3.1264 \(\int \frac{(A+B x) (d+e x)^{5/2}}{\sqrt{b x+c x^2}} \, dx\)

Optimal. Leaf size=460 \[ -\frac{2 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) \left (28 A c e (2 c d-b e)+B \left (24 b^2 e^2-43 b c d e+15 c^2 d^2\right )\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{105 c^{7/2} e \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} \left (28 A c e (2 c d-b e)+B \left (24 b^2 e^2-43 b c d e+15 c^2 d^2\right )\right )}{105 c^3}+\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (7 A c e \left (8 b^2 e^2-23 b c d e+23 c^2 d^2\right )+B \left (128 b^2 c d e^2-48 b^3 e^3-103 b c^2 d^2 e+15 c^3 d^3\right )\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{105 c^{7/2} e \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \sqrt{b x+c x^2} (d+e x)^{3/2} (7 A c e-6 b B e+5 B c d)}{35 c^2}+\frac{2 B \sqrt{b x+c x^2} (d+e x)^{5/2}}{7 c} \]

[Out]

(2*(28*A*c*e*(2*c*d - b*e) + B*(15*c^2*d^2 - 43*b*c*d*e + 24*b^2*e^2))*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])/(105*c
^3) + (2*(5*B*c*d - 6*b*B*e + 7*A*c*e)*(d + e*x)^(3/2)*Sqrt[b*x + c*x^2])/(35*c^2) + (2*B*(d + e*x)^(5/2)*Sqrt
[b*x + c*x^2])/(7*c) + (2*Sqrt[-b]*(7*A*c*e*(23*c^2*d^2 - 23*b*c*d*e + 8*b^2*e^2) + B*(15*c^3*d^3 - 103*b*c^2*
d^2*e + 128*b^2*c*d*e^2 - 48*b^3*e^3))*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[
x])/Sqrt[-b]], (b*e)/(c*d)])/(105*c^(7/2)*e*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (2*Sqrt[-b]*d*(c*d - b*e)*(
28*A*c*e*(2*c*d - b*e) + B*(15*c^2*d^2 - 43*b*c*d*e + 24*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)])/(105*c^(7/2)*e*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.765661, antiderivative size = 460, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {832, 843, 715, 112, 110, 117, 116} \[ \frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} \left (28 A c e (2 c d-b e)+B \left (24 b^2 e^2-43 b c d e+15 c^2 d^2\right )\right )}{105 c^3}-\frac{2 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) \left (28 A c e (2 c d-b e)+B \left (24 b^2 e^2-43 b c d e+15 c^2 d^2\right )\right ) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{105 c^{7/2} e \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (7 A c e \left (8 b^2 e^2-23 b c d e+23 c^2 d^2\right )+B \left (128 b^2 c d e^2-48 b^3 e^3-103 b c^2 d^2 e+15 c^3 d^3\right )\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{105 c^{7/2} e \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \sqrt{b x+c x^2} (d+e x)^{3/2} (7 A c e-6 b B e+5 B c d)}{35 c^2}+\frac{2 B \sqrt{b x+c x^2} (d+e x)^{5/2}}{7 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(28*A*c*e*(2*c*d - b*e) + B*(15*c^2*d^2 - 43*b*c*d*e + 24*b^2*e^2))*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])/(105*c
^3) + (2*(5*B*c*d - 6*b*B*e + 7*A*c*e)*(d + e*x)^(3/2)*Sqrt[b*x + c*x^2])/(35*c^2) + (2*B*(d + e*x)^(5/2)*Sqrt
[b*x + c*x^2])/(7*c) + (2*Sqrt[-b]*(7*A*c*e*(23*c^2*d^2 - 23*b*c*d*e + 8*b^2*e^2) + B*(15*c^3*d^3 - 103*b*c^2*
d^2*e + 128*b^2*c*d*e^2 - 48*b^3*e^3))*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[
x])/Sqrt[-b]], (b*e)/(c*d)])/(105*c^(7/2)*e*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*x^2]) - (2*Sqrt[-b]*d*(c*d - b*e)*(
28*A*c*e*(2*c*d - b*e) + B*(15*c^2*d^2 - 43*b*c*d*e + 24*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)])/(105*c^(7/2)*e*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + 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 + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + 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 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) (d+e x)^{5/2}}{\sqrt{b x+c x^2}} \, dx &=\frac{2 B (d+e x)^{5/2} \sqrt{b x+c x^2}}{7 c}+\frac{2 \int \frac{(d+e x)^{3/2} \left (-\frac{1}{2} (b B-7 A c) d+\frac{1}{2} (5 B c d-6 b B e+7 A c e) x\right )}{\sqrt{b x+c x^2}} \, dx}{7 c}\\ &=\frac{2 (5 B c d-6 b B e+7 A c e) (d+e x)^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 B (d+e x)^{5/2} \sqrt{b x+c x^2}}{7 c}+\frac{4 \int \frac{\sqrt{d+e x} \left (-\frac{1}{4} d \left (10 b B c d-35 A c^2 d-6 b^2 B e+7 A b c e\right )+\frac{1}{4} \left (28 A c e (2 c d-b e)+B \left (15 c^2 d^2-43 b c d e+24 b^2 e^2\right )\right ) x\right )}{\sqrt{b x+c x^2}} \, dx}{35 c^2}\\ &=\frac{2 \left (28 A c e (2 c d-b e)+B \left (15 c^2 d^2-43 b c d e+24 b^2 e^2\right )\right ) \sqrt{d+e x} \sqrt{b x+c x^2}}{105 c^3}+\frac{2 (5 B c d-6 b B e+7 A c e) (d+e x)^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 B (d+e x)^{5/2} \sqrt{b x+c x^2}}{7 c}+\frac{8 \int \frac{\frac{1}{8} d \left (105 A c^3 d^2-24 b^3 B e^2+b^2 c e (61 B d+28 A e)-b c^2 d (45 B d+77 A e)\right )+\frac{1}{8} \left (7 A c e \left (23 c^2 d^2-23 b c d e+8 b^2 e^2\right )+B \left (15 c^3 d^3-103 b c^2 d^2 e+128 b^2 c d e^2-48 b^3 e^3\right )\right ) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{105 c^3}\\ &=\frac{2 \left (28 A c e (2 c d-b e)+B \left (15 c^2 d^2-43 b c d e+24 b^2 e^2\right )\right ) \sqrt{d+e x} \sqrt{b x+c x^2}}{105 c^3}+\frac{2 (5 B c d-6 b B e+7 A c e) (d+e x)^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 B (d+e x)^{5/2} \sqrt{b x+c x^2}}{7 c}-\frac{\left (d (c d-b e) \left (28 A c e (2 c d-b e)+B \left (15 c^2 d^2-43 b c d e+24 b^2 e^2\right )\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{105 c^3 e}+\frac{\left (7 A c e \left (23 c^2 d^2-23 b c d e+8 b^2 e^2\right )+B \left (15 c^3 d^3-103 b c^2 d^2 e+128 b^2 c d e^2-48 b^3 e^3\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{105 c^3 e}\\ &=\frac{2 \left (28 A c e (2 c d-b e)+B \left (15 c^2 d^2-43 b c d e+24 b^2 e^2\right )\right ) \sqrt{d+e x} \sqrt{b x+c x^2}}{105 c^3}+\frac{2 (5 B c d-6 b B e+7 A c e) (d+e x)^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 B (d+e x)^{5/2} \sqrt{b x+c x^2}}{7 c}-\frac{\left (d (c d-b e) \left (28 A c e (2 c d-b e)+B \left (15 c^2 d^2-43 b c d e+24 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}{105 c^3 e \sqrt{b x+c x^2}}+\frac{\left (\left (7 A c e \left (23 c^2 d^2-23 b c d e+8 b^2 e^2\right )+B \left (15 c^3 d^3-103 b c^2 d^2 e+128 b^2 c d e^2-48 b^3 e^3\right )\right ) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{\sqrt{d+e x}}{\sqrt{x} \sqrt{b+c x}} \, dx}{105 c^3 e \sqrt{b x+c x^2}}\\ &=\frac{2 \left (28 A c e (2 c d-b e)+B \left (15 c^2 d^2-43 b c d e+24 b^2 e^2\right )\right ) \sqrt{d+e x} \sqrt{b x+c x^2}}{105 c^3}+\frac{2 (5 B c d-6 b B e+7 A c e) (d+e x)^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 B (d+e x)^{5/2} \sqrt{b x+c x^2}}{7 c}+\frac{\left (\left (7 A c e \left (23 c^2 d^2-23 b c d e+8 b^2 e^2\right )+B \left (15 c^3 d^3-103 b c^2 d^2 e+128 b^2 c d e^2-48 b^3 e^3\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}{105 c^3 e \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{\left (d (c d-b e) \left (28 A c e (2 c d-b e)+B \left (15 c^2 d^2-43 b c d e+24 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}{105 c^3 e \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=\frac{2 \left (28 A c e (2 c d-b e)+B \left (15 c^2 d^2-43 b c d e+24 b^2 e^2\right )\right ) \sqrt{d+e x} \sqrt{b x+c x^2}}{105 c^3}+\frac{2 (5 B c d-6 b B e+7 A c e) (d+e x)^{3/2} \sqrt{b x+c x^2}}{35 c^2}+\frac{2 B (d+e x)^{5/2} \sqrt{b x+c x^2}}{7 c}+\frac{2 \sqrt{-b} \left (7 A c e \left (23 c^2 d^2-23 b c d e+8 b^2 e^2\right )+B \left (15 c^3 d^3-103 b c^2 d^2 e+128 b^2 c d e^2-48 b^3 e^3\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 )}{105 c^{7/2} e \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{2 \sqrt{-b} d (c d-b e) \left (28 A c e (2 c d-b e)+B \left (15 c^2 d^2-43 b c d e+24 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 )}{105 c^{7/2} e \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 4.19065, size = 479, normalized size = 1.04 \[ \frac{2 \sqrt{x} \left (\frac{i x \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} (b e-c d) \left (-8 b^2 c e (7 A e+13 B d)+b c^2 d (133 A e+60 B d)-105 A c^3 d^2+48 b^3 B e^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )}{b}+\sqrt{x} (b+c x) (d+e x) \left (7 A c e (-4 b e+11 c d+3 c e x)+B \left (24 b^2 e^2-b c e (61 d+18 e x)+15 c^2 \left (3 d^2+3 d e x+e^2 x^2\right )\right )\right )+\frac{(b+c x) (d+e x) \left (7 A c e \left (8 b^2 e^2-23 b c d e+23 c^2 d^2\right )+B \left (128 b^2 c d e^2-48 b^3 e^3-103 b c^2 d^2 e+15 c^3 d^3\right )\right )}{c e \sqrt{x}}+i x \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (7 A c e \left (8 b^2 e^2-23 b c d e+23 c^2 d^2\right )+B \left (128 b^2 c d e^2-48 b^3 e^3-103 b c^2 d^2 e+15 c^3 d^3\right )\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )\right )}{105 c^3 \sqrt{x (b+c x)} \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(((7*A*c*e*(23*c^2*d^2 - 23*b*c*d*e + 8*b^2*e^2) + B*(15*c^3*d^3 - 103*b*c^2*d^2*e + 128*b^2*c*d*e^
2 - 48*b^3*e^3))*(b + c*x)*(d + e*x))/(c*e*Sqrt[x]) + Sqrt[x]*(b + c*x)*(d + e*x)*(7*A*c*e*(11*c*d - 4*b*e + 3
*c*e*x) + B*(24*b^2*e^2 - b*c*e*(61*d + 18*e*x) + 15*c^2*(3*d^2 + 3*d*e*x + e^2*x^2))) + I*Sqrt[b/c]*(7*A*c*e*
(23*c^2*d^2 - 23*b*c*d*e + 8*b^2*e^2) + B*(15*c^3*d^3 - 103*b*c^2*d^2*e + 128*b^2*c*d*e^2 - 48*b^3*e^3))*Sqrt[
1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x*EllipticE[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)] + (I*Sqrt[b/c]*(-(c*d) +
b*e)*(-105*A*c^3*d^2 + 48*b^3*B*e^2 - 8*b^2*c*e*(13*B*d + 7*A*e) + b*c^2*d*(60*B*d + 133*A*e))*Sqrt[1 + b/(c*
x)]*Sqrt[1 + d/(e*x)]*x*EllipticF[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)])/b))/(105*c^3*Sqrt[x*(b + c*x)]*S
qrt[d + e*x])

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Maple [B]  time = 0.05, size = 1610, 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((B*x+A)*(e*x+d)^(5/2)/(c*x^2+b*x)^(1/2),x)

[Out]

-2/105*(e*x+d)^(1/2)*(x*(c*x+b))^(1/2)*(3*B*x^4*b*c^4*e^4-58*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*
(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^3*d^3*e+176*B*((c*x+b)/b)^(1/2)*(-(e*x
+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*c*d*e^3+118*B*((c
*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*
b^2*c^3*d^3*e+56*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(
b*e/(b*e-c*d))^(1/2))*b*c^4*d^3*e-217*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*Elliptic
E(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c^2*d*e^3+322*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*
(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^3*d^2*e^2-161*A*((c*x+b)/b)^(1/2)*(-(e
*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^4*d^3*e-24*B*((
c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))
*b^4*c*d*e^3+67*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b
*e/(b*e-c*d))^(1/2))*b^3*c^2*d^2*e^2+28*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*Ellipt
icF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c^2*d*e^3-84*A*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)
*(-c*x/b)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^3*d^2*e^2+61*B*x*b^2*c^3*d^2*e^2-45*B
*x*b*c^4*d^3*e-29*B*x^2*b*c^4*d^2*e^2+28*A*x*b^2*c^3*d*e^3-77*A*x*b*c^4*d^2*e^2+19*B*x^3*b*c^4*d*e^3-70*A*x^2*
b*c^4*d*e^3+55*B*x^2*b^2*c^3*d*e^3-15*B*x^5*c^5*e^4-21*A*x^4*c^5*e^4-24*B*x*b^3*c^2*d*e^3-60*B*x^4*c^5*d*e^3+7
*A*x^3*b*c^4*e^4-98*A*x^3*c^5*d*e^3-6*B*x^3*b^2*c^3*e^4-90*B*x^3*c^5*d^2*e^2+28*A*x^2*b^2*c^3*e^4-77*A*x^2*c^5
*d^2*e^2-24*B*x^2*b^3*c^2*e^4-45*B*x^2*c^5*d^3*e+15*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^
(1/2)*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^4*d^4-15*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d
))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^4*d^4+56*A*((c*x+b)/b)^(1/2)*(-
(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*c*e^4-231*B*E
llipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*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)-48*B*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)*EllipticE(((c*x+b)/b)^(1/
2),(b*e/(b*e-c*d))^(1/2))*b^5*e^4)/e/c^5/x/(c*e*x^2+b*e*x+c*d*x+b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(5/2)/(c*x**2+b*x)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

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