3.1262 \(\int \frac{(A+B x) (b x+c x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=413 \[ \frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} \left (5 A e \left (3 b^2 e^2-16 b c d e+16 c^2 d^2\right )-B d \left (39 b^2 e^2-152 b c d e+128 c^2 d^2\right )\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{15 \sqrt{c} e^5 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (40 A c e (2 c d-b e)-B \left (3 b^2 e^2-88 b c d e+128 c^2 d^2\right )\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 \sqrt{c} e^5 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \left (b x+c x^2\right )^{3/2} (-5 A e+8 B d+3 B e x)}{15 e^2 (d+e x)^{3/2}}-\frac{2 \sqrt{b x+c x^2} (e x (-10 A c e-3 b B e+16 B c d)-5 A e (8 c d-3 b e)+4 B d (16 c d-9 b e))}{15 e^4 \sqrt{d+e x}} \]

[Out]

(-2*(4*B*d*(16*c*d - 9*b*e) - 5*A*e*(8*c*d - 3*b*e) + e*(16*B*c*d - 3*b*B*e - 10*A*c*e)*x)*Sqrt[b*x + c*x^2])/
(15*e^4*Sqrt[d + e*x]) + (2*(8*B*d - 5*A*e + 3*B*e*x)*(b*x + c*x^2)^(3/2))/(15*e^2*(d + e*x)^(3/2)) - (2*Sqrt[
-b]*(40*A*c*e*(2*c*d - b*e) - B*(128*c^2*d^2 - 88*b*c*d*e + 3*b^2*e^2))*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x
]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*Sqrt[c]*e^5*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x
^2]) + (2*Sqrt[-b]*(5*A*e*(16*c^2*d^2 - 16*b*c*d*e + 3*b^2*e^2) - B*d*(128*c^2*d^2 - 152*b*c*d*e + 39*b^2*e^2)
)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*
Sqrt[c]*e^5*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.501875, antiderivative size = 413, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {812, 843, 715, 112, 110, 117, 116} \[ \frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} \left (5 A e \left (3 b^2 e^2-16 b c d e+16 c^2 d^2\right )-B d \left (39 b^2 e^2-152 b c d e+128 c^2 d^2\right )\right ) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 \sqrt{c} e^5 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (40 A c e (2 c d-b e)-B \left (3 b^2 e^2-88 b c d e+128 c^2 d^2\right )\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 \sqrt{c} e^5 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \left (b x+c x^2\right )^{3/2} (-5 A e+8 B d+3 B e x)}{15 e^2 (d+e x)^{3/2}}-\frac{2 \sqrt{b x+c x^2} (e x (-10 A c e-3 b B e+16 B c d)-5 A e (8 c d-3 b e)+4 B d (16 c d-9 b e))}{15 e^4 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

(-2*(4*B*d*(16*c*d - 9*b*e) - 5*A*e*(8*c*d - 3*b*e) + e*(16*B*c*d - 3*b*B*e - 10*A*c*e)*x)*Sqrt[b*x + c*x^2])/
(15*e^4*Sqrt[d + e*x]) + (2*(8*B*d - 5*A*e + 3*B*e*x)*(b*x + c*x^2)^(3/2))/(15*e^2*(d + e*x)^(3/2)) - (2*Sqrt[
-b]*(40*A*c*e*(2*c*d - b*e) - B*(128*c^2*d^2 - 88*b*c*d*e + 3*b^2*e^2))*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x
]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*Sqrt[c]*e^5*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x
^2]) + (2*Sqrt[-b]*(5*A*e*(16*c^2*d^2 - 16*b*c*d*e + 3*b^2*e^2) - B*d*(128*c^2*d^2 - 152*b*c*d*e + 39*b^2*e^2)
)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*
Sqrt[c]*e^5*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 715

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(Sqrt[x]*Sqrt[b + c*x])/Sqrt[
b*x + c*x^2], Int[(d + e*x)^m/(Sqrt[x]*Sqrt[b + c*x]), x], x] /; FreeQ[{b, c, d, e}, x] && NeQ[c*d - b*e, 0] &
& NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4]

Rule 112

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[e + f*x]*Sqrt[
1 + (d*x)/c])/(Sqrt[c + d*x]*Sqrt[1 + (f*x)/e]), Int[Sqrt[1 + (f*x)/e]/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]), x], x] /
; FreeQ[{b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] &&  !(GtQ[c, 0] && GtQ[e, 0])

Rule 110

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[(2*Sqrt[e]*Rt[-(b/d)
, 2]*EllipticE[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/b, x] /; FreeQ[{b, c, d, e, f}, x] &&
NeQ[d*e - c*f, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !LtQ[-(b/d), 0]

Rule 117

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[1 + (d*x)/c]
*Sqrt[1 + (f*x)/e])/(Sqrt[c + d*x]*Sqrt[e + f*x]), Int[1/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]*Sqrt[1 + (f*x)/e]), x],
x] /; FreeQ[{b, c, d, e, f}, x] &&  !(GtQ[c, 0] && GtQ[e, 0])

Rule 116

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-(b/d), 2]*E
llipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/(b*Sqrt[e]), x] /; FreeQ[{b, c, d, e, f}, x]
&& GtQ[c, 0] && GtQ[e, 0] && (PosQ[-(b/d)] || NegQ[-(b/f)])

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx &=\frac{2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac{2 \int \frac{\left (\frac{1}{2} b (8 B d-5 A e)+\frac{1}{2} (16 B c d-3 b B e-10 A c e) x\right ) \sqrt{b x+c x^2}}{(d+e x)^{3/2}} \, dx}{5 e^2}\\ &=-\frac{2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt{b x+c x^2}}{15 e^4 \sqrt{d+e x}}+\frac{2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}+\frac{4 \int \frac{\frac{1}{4} b (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e))-\frac{1}{4} \left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\right ) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{15 e^4}\\ &=-\frac{2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt{b x+c x^2}}{15 e^4 \sqrt{d+e x}}+\frac{2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac{\left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{15 e^5}+\frac{\left (5 A e \left (16 c^2 d^2-16 b c d e+3 b^2 e^2\right )-B d \left (128 c^2 d^2-152 b c d e+39 b^2 e^2\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{15 e^5}\\ &=-\frac{2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt{b x+c x^2}}{15 e^4 \sqrt{d+e x}}+\frac{2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac{\left (\left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\right ) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{\sqrt{d+e x}}{\sqrt{x} \sqrt{b+c x}} \, dx}{15 e^5 \sqrt{b x+c x^2}}+\frac{\left (\left (5 A e \left (16 c^2 d^2-16 b c d e+3 b^2 e^2\right )-B d \left (128 c^2 d^2-152 b c d e+39 b^2 e^2\right )\right ) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x} \sqrt{d+e x}} \, dx}{15 e^5 \sqrt{b x+c x^2}}\\ &=-\frac{2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt{b x+c x^2}}{15 e^4 \sqrt{d+e x}}+\frac{2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac{\left (\left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\right ) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{d+e x}\right ) \int \frac{\sqrt{1+\frac{e x}{d}}}{\sqrt{x} \sqrt{1+\frac{c x}{b}}} \, dx}{15 e^5 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{\left (\left (5 A e \left (16 c^2 d^2-16 b c d e+3 b^2 e^2\right )-B d \left (128 c^2 d^2-152 b c d e+39 b^2 e^2\right )\right ) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{1+\frac{e x}{d}}\right ) \int \frac{1}{\sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{1+\frac{e x}{d}}} \, dx}{15 e^5 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=-\frac{2 (4 B d (16 c d-9 b e)-5 A e (8 c d-3 b e)+e (16 B c d-3 b B e-10 A c e) x) \sqrt{b x+c x^2}}{15 e^4 \sqrt{d+e x}}+\frac{2 (8 B d-5 A e+3 B e x) \left (b x+c x^2\right )^{3/2}}{15 e^2 (d+e x)^{3/2}}-\frac{2 \sqrt{-b} \left (40 A c e (2 c d-b e)-B \left (128 c^2 d^2-88 b c d e+3 b^2 e^2\right )\right ) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{d+e x} E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 \sqrt{c} e^5 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{2 \sqrt{-b} \left (5 A e \left (16 c^2 d^2-16 b c d e+3 b^2 e^2\right )-B d \left (128 c^2 d^2-152 b c d e+39 b^2 e^2\right )\right ) \sqrt{x} \sqrt{1+\frac{c x}{b}} \sqrt{1+\frac{e x}{d}} F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 \sqrt{c} e^5 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 2.222, size = 436, normalized size = 1.06 \[ \frac{2 (x (b+c x))^{3/2} \left (i e x^{3/2} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (5 A c e (8 c d-5 b e)+B \left (-3 b^2 e^2+52 b c d e-64 c^2 d^2\right )\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )-i e x^{3/2} \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (40 A c e (2 c d-b e)+B \left (-3 b^2 e^2+88 b c d e-128 c^2 d^2\right )\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )+\frac{(b+c x) (d+e x) \left (40 A c e (b e-2 c d)+B \left (3 b^2 e^2-88 b c d e+128 c^2 d^2\right )\right )}{c}+\frac{e x (b+c x) \left (5 A e \left (c \left (8 d^2+10 d e x+e^2 x^2\right )-b e (3 d+4 e x)\right )+B \left (b e \left (36 d^2+47 d e x+6 e^2 x^2\right )-c \left (80 d^2 e x+64 d^3+8 d e^2 x^2-3 e^3 x^3\right )\right )\right )}{d+e x}\right )}{15 e^5 x^2 (b+c x)^2 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

(2*(x*(b + c*x))^(3/2)*(((40*A*c*e*(-2*c*d + b*e) + B*(128*c^2*d^2 - 88*b*c*d*e + 3*b^2*e^2))*(b + c*x)*(d + e
*x))/c + (e*x*(b + c*x)*(5*A*e*(-(b*e*(3*d + 4*e*x)) + c*(8*d^2 + 10*d*e*x + e^2*x^2)) + B*(b*e*(36*d^2 + 47*d
*e*x + 6*e^2*x^2) - c*(64*d^3 + 80*d^2*e*x + 8*d*e^2*x^2 - 3*e^3*x^3))))/(d + e*x) - I*Sqrt[b/c]*e*(40*A*c*e*(
2*c*d - b*e) + B*(-128*c^2*d^2 + 88*b*c*d*e - 3*b^2*e^2))*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*Elliptic
E[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)] + I*Sqrt[b/c]*e*(5*A*c*e*(8*c*d - 5*b*e) + B*(-64*c^2*d^2 + 52*b*
c*d*e - 3*b^2*e^2))*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/
(b*e)]))/(15*e^5*x^2*(b + c*x)^2*Sqrt[d + e*x])

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Maple [B]  time = 0.037, size = 2367, normalized size = 5.7 \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+b*x)^(3/2)/(e*x+d)^(5/2),x)

[Out]

-2/15*(x*(c*x+b))^(1/2)*(-40*A*x^2*c^4*d^2*e^2-36*B*x*b^2*c^2*d^2*e^2+64*B*x*b*c^3*d^3*e-40*A*x*b*c^3*d^2*e^2-
47*B*x^2*b^2*c^2*d*e^3+44*B*x^2*b*c^3*d^2*e^2+15*A*x*b^2*c^2*d*e^3-35*A*x^2*b*c^3*d*e^3+3*B*EllipticE(((c*x+b)
/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*d*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-128*B
*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^4*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(
-c*x/b)^(1/2)+128*B*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^4*((c*x+b)/b)^(1/2)*(-(e*x+d)*c
/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-3*B*x^5*c^4*e^4-5*A*x^4*c^4*e^4-39*B*x^3*b*c^3*d*e^3+64*B*x^2*c^4*d^3*e+15*A*
x^3*b*c^3*e^4-50*A*x^3*c^4*d*e^3-6*B*x^3*b^2*c^2*e^4+80*B*x^3*c^4*d^2*e^2+20*A*x^2*b^2*c^2*e^4-9*B*x^4*b*c^3*e
^4+8*B*x^4*c^4*d*e^3+3*B*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^4*e^4*((c*x+b)/b)^(1/2)*(-(e*x
+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-120*A*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c^2*d*e^3
*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+80*A*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d)
)^(1/2))*x*b*c^3*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+80*A*EllipticF(((c*x+b)
/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c^2*d*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)
+80*A*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*
d))^(1/2)*(-c*x/b)^(1/2)-80*A*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^3*e*((c*x+b)/b)^(1/2)
*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-91*B*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*d^2
*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+216*B*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e
-c*d))^(1/2))*b^2*c^2*d^3*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+39*B*EllipticF(((c*x
+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2
)-152*B*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^3*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*
d))^(1/2)*(-c*x/b)^(1/2)+40*A*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*c*e^4*((c*x+b)/b)^(1/2)
*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-15*A*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*c*e
^4*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+40*A*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*
d))^(1/2))*b^3*c*d*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-120*A*EllipticE(((c*x+b)/
b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+
80*A*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^3*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(
1/2)*(-c*x/b)^(1/2)-15*A*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*d*e^3*((c*x+b)/b)^(1/2)*(-(e
*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-80*A*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^3*d^2*e^
2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-91*B*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d
))^(1/2))*x*b^3*c*d*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+216*B*EllipticE(((c*x+b)
/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c^2*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/
2)-128*B*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^3*d^3*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c
*d))^(1/2)*(-c*x/b)^(1/2)+39*B*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*c*d*e^3*((c*x+b)/b)^(1
/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-152*B*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2
*c^2*d^2*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+128*B*EllipticF(((c*x+b)/b)^(1/2),(
b*e/(b*e-c*d))^(1/2))*x*b*c^3*d^3*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2))/(c*x+b)/x/(
e*x+d)^(3/2)/c^2/e^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x\right )}^{\frac{3}{2}}{\left (B x + A\right )}}{{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)*(B*x + A)/(e*x + d)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

integral((B*c*x^3 + A*b*x + (B*b + A*c)*x^2)*sqrt(c*x^2 + b*x)*sqrt(e*x + d)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*
x + d^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x \left (b + c x\right )\right )^{\frac{3}{2}} \left (A + B x\right )}{\left (d + e x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(3/2)/(e*x+d)**(5/2),x)

[Out]

Integral((x*(b + c*x))**(3/2)*(A + B*x)/(d + e*x)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x\right )}^{\frac{3}{2}}{\left (B x + A\right )}}{{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^(3/2)*(B*x + A)/(e*x + d)^(5/2), x)