### 3.402 $$\int \frac{(b x+c x^2)^{5/2}}{(d+e x)^{5/2}} \, dx$$

Optimal. Leaf size=401 $\frac{4 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) \left (27 b^2 e^2-128 b c d e+128 c^2 d^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{21 \sqrt{c} e^6 \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} \left (51 b^2 e^2-48 c e x (2 c d-b e)-176 b c d e+128 c^2 d^2\right )}{21 e^5}-\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} (2 c d-b e) \left (3 b^2 e^2-128 b c d e+128 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{21 \sqrt{c} e^6 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{10 \left (b x+c x^2\right )^{3/2} (-7 b e+16 c d+2 c e x)}{21 e^3 \sqrt{d+e x}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^{3/2}}$

[Out]

(2*Sqrt[d + e*x]*(128*c^2*d^2 - 176*b*c*d*e + 51*b^2*e^2 - 48*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(21*e^5)
+ (10*(16*c*d - 7*b*e + 2*c*e*x)*(b*x + c*x^2)^(3/2))/(21*e^3*Sqrt[d + e*x]) - (2*(b*x + c*x^2)^(5/2))/(3*e*(
d + e*x)^(3/2)) - (2*Sqrt[-b]*(2*c*d - b*e)*(128*c^2*d^2 - 128*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)])/(21*Sqrt[c]*e^6*Sqrt[1 + (e*x)/d]*Sq
rt[b*x + c*x^2]) + (4*Sqrt[-b]*d*(c*d - b*e)*(128*c^2*d^2 - 128*b*c*d*e + 27*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)])/(21*Sqrt[c]*e^6*Sqrt[d + e*x]*
Sqrt[b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.476616, antiderivative size = 401, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.391, Rules used = {732, 812, 814, 843, 715, 112, 110, 117, 116} $\frac{2 \sqrt{b x+c x^2} \sqrt{d+e x} \left (51 b^2 e^2-48 c e x (2 c d-b e)-176 b c d e+128 c^2 d^2\right )}{21 e^5}+\frac{4 \sqrt{-b} d \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (c d-b e) \left (27 b^2 e^2-128 b c d e+128 c^2 d^2\right ) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{21 \sqrt{c} e^6 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} (2 c d-b e) \left (3 b^2 e^2-128 b c d e+128 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{21 \sqrt{c} e^6 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{10 \left (b x+c x^2\right )^{3/2} (-7 b e+16 c d+2 c e x)}{21 e^3 \sqrt{d+e x}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(128*c^2*d^2 - 176*b*c*d*e + 51*b^2*e^2 - 48*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(21*e^5)
+ (10*(16*c*d - 7*b*e + 2*c*e*x)*(b*x + c*x^2)^(3/2))/(21*e^3*Sqrt[d + e*x]) - (2*(b*x + c*x^2)^(5/2))/(3*e*(
d + e*x)^(3/2)) - (2*Sqrt[-b]*(2*c*d - b*e)*(128*c^2*d^2 - 128*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)])/(21*Sqrt[c]*e^6*Sqrt[1 + (e*x)/d]*Sq
rt[b*x + c*x^2]) + (4*Sqrt[-b]*d*(c*d - b*e)*(128*c^2*d^2 - 128*b*c*d*e + 27*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)])/(21*Sqrt[c]*e^6*Sqrt[d + e*x]*
Sqrt[b*x + c*x^2])

Rule 732

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

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 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 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{\left (b x+c x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^{3/2}}+\frac{5 \int \frac{(b+2 c x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx}{3 e}\\ &=\frac{10 (16 c d-7 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{21 e^3 \sqrt{d+e x}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^{3/2}}-\frac{10 \int \frac{\left (\frac{1}{2} b (16 c d-7 b e)+8 c (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{\sqrt{d+e x}} \, dx}{7 e^3}\\ &=\frac{2 \sqrt{d+e x} \left (128 c^2 d^2-176 b c d e+51 b^2 e^2-48 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{21 e^5}+\frac{10 (16 c d-7 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{21 e^3 \sqrt{d+e x}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^{3/2}}+\frac{4 \int \frac{-\frac{1}{4} b c d \left (128 c^2 d^2-176 b c d e+51 b^2 e^2\right )-\frac{1}{4} c (2 c d-b e) \left (128 c^2 d^2-128 b c d e+3 b^2 e^2\right ) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{21 c e^5}\\ &=\frac{2 \sqrt{d+e x} \left (128 c^2 d^2-176 b c d e+51 b^2 e^2-48 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{21 e^5}+\frac{10 (16 c d-7 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{21 e^3 \sqrt{d+e x}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^{3/2}}-\frac{\left ((2 c d-b e) \left (128 c^2 d^2-128 b c d e+3 b^2 e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{21 e^6}+\frac{\left (2 d (c d-b e) \left (128 c^2 d^2-128 b c d e+27 b^2 e^2\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{21 e^6}\\ &=\frac{2 \sqrt{d+e x} \left (128 c^2 d^2-176 b c d e+51 b^2 e^2-48 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{21 e^5}+\frac{10 (16 c d-7 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{21 e^3 \sqrt{d+e x}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^{3/2}}-\frac{\left ((2 c d-b e) \left (128 c^2 d^2-128 b c d e+3 b^2 e^2\right ) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{\sqrt{d+e x}}{\sqrt{x} \sqrt{b+c x}} \, dx}{21 e^6 \sqrt{b x+c x^2}}+\frac{\left (2 d (c d-b e) \left (128 c^2 d^2-128 b c d e+27 b^2 e^2\right ) \sqrt{x} \sqrt{b+c x}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x} \sqrt{d+e x}} \, dx}{21 e^6 \sqrt{b x+c x^2}}\\ &=\frac{2 \sqrt{d+e x} \left (128 c^2 d^2-176 b c d e+51 b^2 e^2-48 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{21 e^5}+\frac{10 (16 c d-7 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{21 e^3 \sqrt{d+e x}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^{3/2}}-\frac{\left ((2 c d-b e) \left (128 c^2 d^2-128 b c d e+3 b^2 e^2\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}{21 e^6 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{\left (2 d (c d-b e) \left (128 c^2 d^2-128 b c d e+27 b^2 e^2\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}{21 e^6 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=\frac{2 \sqrt{d+e x} \left (128 c^2 d^2-176 b c d e+51 b^2 e^2-48 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{21 e^5}+\frac{10 (16 c d-7 b e+2 c e x) \left (b x+c x^2\right )^{3/2}}{21 e^3 \sqrt{d+e x}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^{3/2}}-\frac{2 \sqrt{-b} (2 c d-b e) \left (128 c^2 d^2-128 b c d e+3 b^2 e^2\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 )}{21 \sqrt{c} e^6 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}+\frac{4 \sqrt{-b} d (c d-b e) \left (128 c^2 d^2-128 b c d e+27 b^2 e^2\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 )}{21 \sqrt{c} e^6 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 2.20526, size = 442, normalized size = 1.1 $\frac{2 (x (b+c x))^{5/2} \left (i e x \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (83 b^2 c d e^2-3 b^3 e^3-208 b c^2 d^2 e+128 c^3 d^3\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )+\frac{e \sqrt{x} (b+c x) \left (b^2 e^2 \left (51 d^2+67 d e x+9 e^2 x^2\right )+b c e \left (-224 d^2 e x-176 d^3-25 d e^2 x^2+9 e^3 x^3\right )+c^2 \left (16 d^2 e^2 x^2+160 d^3 e x+128 d^4-6 d e^3 x^3+3 e^4 x^4\right )\right )}{d+e x}-\frac{(b+c x) (d+e x) \left (134 b^2 c d e^2-3 b^3 e^3-384 b c^2 d^2 e+256 c^3 d^3\right )}{c \sqrt{x}}+i e x \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (-134 b^2 c d e^2+3 b^3 e^3+384 b c^2 d^2 e-256 c^3 d^3\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )\right )}{21 e^6 x^{5/2} (b+c x)^3 \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(x*(b + c*x))^(5/2)*(-(((256*c^3*d^3 - 384*b*c^2*d^2*e + 134*b^2*c*d*e^2 - 3*b^3*e^3)*(b + c*x)*(d + e*x))/
(c*Sqrt[x])) + (e*Sqrt[x]*(b + c*x)*(b^2*e^2*(51*d^2 + 67*d*e*x + 9*e^2*x^2) + b*c*e*(-176*d^3 - 224*d^2*e*x -
25*d*e^2*x^2 + 9*e^3*x^3) + c^2*(128*d^4 + 160*d^3*e*x + 16*d^2*e^2*x^2 - 6*d*e^3*x^3 + 3*e^4*x^4)))/(d + e*x
) + I*Sqrt[b/c]*e*(-256*c^3*d^3 + 384*b*c^2*d^2*e - 134*b^2*c*d*e^2 + 3*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]*e*(128*c^3*d^3 - 208*b*c^2*d^2*e +
83*b^2*c*d*e^2 - 3*b^3*e^3)*Sqrt[1 + b/(c*x)]*Sqrt[1 + d/(e*x)]*x*EllipticF[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*
d)/(b*e)]))/(21*e^6*x^(5/2)*(b + c*x)^3*Sqrt[d + e*x])

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Maple [B]  time = 0.3, size = 1692, normalized size = 4.2 \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+b*x)^(5/2)/(e*x+d)^(5/2),x)

[Out]

-2/21*(x*(c*x+b))^(1/2)*(-640*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^3*d^4*e*((c*x+b)/b)^(1/
2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+54*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*c*d^2
*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-310*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c
*d))^(1/2))*b^3*c^2*d^3*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+512*EllipticF(((c*x+
b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^3*d^4*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)
-12*x^5*b*c^4*e^5+6*x^5*c^5*d*e^4-18*x^4*b^2*c^3*e^5-16*x^4*c^5*d^2*e^3-9*x^3*b^3*c^2*e^5-160*x^3*c^5*d^3*e^2-
128*x^2*c^5*d^4*e-137*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*c*d^2*e^3*((c*x+b)/b)^(1/2)*(-(e*
x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-137*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^4*c*d*e^4*((
c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+518*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1
/2))*x*b^3*c^2*d^2*e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-640*EllipticE(((c*x+b)/b)
^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c^3*d^3*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+
256*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^4*d^4*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^
(1/2)*(-c*x/b)^(1/2)+54*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^4*c*d*e^4*((c*x+b)/b)^(1/2)*(-(
e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-310*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*c^2*d^2*
e^3*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+512*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*
d))^(1/2))*x*b^2*c^3*d^3*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-256*EllipticF(((c*x
+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^4*d^4*e*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2
)-128*x*b*c^4*d^4*e+31*x^4*b*c^4*d*e^4-42*x^3*b^2*c^3*d*e^4+208*x^3*b*c^4*d^2*e^3-67*x^2*b^3*c^2*d*e^4+173*x^2
*b^2*c^3*d^2*e^3+16*x^2*b*c^4*d^3*e^2-51*x*b^3*c^2*d^2*e^3+518*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/
2))*b^3*c^2*d^3*e^2*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+176*x*b^2*c^3*d^3*e^2-3*x^6*
c^5*e^5+3*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^5*e^5*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d)
)^(1/2)*(-c*x/b)^(1/2)+3*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^5*d*e^4*((c*x+b)/b)^(1/2)*(-(e*x
+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+256*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^4*d^5*((c*x+b
)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)-256*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*
b*c^4*d^5*((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^6

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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{5}{2}}}{{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x)^(5/2)/(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 (c^{2} x^{4} + 2 \, b c x^{3} + b^{2} 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((c*x^2+b*x)^(5/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + b^2*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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{5}{2}}}{{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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