### 3.403 $$\int \frac{(b x+c x^2)^{5/2}}{(d+e x)^{7/2}} \, dx$$

Optimal. Leaf size=392 $-\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (2 c d-b e) \left (15 b^2 e^2-128 b c d e+128 c^2 d^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right ),\frac{b e}{c d}\right )}{15 \sqrt{c} e^6 \sqrt{b x+c x^2} \sqrt{d+e x}}-\frac{2 \sqrt{b x+c x^2} \left (15 b^2 e^2+16 c e x (2 c d-b e)-112 b c d e+128 c^2 d^2\right )}{15 e^5 \sqrt{d+e x}}+\frac{4 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (23 b^2 e^2-128 b c d e+128 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 e^6 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \left (b x+c x^2\right )^{3/2} (-5 b e+16 c d+6 c e x)}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}$

[Out]

(-2*(128*c^2*d^2 - 112*b*c*d*e + 15*b^2*e^2 + 16*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(15*e^5*Sqrt[d + e*x]
) + (2*(16*c*d - 5*b*e + 6*c*e*x)*(b*x + c*x^2)^(3/2))/(15*e^3*(d + e*x)^(3/2)) - (2*(b*x + c*x^2)^(5/2))/(5*e
*(d + e*x)^(5/2)) + (4*Sqrt[-b]*Sqrt[c]*(128*c^2*d^2 - 128*b*c*d*e + 23*b^2*e^2)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqr
t[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*e^6*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*
x^2]) - (2*Sqrt[-b]*(2*c*d - b*e)*(128*c^2*d^2 - 128*b*c*d*e + 15*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^6*Sqrt[d + e*x]*Sqrt[b*x +
c*x^2])

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Rubi [A]  time = 0.429493, antiderivative size = 392, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.348, Rules used = {732, 812, 843, 715, 112, 110, 117, 116} $-\frac{2 \sqrt{b x+c x^2} \left (15 b^2 e^2+16 c e x (2 c d-b e)-112 b c d e+128 c^2 d^2\right )}{15 e^5 \sqrt{d+e x}}-\frac{2 \sqrt{-b} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{\frac{e x}{d}+1} (2 c d-b e) \left (15 b^2 e^2-128 b c d e+128 c^2 d^2\right ) F\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 \sqrt{c} e^6 \sqrt{b x+c x^2} \sqrt{d+e x}}+\frac{4 \sqrt{-b} \sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sqrt{d+e x} \left (23 b^2 e^2-128 b c d e+128 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{-b}}\right )|\frac{b e}{c d}\right )}{15 e^6 \sqrt{b x+c x^2} \sqrt{\frac{e x}{d}+1}}+\frac{2 \left (b x+c x^2\right )^{3/2} (-5 b e+16 c d+6 c e x)}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*x + c*x^2)^(5/2)/(d + e*x)^(7/2),x]

[Out]

(-2*(128*c^2*d^2 - 112*b*c*d*e + 15*b^2*e^2 + 16*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(15*e^5*Sqrt[d + e*x]
) + (2*(16*c*d - 5*b*e + 6*c*e*x)*(b*x + c*x^2)^(3/2))/(15*e^3*(d + e*x)^(3/2)) - (2*(b*x + c*x^2)^(5/2))/(5*e
*(d + e*x)^(5/2)) + (4*Sqrt[-b]*Sqrt[c]*(128*c^2*d^2 - 128*b*c*d*e + 23*b^2*e^2)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqr
t[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/(15*e^6*Sqrt[1 + (e*x)/d]*Sqrt[b*x + c*
x^2]) - (2*Sqrt[-b]*(2*c*d - b*e)*(128*c^2*d^2 - 128*b*c*d*e + 15*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^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 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)^{7/2}} \, dx &=-\frac{2 \left (b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{\int \frac{(b+2 c x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx}{e}\\ &=\frac{2 (16 c d-5 b e+6 c e x) \left (b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac{2 \int \frac{\left (\frac{1}{2} b (16 c d-5 b e)+8 c (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{(d+e x)^{3/2}} \, dx}{5 e^3}\\ &=-\frac{2 \left (128 c^2 d^2-112 b c d e+15 b^2 e^2+16 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{4 \int \frac{\frac{1}{4} b \left (128 c^2 d^2-112 b c d e+15 b^2 e^2\right )+\frac{1}{2} c \left (128 c^2 d^2-128 b c d e+23 b^2 e^2\right ) x}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{15 e^5}\\ &=-\frac{2 \left (128 c^2 d^2-112 b c d e+15 b^2 e^2+16 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac{\left ((2 c d-b e) \left (128 c^2 d^2-128 b c d e+15 b^2 e^2\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{b x+c x^2}} \, dx}{15 e^6}+\frac{\left (2 c \left (128 c^2 d^2-128 b c d e+23 b^2 e^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{b x+c x^2}} \, dx}{15 e^6}\\ &=-\frac{2 \left (128 c^2 d^2-112 b c d e+15 b^2 e^2+16 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac{\left ((2 c d-b e) \left (128 c^2 d^2-128 b c d e+15 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}{15 e^6 \sqrt{b x+c x^2}}+\frac{\left (2 c \left (128 c^2 d^2-128 b c d e+23 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}{15 e^6 \sqrt{b x+c x^2}}\\ &=-\frac{2 \left (128 c^2 d^2-112 b c d e+15 b^2 e^2+16 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{\left (2 c \left (128 c^2 d^2-128 b c d e+23 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}{15 e^6 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{\left ((2 c d-b e) \left (128 c^2 d^2-128 b c d e+15 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}{15 e^6 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ &=-\frac{2 \left (128 c^2 d^2-112 b c d e+15 b^2 e^2+16 c e (2 c d-b e) x\right ) \sqrt{b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{4 \sqrt{-b} \sqrt{c} \left (128 c^2 d^2-128 b c d e+23 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 )}{15 e^6 \sqrt{1+\frac{e x}{d}} \sqrt{b x+c x^2}}-\frac{2 \sqrt{-b} (2 c d-b e) \left (128 c^2 d^2-128 b c d e+15 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 )}{15 \sqrt{c} e^6 \sqrt{d+e x} \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [C]  time = 1.62915, size = 401, normalized size = 1.02 $\frac{2 (x (b+c x))^{5/2} \left (-i c e x \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (31 b^2 e^2-144 b c d e+128 c^2 d^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right ),\frac{c d}{b e}\right )-\frac{e \sqrt{x} (b+c x) \left (b^2 e^2 \left (15 d^2+35 d e x+23 e^2 x^2\right )-b c e \left (256 d^2 e x+112 d^3+161 d e^2 x^2+11 e^3 x^3\right )+c^2 \left (176 d^2 e^2 x^2+288 d^3 e x+128 d^4+10 d e^3 x^3-3 e^4 x^4\right )\right )}{(d+e x)^2}+\frac{2 (b+c x) (d+e x) \left (23 b^2 e^2-128 b c d e+128 c^2 d^2\right )}{\sqrt{x}}+2 i c e x \sqrt{\frac{b}{c}} \sqrt{\frac{b}{c x}+1} \sqrt{\frac{d}{e x}+1} \left (23 b^2 e^2-128 b c d e+128 c^2 d^2\right ) E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{b}{c}}}{\sqrt{x}}\right )|\frac{c d}{b e}\right )\right )}{15 e^6 x^{5/2} (b+c x)^3 \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*x + c*x^2)^(5/2)/(d + e*x)^(7/2),x]

[Out]

(2*(x*(b + c*x))^(5/2)*((2*(128*c^2*d^2 - 128*b*c*d*e + 23*b^2*e^2)*(b + c*x)*(d + e*x))/Sqrt[x] - (e*Sqrt[x]*
(b + c*x)*(b^2*e^2*(15*d^2 + 35*d*e*x + 23*e^2*x^2) - b*c*e*(112*d^3 + 256*d^2*e*x + 161*d*e^2*x^2 + 11*e^3*x^
3) + c^2*(128*d^4 + 288*d^3*e*x + 176*d^2*e^2*x^2 + 10*d*e^3*x^3 - 3*e^4*x^4)))/(d + e*x)^2 + (2*I)*Sqrt[b/c]*
c*e*(128*c^2*d^2 - 128*b*c*d*e + 23*b^2*e^2)*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*e*(128*c^2*d^2 - 144*b*c*d*e + 31*b^2*e^2)*Sqrt[1 + b/(c*x)]*Sqrt[1
+ d/(e*x)]*x*EllipticF[I*ArcSinh[Sqrt[b/c]/Sqrt[x]], (c*d)/(b*e)]))/(15*e^6*x^(5/2)*(b + c*x)^3*Sqrt[d + e*x])

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Maple [B]  time = 0.301, size = 2170, normalized size = 5.5 \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+b*x)^(5/2)/(e*x+d)^(7/2),x)

[Out]

2/15*(x*(c*x+b))^(1/2)*(126*x^3*b^2*c^2*d*e^4+80*x^3*b*c^3*d^2*e^3-35*x^2*b^3*c*d*e^4+241*x^2*b^2*c^2*d^2*e^3-
176*x^2*b*c^3*d^3*e^2+112*x*b^2*c^2*d^3*e^2-512*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^2*c^2
*d^2*e^3*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+256*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b
*e-c*d))^(1/2))*x^2*b*c^3*d^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-316*EllipticF(
((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^3*c*d^2*e^3*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d
))^(1/2)+768*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^2*c^2*d^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(
1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-512*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^3*d^4*e*(-c*x/b
)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+604*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*
x*b^3*c*d^2*e^3*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-1024*EllipticE(((c*x+b)/b)^(1/2)
,(b*e/(b*e-c*d))^(1/2))*x*b^2*c^2*d^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+512*El
lipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b*c^3*d^4*e*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b
*e-c*d))^(1/2)-158*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^3*c*d*e^4*(-c*x/b)^(1/2)*((c*x+b)/
b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+384*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^2*c^2*d^2*e
^3*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-15*x*b^3*c*d^2*e^3-128*x*b*c^3*d^4*e-46*Ellip
ticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*d^2*e^3*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*
d))^(1/2)+3*x^6*c^4*e^5+15*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^4*e^5*((c*x+b)/b)^(1/2)*(-
(e*x+d)*c/(b*e-c*d))^(1/2)*(-c*x/b)^(1/2)+151*x^4*b*c^3*d*e^4-158*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^
(1/2))*b^3*c*d^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+384*EllipticF(((c*x+b)/b)^(
1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^4*e*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+302*El
lipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^3*c*d^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b
*e-c*d))^(1/2)-512*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^2*c^2*d^4*e*(-c*x/b)^(1/2)*((c*x+b)/b)
^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-12*x^4*b^2*c^2*e^5-176*x^4*c^4*d^2*e^3-23*x^3*b^3*c*e^5-288*x^3*c^4*d^3*e^
2-128*x^2*c^4*d^4*e+15*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b^4*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)-92*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x*b^4*d*e^4*(-c*x/
b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+14*x^5*b*c^3*e^5-10*x^5*c^4*d*e^4-256*EllipticF(((c*x+
b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b*c^3*d^3*e^2*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(
1/2)+302*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^3*c*d*e^4*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(
-(e*x+d)*c/(b*e-c*d))^(1/2)-46*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*x^2*b^4*e^5*(-c*x/b)^(1/2)*(
(c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)-256*EllipticF(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))^(1/2))*b*c^3*d^5
*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+256*EllipticE(((c*x+b)/b)^(1/2),(b*e/(b*e-c*d))
^(1/2))*b*c^3*d^5*(-c*x/b)^(1/2)*((c*x+b)/b)^(1/2)*(-(e*x+d)*c/(b*e-c*d))^(1/2)+30*EllipticF(((c*x+b)/b)^(1/2)
,(b*e/(b*e-c*d))^(1/2))*x*b^4*d*e^4*((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)^(5/2)/e^6/c

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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{7}{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)^(7/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(5/2)/(e*x + d)^(7/2), 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^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}, 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)^(7/2),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^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^
2 + 4*d^3*e*x + d^4), 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)**(7/2),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{7}{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)^(7/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^(5/2)/(e*x + d)^(7/2), x)