right )+\sqrt{c} (f+g x)^{3/2} \left (-3 a^{3/2} g^3+\sqrt{a} c f^2 g+3 i a \sqrt{c} f g^2-i c^{3/2} f^3\right ) \sqrt{\frac{g \left (x+\frac{i \sqrt{a}}{\sqrt{c}}\right )}{f+g x}} \sqrt{-\frac{-g x+\frac{i \sqrt{a} g}{\sqrt{c}}}{f+g x}} E\left (i \sinh ^{-1}\left (\frac{\sqrt{-f-\frac{i \sqrt{a} g}{\sqrt{c}}}}{\sqrt{f+g x}}\right )|\frac{\sqrt{c} f-i \sqrt{a} g}{\sqrt{c} f+i \sqrt{a} g}\right )\right )}{c g^3 (f+g x) \sqrt{-f-\frac{i \sqrt{a} g}{\sqrt{c}}}}\right )}{15 \sqrt{a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[f + g*x]*Sqrt[a + c*x^2],x]

[Out]

(Sqrt[f + g*x]*((2*(f + 3*g*x)*(a + c*x^2))/g - (4*(g^2*Sqrt[-f - (I*Sqrt[a]*g)/Sqrt[c]]*(-3*a^2*g^2 + c^2*f^2
*x^2 + a*c*(f^2 - 3*g^2*x^2)) + Sqrt[c]*((-I)*c^(3/2)*f^3 + Sqrt[a]*c*f^2*g + (3*I)*a*Sqrt[c]*f*g^2 - 3*a^(3/2
)*g^3)*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*(c*f^2 + (4*I)*Sqrt[a]*Sqrt[c]*f*g - 3*a*g^2)*Sqrt[(g*((I*Sqrt[a])/Sqr
t[c] + x))/(f + g*x)]*Sqrt[-(((I*Sqrt[a]*g)/Sqrt[c] - g*x)/(f + g*x))]*(f + g*x)^(3/2)*EllipticF[I*ArcSinh[Sqr
t[-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)]))/(c*g^3*Sq
rt[-f - (I*Sqrt[a]*g)/Sqrt[c]]*(f + g*x))))/(15*Sqrt[a + c*x^2])

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Maple [B]  time = 0.29, size = 1162, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(g*x+f)^(1/2)*(c*x^2+a)^(1/2)*(6*a^2*(-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2)*((-c*x+(-a*c)^(1/2))*g/((-a*
c)^(1/2)*g+c*f))^(1/2)*((c*x+(-a*c)^(1/2))*g/((-a*c)^(1/2)*g-c*f))^(1/2)*EllipticF((-(g*x+f)*c/((-a*c)^(1/2)*g
-c*f))^(1/2),(-((-a*c)^(1/2)*g-c*f)/((-a*c)^(1/2)*g+c*f))^(1/2))*g^4+6*(-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2)
*((-c*x+(-a*c)^(1/2))*g/((-a*c)^(1/2)*g+c*f))^(1/2)*((c*x+(-a*c)^(1/2))*g/((-a*c)^(1/2)*g-c*f))^(1/2)*Elliptic
F((-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2),(-((-a*c)^(1/2)*g-c*f)/((-a*c)^(1/2)*g+c*f))^(1/2))*a*c*f^2*g^2-2*(-
a*c)^(1/2)*(-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2)*((-c*x+(-a*c)^(1/2))*g/((-a*c)^(1/2)*g+c*f))^(1/2)*((c*x+(-
a*c)^(1/2))*g/((-a*c)^(1/2)*g-c*f))^(1/2)*EllipticF((-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2),(-((-a*c)^(1/2)*g-
c*f)/((-a*c)^(1/2)*g+c*f))^(1/2))*a*f*g^3-2*(-a*c)^(1/2)*(-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2)*((-c*x+(-a*c)
^(1/2))*g/((-a*c)^(1/2)*g+c*f))^(1/2)*((c*x+(-a*c)^(1/2))*g/((-a*c)^(1/2)*g-c*f))^(1/2)*EllipticF((-(g*x+f)*c/
((-a*c)^(1/2)*g-c*f))^(1/2),(-((-a*c)^(1/2)*g-c*f)/((-a*c)^(1/2)*g+c*f))^(1/2))*c*f^3*g-6*a^2*(-(g*x+f)*c/((-a
*c)^(1/2)*g-c*f))^(1/2)*((-c*x+(-a*c)^(1/2))*g/((-a*c)^(1/2)*g+c*f))^(1/2)*((c*x+(-a*c)^(1/2))*g/((-a*c)^(1/2)
*g-c*f))^(1/2)*EllipticE((-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2),(-((-a*c)^(1/2)*g-c*f)/((-a*c)^(1/2)*g+c*f))^
(1/2))*g^4-4*(-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2)*((-c*x+(-a*c)^(1/2))*g/((-a*c)^(1/2)*g+c*f))^(1/2)*((c*x+
(-a*c)^(1/2))*g/((-a*c)^(1/2)*g-c*f))^(1/2)*EllipticE((-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2),(-((-a*c)^(1/2)*
g-c*f)/((-a*c)^(1/2)*g+c*f))^(1/2))*a*c*f^2*g^2+2*(-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2)*((-c*x+(-a*c)^(1/2))
*g/((-a*c)^(1/2)*g+c*f))^(1/2)*((c*x+(-a*c)^(1/2))*g/((-a*c)^(1/2)*g-c*f))^(1/2)*EllipticE((-(g*x+f)*c/((-a*c)
^(1/2)*g-c*f))^(1/2),(-((-a*c)^(1/2)*g-c*f)/((-a*c)^(1/2)*g+c*f))^(1/2))*c^2*f^4+3*x^4*c^2*g^4+4*x^3*c^2*f*g^3
+3*x^2*a*c*g^4+x^2*c^2*f^2*g^2+4*x*a*c*f*g^3+a*c*f^2*g^2)/c/(c*g*x^3+c*f*x^2+a*g*x+a*f)/g^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c x^{2} + a} \sqrt{g x + f}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + c x^{2}} \sqrt{f + g x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c x^{2} + a} \sqrt{g x + f}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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