3.329 \(\int \frac{(d+e x)^4}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=208 \[ -\frac{2 e \sqrt{b x+c x^2} (2 c d-b e) \left (-3 b^2 e^2-8 b c d e+8 c^2 d^2\right )}{3 b^4 c^2}+\frac{4 (d+e x) \left (x (2 c d-b e) \left (-b^2 e^2-4 b c d e+4 c^2 d^2\right )+b c d^2 (4 c d-5 b e)\right )}{3 b^4 c \sqrt{b x+c x^2}}-\frac{2 (d+e x)^3 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{2 e^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}} \]

[Out]

(-2*(d + e*x)^3*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (4*(d + e*x)*(b*c*d^2*(4*c*d - 5*b*e) +
(2*c*d - b*e)*(4*c^2*d^2 - 4*b*c*d*e - b^2*e^2)*x))/(3*b^4*c*Sqrt[b*x + c*x^2]) - (2*e*(2*c*d - b*e)*(8*c^2*d
^2 - 8*b*c*d*e - 3*b^2*e^2)*Sqrt[b*x + c*x^2])/(3*b^4*c^2) + (2*e^4*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^
(5/2)

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Rubi [A]  time = 0.198054, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {738, 818, 640, 620, 206} \[ -\frac{2 e \sqrt{b x+c x^2} (2 c d-b e) \left (-3 b^2 e^2-8 b c d e+8 c^2 d^2\right )}{3 b^4 c^2}+\frac{4 (d+e x) \left (x (2 c d-b e) \left (-b^2 e^2-4 b c d e+4 c^2 d^2\right )+b c d^2 (4 c d-5 b e)\right )}{3 b^4 c \sqrt{b x+c x^2}}-\frac{2 (d+e x)^3 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{2 e^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(d + e*x)^3*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (4*(d + e*x)*(b*c*d^2*(4*c*d - 5*b*e) +
(2*c*d - b*e)*(4*c^2*d^2 - 4*b*c*d*e - b^2*e^2)*x))/(3*b^4*c*Sqrt[b*x + c*x^2]) - (2*e*(2*c*d - b*e)*(8*c^2*d
^2 - 8*b*c*d*e - 3*b^2*e^2)*Sqrt[b*x + c*x^2])/(3*b^4*c^2) + (2*e^4*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^
(5/2)

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (d+e x)^3 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}-\frac{2 \int \frac{(d+e x)^2 (d (4 c d-5 b e)-e (2 c d-b e) x)}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b^2}\\ &=-\frac{2 (d+e x)^3 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{4 (d+e x) \left (b c d^2 (4 c d-5 b e)+(2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt{b x+c x^2}}-\frac{4 \int \frac{\frac{1}{2} b d e \left (8 c^2 d^2-12 b c d e+b^2 e^2\right )+\frac{1}{2} e (2 c d-b e) \left (8 c^2 d^2-8 b c d e-3 b^2 e^2\right ) x}{\sqrt{b x+c x^2}} \, dx}{3 b^4 c}\\ &=-\frac{2 (d+e x)^3 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{4 (d+e x) \left (b c d^2 (4 c d-5 b e)+(2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt{b x+c x^2}}-\frac{2 e (2 c d-b e) \left (8 c^2 d^2-8 b c d e-3 b^2 e^2\right ) \sqrt{b x+c x^2}}{3 b^4 c^2}+\frac{e^4 \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{c^2}\\ &=-\frac{2 (d+e x)^3 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{4 (d+e x) \left (b c d^2 (4 c d-5 b e)+(2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt{b x+c x^2}}-\frac{2 e (2 c d-b e) \left (8 c^2 d^2-8 b c d e-3 b^2 e^2\right ) \sqrt{b x+c x^2}}{3 b^4 c^2}+\frac{\left (2 e^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{c^2}\\ &=-\frac{2 (d+e x)^3 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{4 (d+e x) \left (b c d^2 (4 c d-5 b e)+(2 c d-b e) \left (4 c^2 d^2-4 b c d e-b^2 e^2\right ) x\right )}{3 b^4 c \sqrt{b x+c x^2}}-\frac{2 e (2 c d-b e) \left (8 c^2 d^2-8 b c d e-3 b^2 e^2\right ) \sqrt{b x+c x^2}}{3 b^4 c^2}+\frac{2 e^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}}\\ \end{align*}

Mathematica [C]  time = 2.01076, size = 389, normalized size = 1.87 \[ -\frac{\sqrt{\frac{c x}{b}+1} \left (-33792 b^2 c x (d+e x)^4 \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},2,2,2,\frac{7}{2}\right \},\left \{1,1,1,\frac{9}{2}\right \},-\frac{c x}{b}\right )-\frac{77 \left (\sqrt{-\frac{c x (b+c x)}{b^2}} \left (2 b^2 c x \left (3810 d^2 e^2 x^2+5060 d^3 e x+1895 d^4+20 d e^3 x^3-241 e^4 x^4\right )-3 b^3 \left (3810 d^2 e^2 x^2+5060 d^3 e x+1895 d^4+20 d e^3 x^3-241 e^4 x^4\right )+8 b c^2 x^2 \left (102 d^2 e^2 x^2-1588 d^3 e x-427 d^4+188 d e^3 x^3+77 e^4 x^4\right )-48 c^3 x^3 \left (138 d^2 e^2 x^2+84 d^3 e x-109 d^4+100 d e^3 x^3+27 e^4 x^4\right )\right )+3 b^3 \left (3810 d^2 e^2 x^2+5060 d^3 e x+1895 d^4+20 d e^3 x^3-241 e^4 x^4\right ) \sin ^{-1}\left (\sqrt{-\frac{c x}{b}}\right )\right )}{\left (-\frac{c x}{b}\right )^{5/2}}+21504 c^3 d x^3 (d+e x)^3 \, _2F_1\left (\frac{3}{2},\frac{11}{2};\frac{13}{2};-\frac{c x}{b}\right )\right )}{44352 b^5 x \sqrt{x (b+c x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d + e*x)^4/(b*x + c*x^2)^(5/2),x]

[Out]

-(Sqrt[1 + (c*x)/b]*((-77*(Sqrt[-((c*x*(b + c*x))/b^2)]*(-3*b^3*(1895*d^4 + 5060*d^3*e*x + 3810*d^2*e^2*x^2 +
20*d*e^3*x^3 - 241*e^4*x^4) + 2*b^2*c*x*(1895*d^4 + 5060*d^3*e*x + 3810*d^2*e^2*x^2 + 20*d*e^3*x^3 - 241*e^4*x
^4) - 48*c^3*x^3*(-109*d^4 + 84*d^3*e*x + 138*d^2*e^2*x^2 + 100*d*e^3*x^3 + 27*e^4*x^4) + 8*b*c^2*x^2*(-427*d^
4 - 1588*d^3*e*x + 102*d^2*e^2*x^2 + 188*d*e^3*x^3 + 77*e^4*x^4)) + 3*b^3*(1895*d^4 + 5060*d^3*e*x + 3810*d^2*
e^2*x^2 + 20*d*e^3*x^3 - 241*e^4*x^4)*ArcSin[Sqrt[-((c*x)/b)]]))/(-((c*x)/b))^(5/2) + 21504*c^3*d*x^3*(d + e*x
)^3*Hypergeometric2F1[3/2, 11/2, 13/2, -((c*x)/b)] - 33792*b^2*c*x*(d + e*x)^4*HypergeometricPFQ[{-1/2, 2, 2,
2, 7/2}, {1, 1, 1, 9/2}, -((c*x)/b)]))/(44352*b^5*x*Sqrt[x*(b + c*x)])

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Maple [B]  time = 0.055, size = 447, normalized size = 2.2 \begin{align*} -{\frac{{e}^{4}{x}^{3}}{3\,c} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{{e}^{4}b{x}^{2}}{2\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{{e}^{4}{b}^{2}x}{6\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{7\,{e}^{4}x}{3\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{{e}^{4}b}{6\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{{e}^{4}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}-4\,{\frac{d{e}^{3}{x}^{2}}{c \left ( c{x}^{2}+bx \right ) ^{3/2}}}-{\frac{4\,d{e}^{3}bx}{3\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{8\,d{e}^{3}x}{3\,bc}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{4\,d{e}^{3}}{3\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-4\,{\frac{{d}^{2}{e}^{2}x}{c \left ( c{x}^{2}+bx \right ) ^{3/2}}}+8\,{\frac{{d}^{2}{e}^{2}x}{{b}^{2}\sqrt{c{x}^{2}+bx}}}+4\,{\frac{{d}^{2}{e}^{2}}{bc\sqrt{c{x}^{2}+bx}}}+{\frac{8\,{d}^{3}ex}{3\,b} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{64\,{d}^{3}ecx}{3\,{b}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{32\,{d}^{3}e}{3\,{b}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{4\,{d}^{4}xc}{3\,{b}^{2}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,{d}^{4}}{3\,b} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}}+{\frac{32\,{c}^{2}{d}^{4}x}{3\,{b}^{4}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{16\,{d}^{4}c}{3\,{b}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(c*x^2+b*x)^(5/2),x)

[Out]

-1/3*e^4*x^3/c/(c*x^2+b*x)^(3/2)+1/2*e^4*b/c^2*x^2/(c*x^2+b*x)^(3/2)+1/6*e^4*b^2/c^3/(c*x^2+b*x)^(3/2)*x-7/3*e
^4/c^2/(c*x^2+b*x)^(1/2)*x-1/6*e^4*b/c^3/(c*x^2+b*x)^(1/2)+e^4/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2
))-4*d*e^3*x^2/c/(c*x^2+b*x)^(3/2)-4/3*d*e^3*b/c^2/(c*x^2+b*x)^(3/2)*x+8/3*d*e^3/b/c/(c*x^2+b*x)^(1/2)*x+4/3*d
*e^3/c^2/(c*x^2+b*x)^(1/2)-4*d^2*e^2/c/(c*x^2+b*x)^(3/2)*x+8*d^2*e^2/b^2/(c*x^2+b*x)^(1/2)*x+4*d^2*e^2/b/c/(c*
x^2+b*x)^(1/2)+8/3*d^3*e/b/(c*x^2+b*x)^(3/2)*x-64/3*d^3*e/b^3*c/(c*x^2+b*x)^(1/2)*x-32/3*d^3*e/b^2/(c*x^2+b*x)
^(1/2)-4/3*d^4/b^2/(c*x^2+b*x)^(3/2)*x*c-2/3*d^4/b/(c*x^2+b*x)^(3/2)+32/3*d^4*c^2/b^4/(c*x^2+b*x)^(1/2)*x+16/3
*d^4*c/b^3/(c*x^2+b*x)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.05773, size = 1050, normalized size = 5.05 \begin{align*} \left [\frac{3 \,{\left (b^{4} c^{2} e^{4} x^{4} + 2 \, b^{5} c e^{4} x^{3} + b^{6} e^{4} x^{2}\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (b^{3} c^{3} d^{4} - 4 \,{\left (4 \, c^{6} d^{4} - 8 \, b c^{5} d^{3} e + 3 \, b^{2} c^{4} d^{2} e^{2} + b^{3} c^{3} d e^{3} - b^{4} c^{2} e^{4}\right )} x^{3} - 3 \,{\left (8 \, b c^{5} d^{4} - 16 \, b^{2} c^{4} d^{3} e + 6 \, b^{3} c^{3} d^{2} e^{2} - b^{5} c e^{4}\right )} x^{2} - 6 \,{\left (b^{2} c^{4} d^{4} - 2 \, b^{3} c^{3} d^{3} e\right )} x\right )} \sqrt{c x^{2} + b x}}{3 \,{\left (b^{4} c^{5} x^{4} + 2 \, b^{5} c^{4} x^{3} + b^{6} c^{3} x^{2}\right )}}, -\frac{2 \,{\left (3 \,{\left (b^{4} c^{2} e^{4} x^{4} + 2 \, b^{5} c e^{4} x^{3} + b^{6} e^{4} x^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (b^{3} c^{3} d^{4} - 4 \,{\left (4 \, c^{6} d^{4} - 8 \, b c^{5} d^{3} e + 3 \, b^{2} c^{4} d^{2} e^{2} + b^{3} c^{3} d e^{3} - b^{4} c^{2} e^{4}\right )} x^{3} - 3 \,{\left (8 \, b c^{5} d^{4} - 16 \, b^{2} c^{4} d^{3} e + 6 \, b^{3} c^{3} d^{2} e^{2} - b^{5} c e^{4}\right )} x^{2} - 6 \,{\left (b^{2} c^{4} d^{4} - 2 \, b^{3} c^{3} d^{3} e\right )} x\right )} \sqrt{c x^{2} + b x}\right )}}{3 \,{\left (b^{4} c^{5} x^{4} + 2 \, b^{5} c^{4} x^{3} + b^{6} c^{3} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*(b^4*c^2*e^4*x^4 + 2*b^5*c*e^4*x^3 + b^6*e^4*x^2)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))
- 2*(b^3*c^3*d^4 - 4*(4*c^6*d^4 - 8*b*c^5*d^3*e + 3*b^2*c^4*d^2*e^2 + b^3*c^3*d*e^3 - b^4*c^2*e^4)*x^3 - 3*(8
*b*c^5*d^4 - 16*b^2*c^4*d^3*e + 6*b^3*c^3*d^2*e^2 - b^5*c*e^4)*x^2 - 6*(b^2*c^4*d^4 - 2*b^3*c^3*d^3*e)*x)*sqrt
(c*x^2 + b*x))/(b^4*c^5*x^4 + 2*b^5*c^4*x^3 + b^6*c^3*x^2), -2/3*(3*(b^4*c^2*e^4*x^4 + 2*b^5*c*e^4*x^3 + b^6*e
^4*x^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (b^3*c^3*d^4 - 4*(4*c^6*d^4 - 8*b*c^5*d^3*e + 3*b^
2*c^4*d^2*e^2 + b^3*c^3*d*e^3 - b^4*c^2*e^4)*x^3 - 3*(8*b*c^5*d^4 - 16*b^2*c^4*d^3*e + 6*b^3*c^3*d^2*e^2 - b^5
*c*e^4)*x^2 - 6*(b^2*c^4*d^4 - 2*b^3*c^3*d^3*e)*x)*sqrt(c*x^2 + b*x))/(b^4*c^5*x^4 + 2*b^5*c^4*x^3 + b^6*c^3*x
^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{4}}{\left (x \left (b + c x\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((d + e*x)**4/(x*(b + c*x))**(5/2), x)

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Giac [A]  time = 1.93715, size = 282, normalized size = 1.36 \begin{align*} -\frac{2 \,{\left (\frac{d^{4}}{b} -{\left (x{\left (\frac{4 \,{\left (4 \, c^{5} d^{4} - 8 \, b c^{4} d^{3} e + 3 \, b^{2} c^{3} d^{2} e^{2} + b^{3} c^{2} d e^{3} - b^{4} c e^{4}\right )} x}{b^{4} c^{2}} + \frac{3 \,{\left (8 \, b c^{4} d^{4} - 16 \, b^{2} c^{3} d^{3} e + 6 \, b^{3} c^{2} d^{2} e^{2} - b^{5} e^{4}\right )}}{b^{4} c^{2}}\right )} + \frac{6 \,{\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e\right )}}{b^{4} c^{2}}\right )} x\right )}}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}}} - \frac{e^{4} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

-2/3*(d^4/b - (x*(4*(4*c^5*d^4 - 8*b*c^4*d^3*e + 3*b^2*c^3*d^2*e^2 + b^3*c^2*d*e^3 - b^4*c*e^4)*x/(b^4*c^2) +
3*(8*b*c^4*d^4 - 16*b^2*c^3*d^3*e + 6*b^3*c^2*d^2*e^2 - b^5*e^4)/(b^4*c^2)) + 6*(b^2*c^3*d^4 - 2*b^3*c^2*d^3*e
)/(b^4*c^2))*x)/(c*x^2 + b*x)^(3/2) - e^4*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2)