### 3.665 $$\int \frac{(a+c x^2)^{3/2}}{\sqrt{d+e x}} \, dx$$

Optimal. Leaf size=393 $-\frac{8 \sqrt{-a} \sqrt{\frac{c x^2}{a}+1} \left (a e^2+c d^2\right ) \left (5 a e^2+4 c d^2\right ) \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right ),-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{35 \sqrt{c} e^4 \sqrt{a+c x^2} \sqrt{d+e x}}+\frac{4 \sqrt{a+c x^2} \sqrt{d+e x} \left (5 a e^2+4 c d^2-3 c d e x\right )}{35 e^3}+\frac{32 \sqrt{-a} \sqrt{c} d \sqrt{\frac{c x^2}{a}+1} \sqrt{d+e x} \left (2 a e^2+c d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{35 e^4 \sqrt{a+c x^2} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}}}+\frac{2 \left (a+c x^2\right )^{3/2} \sqrt{d+e x}}{7 e}$

[Out]

(4*Sqrt[d + e*x]*(4*c*d^2 + 5*a*e^2 - 3*c*d*e*x)*Sqrt[a + c*x^2])/(35*e^3) + (2*Sqrt[d + e*x]*(a + c*x^2)^(3/2
))/(7*e) + (32*Sqrt[-a]*Sqrt[c]*d*(c*d^2 + 2*a*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)])/(35*e^4*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt
[c]*d + Sqrt[-a]*e)]*Sqrt[a + c*x^2]) - (8*Sqrt[-a]*(c*d^2 + a*e^2)*(4*c*d^2 + 5*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)])/(35*Sqrt[c]*e^4*Sqrt[d + e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.357265, antiderivative size = 393, 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, 815, 844, 719, 424, 419} $\frac{4 \sqrt{a+c x^2} \sqrt{d+e x} \left (5 a e^2+4 c d^2-3 c d e x\right )}{35 e^3}-\frac{8 \sqrt{-a} \sqrt{\frac{c x^2}{a}+1} \left (a e^2+c d^2\right ) \left (5 a e^2+4 c d^2\right ) \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}} F\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{35 \sqrt{c} e^4 \sqrt{a+c x^2} \sqrt{d+e x}}+\frac{32 \sqrt{-a} \sqrt{c} d \sqrt{\frac{c x^2}{a}+1} \sqrt{d+e x} \left (2 a e^2+c d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{1-\frac{\sqrt{c} x}{\sqrt{-a}}}}{\sqrt{2}}\right )|-\frac{2 a e}{\sqrt{-a} \sqrt{c} d-a e}\right )}{35 e^4 \sqrt{a+c x^2} \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{-a} e+\sqrt{c} d}}}+\frac{2 \left (a+c x^2\right )^{3/2} \sqrt{d+e x}}{7 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*Sqrt[d + e*x]*(4*c*d^2 + 5*a*e^2 - 3*c*d*e*x)*Sqrt[a + c*x^2])/(35*e^3) + (2*Sqrt[d + e*x]*(a + c*x^2)^(3/2
))/(7*e) + (32*Sqrt[-a]*Sqrt[c]*d*(c*d^2 + 2*a*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)])/(35*e^4*Sqrt[(Sqrt[c]*(d + e*x))/(Sqrt
[c]*d + Sqrt[-a]*e)]*Sqrt[a + c*x^2]) - (8*Sqrt[-a]*(c*d^2 + a*e^2)*(4*c*d^2 + 5*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)])/(35*Sqrt[c]*e^4*Sqrt[d + e*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 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{\left (a+c x^2\right )^{3/2}}{\sqrt{d+e x}} \, dx &=\frac{2 \sqrt{d+e x} \left (a+c x^2\right )^{3/2}}{7 e}+\frac{6 \int \frac{(a e-c d x) \sqrt{a+c x^2}}{\sqrt{d+e x}} \, dx}{7 e}\\ &=\frac{4 \sqrt{d+e x} \left (4 c d^2+5 a e^2-3 c d e x\right ) \sqrt{a+c x^2}}{35 e^3}+\frac{2 \sqrt{d+e x} \left (a+c x^2\right )^{3/2}}{7 e}+\frac{8 \int \frac{\frac{1}{2} a c e \left (c d^2+5 a e^2\right )-2 c^2 d \left (c d^2+2 a e^2\right ) x}{\sqrt{d+e x} \sqrt{a+c x^2}} \, dx}{35 c e^3}\\ &=\frac{4 \sqrt{d+e x} \left (4 c d^2+5 a e^2-3 c d e x\right ) \sqrt{a+c x^2}}{35 e^3}+\frac{2 \sqrt{d+e x} \left (a+c x^2\right )^{3/2}}{7 e}-\frac{\left (16 c d \left (c d^2+2 a e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{a+c x^2}} \, dx}{35 e^4}+\frac{\left (4 \left (c d^2+a e^2\right ) \left (4 c d^2+5 a e^2\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{a+c x^2}} \, dx}{35 e^4}\\ &=\frac{4 \sqrt{d+e x} \left (4 c d^2+5 a e^2-3 c d e x\right ) \sqrt{a+c x^2}}{35 e^3}+\frac{2 \sqrt{d+e x} \left (a+c x^2\right )^{3/2}}{7 e}-\frac{\left (32 a \sqrt{c} d \left (c d^2+2 a 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 )}{35 \sqrt{-a} e^4 \sqrt{\frac{c (d+e x)}{c d-\frac{a \sqrt{c} e}{\sqrt{-a}}}} \sqrt{a+c x^2}}+\frac{\left (8 a \left (c d^2+a e^2\right ) \left (4 c d^2+5 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 )}{35 \sqrt{-a} \sqrt{c} e^4 \sqrt{d+e x} \sqrt{a+c x^2}}\\ &=\frac{4 \sqrt{d+e x} \left (4 c d^2+5 a e^2-3 c d e x\right ) \sqrt{a+c x^2}}{35 e^3}+\frac{2 \sqrt{d+e x} \left (a+c x^2\right )^{3/2}}{7 e}+\frac{32 \sqrt{-a} \sqrt{c} d \left (c d^2+2 a 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 )}{35 e^4 \sqrt{\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}} \sqrt{a+c x^2}}-\frac{8 \sqrt{-a} \left (c d^2+a e^2\right ) \left (4 c d^2+5 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 )}{35 \sqrt{c} e^4 \sqrt{d+e x} \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [C]  time = 3.24404, size = 575, normalized size = 1.46 $\frac{\sqrt{d+e x} \left (\frac{2 \left (a+c x^2\right ) \left (15 a e^2+c \left (8 d^2-6 d e x+5 e^2 x^2\right )\right )}{e^3}-\frac{8 \left (-\sqrt{a} e (d+e x)^{3/2} \left (5 i a^{3/2} e^3+i \sqrt{a} c d^2 e+8 a \sqrt{c} d e^2+4 c^{3/2} d^3\right ) \sqrt{\frac{e \left (x+\frac{i \sqrt{a}}{\sqrt{c}}\right )}{d+e x}} \sqrt{-\frac{-e x+\frac{i \sqrt{a} e}{\sqrt{c}}}{d+e x}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}{\sqrt{d+e x}}\right ),\frac{\sqrt{c} d-i \sqrt{a} e}{\sqrt{c} d+i \sqrt{a} e}\right )+4 d e^2 \sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}} \left (2 a^2 e^2+a c \left (d^2+2 e^2 x^2\right )+c^2 d^2 x^2\right )+4 \sqrt{c} d (d+e x)^{3/2} \left (2 a^{3/2} e^3+\sqrt{a} c d^2 e-2 i a \sqrt{c} d e^2-i c^{3/2} d^3\right ) \sqrt{\frac{e \left (x+\frac{i \sqrt{a}}{\sqrt{c}}\right )}{d+e x}} \sqrt{-\frac{-e x+\frac{i \sqrt{a} e}{\sqrt{c}}}{d+e x}} E\left (i \sinh ^{-1}\left (\frac{\sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}{\sqrt{d+e x}}\right )|\frac{\sqrt{c} d-i \sqrt{a} e}{\sqrt{c} d+i \sqrt{a} e}\right )\right )}{e^5 (d+e x) \sqrt{-d-\frac{i \sqrt{a} e}{\sqrt{c}}}}\right )}{35 \sqrt{a+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[d + e*x]*((2*(a + c*x^2)*(15*a*e^2 + c*(8*d^2 - 6*d*e*x + 5*e^2*x^2)))/e^3 - (8*(4*d*e^2*Sqrt[-d - (I*Sq
rt[a]*e)/Sqrt[c]]*(2*a^2*e^2 + c^2*d^2*x^2 + a*c*(d^2 + 2*e^2*x^2)) + 4*Sqrt[c]*d*((-I)*c^(3/2)*d^3 + Sqrt[a]*
c*d^2*e - (2*I)*a*Sqrt[c]*d*e^2 + 2*a^(3/2)*e^3)*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]*e*(4*c^(3/2)*d^3 + I*Sqrt[a]*c*d^2*e +
8*a*Sqrt[c]*d*e^2 + (5*I)*a^(3/2)*e^3)*Sqrt[(e*((I*Sqrt[a])/Sqrt[c] + x))/(d + e*x)]*Sqrt[-(((I*Sqrt[a]*e)/Sq
rt[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)]))/(e^5*Sqrt[-d - (I*Sqrt[a]*e)/Sqrt[c]]*(d + e*x))))/(35*
Sqrt[a + c*x^2])

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Maple [B]  time = 0.25, size = 1385, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(c*x^2+a)^(1/2)*(e*x+d)^(1/2)*(-5*x^5*c^3*e^5+20*(-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)*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^5+36*(-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)*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^3+16*(-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)*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^2*d^4*e+12*a^2*c*(-(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))*d*e^4+12*a*c^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)*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))*d^3*e^2-32*a^2*c*(-(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))*d*e^4-48*a*c^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)*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))*d^3*e^2-16*(-(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^3*d^5+x^4*c^3*d*e^4-20*x^3*a*c^2*e^5-2*x^3*c^3*d^2*e^3-14
*x^2*a*c^2*d*e^4-8*x^2*c^3*d^3*e^2-15*x*a^2*c*e^5-2*x*a*c^2*d^2*e^3-15*a^2*c*d*e^4-8*a*c^2*d^3*e^2)/c/e^5/(c*e
*x^3+c*d*x^2+a*e*x+a*d)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + a)^(3/2)/sqrt(e*x + d), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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