3.1206 \(\int \frac{(A+B x) (d+e x)^4}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=341 \[ -\frac{e \sqrt{b x+c x^2} \left (4 b^2 c^2 d e (A e+2 B d)+2 b^3 c e^2 (3 A e+7 B d)-16 b c^3 d^2 (3 A e+B d)+32 A c^4 d^3-15 b^4 B e^3\right )}{3 b^4 c^3}-\frac{2 (d+e x) \left (b c d^2 \left (10 A b c e-8 A c^2 d+b^2 (-B) e+4 b B c d\right )-x \left (4 b^2 c^2 d e (A e+B d)+2 b^3 c e^2 (A e+3 B d)-8 b c^3 d^2 (3 A e+B d)+16 A c^4 d^3-5 b^4 B e^3\right )\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}-\frac{2 (d+e x)^3 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{e^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) (2 A c e-5 b B e+8 B c d)}{c^{7/2}} \]

[Out]

(-2*(d + e*x)^3*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(3*b^2*c*(b*x + c*x^2)^(3/2)) - (2*(d +
e*x)*(b*c*d^2*(4*b*B*c*d - 8*A*c^2*d - b^2*B*e + 10*A*b*c*e) - (16*A*c^4*d^3 - 5*b^4*B*e^3 + 4*b^2*c^2*d*e*(B
*d + A*e) + 2*b^3*c*e^2*(3*B*d + A*e) - 8*b*c^3*d^2*(B*d + 3*A*e))*x))/(3*b^4*c^2*Sqrt[b*x + c*x^2]) - (e*(32*
A*c^4*d^3 - 15*b^4*B*e^3 + 4*b^2*c^2*d*e*(2*B*d + A*e) - 16*b*c^3*d^2*(B*d + 3*A*e) + 2*b^3*c*e^2*(7*B*d + 3*A
*e))*Sqrt[b*x + c*x^2])/(3*b^4*c^3) + (e^3*(8*B*c*d - 5*b*B*e + 2*A*c*e)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]
])/c^(7/2)

________________________________________________________________________________________

Rubi [A]  time = 0.457871, antiderivative size = 341, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {818, 640, 620, 206} \[ -\frac{e \sqrt{b x+c x^2} \left (4 b^2 c^2 d e (A e+2 B d)+2 b^3 c e^2 (3 A e+7 B d)-16 b c^3 d^2 (3 A e+B d)+32 A c^4 d^3-15 b^4 B e^3\right )}{3 b^4 c^3}-\frac{2 (d+e x) \left (b c d^2 \left (10 A b c e-8 A c^2 d+b^2 (-B) e+4 b B c d\right )-x \left (4 b^2 c^2 d e (A e+B d)+2 b^3 c e^2 (A e+3 B d)-8 b c^3 d^2 (3 A e+B d)+16 A c^4 d^3-5 b^4 B e^3\right )\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}-\frac{2 (d+e x)^3 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{e^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) (2 A c e-5 b B e+8 B c d)}{c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^3*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(3*b^2*c*(b*x + c*x^2)^(3/2)) - (2*(d +
e*x)*(b*c*d^2*(4*b*B*c*d - 8*A*c^2*d - b^2*B*e + 10*A*b*c*e) - (16*A*c^4*d^3 - 5*b^4*B*e^3 + 4*b^2*c^2*d*e*(B
*d + A*e) + 2*b^3*c*e^2*(3*B*d + A*e) - 8*b*c^3*d^2*(B*d + 3*A*e))*x))/(3*b^4*c^2*Sqrt[b*x + c*x^2]) - (e*(32*
A*c^4*d^3 - 15*b^4*B*e^3 + 4*b^2*c^2*d*e*(2*B*d + A*e) - 16*b*c^3*d^2*(B*d + 3*A*e) + 2*b^3*c*e^2*(7*B*d + 3*A
*e))*Sqrt[b*x + c*x^2])/(3*b^4*c^3) + (e^3*(8*B*c*d - 5*b*B*e + 2*A*c*e)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]
])/c^(7/2)

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{(A+B x) (d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (d+e x)^3 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{2 \int \frac{(d+e x)^2 \left (\frac{1}{2} d \left (4 b B c d-8 A c^2 d-b^2 B e+10 A b c e\right )+\frac{1}{2} e \left (4 A c^2 d+5 b^2 B e-2 b c (B d+A e)\right ) x\right )}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b^2 c}\\ &=-\frac{2 (d+e x)^3 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}-\frac{2 (d+e x) \left (b c d^2 \left (4 b B c d-8 A c^2 d-b^2 B e+10 A b c e\right )-\left (16 A c^4 d^3-5 b^4 B e^3+4 b^2 c^2 d e (B d+A e)+2 b^3 c e^2 (3 B d+A e)-8 b c^3 d^2 (B d+3 A e)\right ) x\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}+\frac{4 \int \frac{-\frac{1}{4} b d e \left (16 A c^3 d^2-5 b^3 B e^2+2 b^2 c e (2 B d+A e)-8 b c^2 d (B d+3 A e)\right )-\frac{1}{4} e \left (32 A c^4 d^3-15 b^4 B e^3+4 b^2 c^2 d e (2 B d+A e)-16 b c^3 d^2 (B d+3 A e)+2 b^3 c e^2 (7 B d+3 A e)\right ) x}{\sqrt{b x+c x^2}} \, dx}{3 b^4 c^2}\\ &=-\frac{2 (d+e x)^3 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}-\frac{2 (d+e x) \left (b c d^2 \left (4 b B c d-8 A c^2 d-b^2 B e+10 A b c e\right )-\left (16 A c^4 d^3-5 b^4 B e^3+4 b^2 c^2 d e (B d+A e)+2 b^3 c e^2 (3 B d+A e)-8 b c^3 d^2 (B d+3 A e)\right ) x\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}-\frac{e \left (32 A c^4 d^3-15 b^4 B e^3+4 b^2 c^2 d e (2 B d+A e)-16 b c^3 d^2 (B d+3 A e)+2 b^3 c e^2 (7 B d+3 A e)\right ) \sqrt{b x+c x^2}}{3 b^4 c^3}+\frac{\left (e^3 (8 B c d-5 b B e+2 A c e)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{2 c^3}\\ &=-\frac{2 (d+e x)^3 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}-\frac{2 (d+e x) \left (b c d^2 \left (4 b B c d-8 A c^2 d-b^2 B e+10 A b c e\right )-\left (16 A c^4 d^3-5 b^4 B e^3+4 b^2 c^2 d e (B d+A e)+2 b^3 c e^2 (3 B d+A e)-8 b c^3 d^2 (B d+3 A e)\right ) x\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}-\frac{e \left (32 A c^4 d^3-15 b^4 B e^3+4 b^2 c^2 d e (2 B d+A e)-16 b c^3 d^2 (B d+3 A e)+2 b^3 c e^2 (7 B d+3 A e)\right ) \sqrt{b x+c x^2}}{3 b^4 c^3}+\frac{\left (e^3 (8 B c d-5 b B e+2 A c e)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{c^3}\\ &=-\frac{2 (d+e x)^3 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}-\frac{2 (d+e x) \left (b c d^2 \left (4 b B c d-8 A c^2 d-b^2 B e+10 A b c e\right )-\left (16 A c^4 d^3-5 b^4 B e^3+4 b^2 c^2 d e (B d+A e)+2 b^3 c e^2 (3 B d+A e)-8 b c^3 d^2 (B d+3 A e)\right ) x\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}-\frac{e \left (32 A c^4 d^3-15 b^4 B e^3+4 b^2 c^2 d e (2 B d+A e)-16 b c^3 d^2 (B d+3 A e)+2 b^3 c e^2 (7 B d+3 A e)\right ) \sqrt{b x+c x^2}}{3 b^4 c^3}+\frac{e^3 (8 B c d-5 b B e+2 A c e) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{7/2}}\\ \end{align*}

Mathematica [C]  time = 4.29732, size = 597, normalized size = 1.75 \[ \frac{\frac{A (b+c x) \sqrt{\frac{c x}{b}+1} \left (-33792 b^3 \left (-\frac{c x}{b}\right )^{7/2} (d+e x)^4 \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},2,2,2,\frac{7}{2}\right \},\left \{1,1,1,\frac{9}{2}\right \},-\frac{c x}{b}\right )+77 \left (\sqrt{-\frac{c x (b+c x)}{b^2}} \left (2 b^2 c x \left (3810 d^2 e^2 x^2+5060 d^3 e x+1895 d^4+20 d e^3 x^3-241 e^4 x^4\right )-3 b^3 \left (3810 d^2 e^2 x^2+5060 d^3 e x+1895 d^4+20 d e^3 x^3-241 e^4 x^4\right )+8 b c^2 x^2 \left (102 d^2 e^2 x^2-1588 d^3 e x-427 d^4+188 d e^3 x^3+77 e^4 x^4\right )-48 c^3 x^3 \left (138 d^2 e^2 x^2+84 d^3 e x-109 d^4+100 d e^3 x^3+27 e^4 x^4\right )\right )+3 b^3 \left (3810 d^2 e^2 x^2+5060 d^3 e x+1895 d^4+20 d e^3 x^3-241 e^4 x^4\right ) \sin ^{-1}\left (\sqrt{-\frac{c x}{b}}\right )\right )-21504 c^3 d x^3 \left (-\frac{c x}{b}\right )^{5/2} (d+e x)^3 \, _2F_1\left (\frac{3}{2},\frac{11}{2};\frac{13}{2};-\frac{c x}{b}\right )\right )}{\left (-\frac{c x}{b}\right )^{5/2}}+\frac{14784 b^2 B x \left (\sqrt{c} \left (-6 b^2 c^3 d^2 \left (d^2-4 d e x-2 e^2 x^2\right )+b^3 c^2 e^3 x^2 (3 e x-32 d)+4 b^4 c e^3 x (5 e x-6 d)+15 b^5 e^4 x+8 b c^4 d^3 x (2 e x-3 d)-16 c^5 d^4 x^2\right )-3 b^{7/2} e^3 \sqrt{x} (b+c x) \sqrt{\frac{c x}{b}+1} (5 b e-8 c d) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )\right )}{c^{7/2}}}{44352 b^5 (x (b+c x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((14784*b^2*B*x*(Sqrt[c]*(15*b^5*e^4*x - 16*c^5*d^4*x^2 + 8*b*c^4*d^3*x*(-3*d + 2*e*x) + b^3*c^2*e^3*x^2*(-32*
d + 3*e*x) + 4*b^4*c*e^3*x*(-6*d + 5*e*x) - 6*b^2*c^3*d^2*(d^2 - 4*d*e*x - 2*e^2*x^2)) - 3*b^(7/2)*e^3*(-8*c*d
+ 5*b*e)*Sqrt[x]*(b + c*x)*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]]))/c^(7/2) + (A*(b + c*x)*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)]]) - 21504*c^3*d*x^3*(-((c*x)/b))^(5/2)*(d + e*x)^3*Hyper
geometric2F1[3/2, 11/2, 13/2, -((c*x)/b)] - 33792*b^3*(-((c*x)/b))^(7/2)*(d + e*x)^4*HypergeometricPFQ[{-1/2,
2, 2, 2, 7/2}, {1, 1, 1, 9/2}, -((c*x)/b)]))/(-((c*x)/b))^(5/2))/(44352*b^5*(x*(b + c*x))^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.014, size = 1026, normalized size = 3. \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)^4/(c*x^2+b*x)^(5/2),x)

[Out]

-4/3*b/c^2/(c*x^2+b*x)^(3/2)*x*A*d*e^3-2*b/c^2/(c*x^2+b*x)^(3/2)*x*B*d^2*e^2+8/3/b/c/(c*x^2+b*x)^(1/2)*x*A*d*e
^3+4/b/c/(c*x^2+b*x)^(1/2)*x*B*d^2*e^2-64/3/b^3*c/(c*x^2+b*x)^(1/2)*x*A*d^3*e+2/3*b^2/c^3/(c*x^2+b*x)^(3/2)*x*
B*d*e^3+2*b/c^2*x^2/(c*x^2+b*x)^(3/2)*B*d*e^3-6*x^2/c/(c*x^2+b*x)^(3/2)*B*d^2*e^2-4/c/(c*x^2+b*x)^(3/2)*x*A*d^
2*e^2-8/3/c/(c*x^2+b*x)^(3/2)*x*B*d^3*e+8/b^2/(c*x^2+b*x)^(1/2)*x*A*d^2*e^2+16/3/b^2/(c*x^2+b*x)^(1/2)*x*B*d^3
*e+4/b/c/(c*x^2+b*x)^(1/2)*A*d^2*e^2+8/3/b/c/(c*x^2+b*x)^(1/2)*B*d^3*e-5/12*B*e^4*b^3/c^4/(c*x^2+b*x)^(3/2)*x+
35/6*B*e^4*b/c^3/(c*x^2+b*x)^(1/2)*x+5/6*B*e^4*b/c^2*x^3/(c*x^2+b*x)^(3/2)-5/4*B*e^4*b^2/c^3*x^2/(c*x^2+b*x)^(
3/2)+32/3*A*d^4*c^2/b^4/(c*x^2+b*x)^(1/2)*x-4/3*x^3/c/(c*x^2+b*x)^(3/2)*B*d*e^3+1/2*b/c^2*x^2/(c*x^2+b*x)^(3/2
)*A*e^4+1/6*b^2/c^3/(c*x^2+b*x)^(3/2)*x*A*e^4-28/3/c^2/(c*x^2+b*x)^(1/2)*x*B*d*e^3-2/3*b/c^3/(c*x^2+b*x)^(1/2)
*B*d*e^3-4*x^2/c/(c*x^2+b*x)^(3/2)*A*d*e^3+8/3/b/(c*x^2+b*x)^(3/2)*x*A*d^3*e-16/3/b^3*c/(c*x^2+b*x)^(1/2)*x*B*
d^4-4/3*A*d^4/b^2/(c*x^2+b*x)^(3/2)*x*c-8/3/b^2/(c*x^2+b*x)^(1/2)*B*d^4-2/3*A*d^4/b/(c*x^2+b*x)^(3/2)+1/c^(5/2
)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*A*e^4-5/2*B*e^4*b/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2)
)-1/3*x^3/c/(c*x^2+b*x)^(3/2)*A*e^4+16/3*A*d^4*c/b^3/(c*x^2+b*x)^(1/2)-7/3/c^2/(c*x^2+b*x)^(1/2)*x*A*e^4-1/6*b
/c^3/(c*x^2+b*x)^(1/2)*A*e^4+4/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d*e^3+4/3/c^2/(c*x^2+b*x)^(
1/2)*A*d*e^3+2/c^2/(c*x^2+b*x)^(1/2)*B*d^2*e^2+B*e^4*x^4/c/(c*x^2+b*x)^(3/2)+5/12*B*e^4*b^2/c^4/(c*x^2+b*x)^(1
/2)-32/3/b^2/(c*x^2+b*x)^(1/2)*A*d^3*e+2/3/b/(c*x^2+b*x)^(3/2)*x*B*d^4

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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((B*x+A)*(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 = 2.0551, size = 1963, normalized size = 5.76 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*((8*B*b^4*c^3*d*e^3 - (5*B*b^5*c^2 - 2*A*b^4*c^3)*e^4)*x^4 + 2*(8*B*b^5*c^2*d*e^3 - (5*B*b^6*c - 2*A*b
^5*c^2)*e^4)*x^3 + (8*B*b^6*c*d*e^3 - (5*B*b^7 - 2*A*b^6*c)*e^4)*x^2)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b
*x)*sqrt(c)) + 2*(3*B*b^4*c^3*e^4*x^4 - 2*A*b^3*c^4*d^4 - 4*(4*(B*b*c^6 - 2*A*c^7)*d^4 - 4*(B*b^2*c^5 - 4*A*b*
c^6)*d^3*e - 3*(B*b^3*c^4 + 2*A*b^2*c^5)*d^2*e^2 + 2*(4*B*b^4*c^3 - A*b^3*c^4)*d*e^3 - (5*B*b^5*c^2 - 2*A*b^4*
c^3)*e^4)*x^3 + 3*(12*A*b^3*c^4*d^2*e^2 - 8*B*b^5*c^2*d*e^3 - 8*(B*b^2*c^5 - 2*A*b*c^6)*d^4 + 8*(B*b^3*c^4 - 4
*A*b^2*c^5)*d^3*e + (5*B*b^6*c - 2*A*b^5*c^2)*e^4)*x^2 - 6*(4*A*b^3*c^4*d^3*e + (B*b^3*c^4 - 2*A*b^2*c^5)*d^4)
*x)*sqrt(c*x^2 + b*x))/(b^4*c^6*x^4 + 2*b^5*c^5*x^3 + b^6*c^4*x^2), -1/3*(3*((8*B*b^4*c^3*d*e^3 - (5*B*b^5*c^2
- 2*A*b^4*c^3)*e^4)*x^4 + 2*(8*B*b^5*c^2*d*e^3 - (5*B*b^6*c - 2*A*b^5*c^2)*e^4)*x^3 + (8*B*b^6*c*d*e^3 - (5*B
*b^7 - 2*A*b^6*c)*e^4)*x^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (3*B*b^4*c^3*e^4*x^4 - 2*A*b^3
*c^4*d^4 - 4*(4*(B*b*c^6 - 2*A*c^7)*d^4 - 4*(B*b^2*c^5 - 4*A*b*c^6)*d^3*e - 3*(B*b^3*c^4 + 2*A*b^2*c^5)*d^2*e^
2 + 2*(4*B*b^4*c^3 - A*b^3*c^4)*d*e^3 - (5*B*b^5*c^2 - 2*A*b^4*c^3)*e^4)*x^3 + 3*(12*A*b^3*c^4*d^2*e^2 - 8*B*b
^5*c^2*d*e^3 - 8*(B*b^2*c^5 - 2*A*b*c^6)*d^4 + 8*(B*b^3*c^4 - 4*A*b^2*c^5)*d^3*e + (5*B*b^6*c - 2*A*b^5*c^2)*e
^4)*x^2 - 6*(4*A*b^3*c^4*d^3*e + (B*b^3*c^4 - 2*A*b^2*c^5)*d^4)*x)*sqrt(c*x^2 + b*x))/(b^4*c^6*x^4 + 2*b^5*c^5
*x^3 + b^6*c^4*x^2)]

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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 )^{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((B*x+A)*(e*x+d)**4/(c*x**2+b*x)**(5/2),x)

[Out]

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

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Giac [A]  time = 1.74288, size = 500, normalized size = 1.47 \begin{align*} -\frac{\frac{2 \, A d^{4}}{b} -{\left ({\left ({\left (\frac{3 \, B x e^{4}}{c} - \frac{4 \,{\left (4 \, B b c^{5} d^{4} - 8 \, A c^{6} d^{4} - 4 \, B b^{2} c^{4} d^{3} e + 16 \, A b c^{5} d^{3} e - 3 \, B b^{3} c^{3} d^{2} e^{2} - 6 \, A b^{2} c^{4} d^{2} e^{2} + 8 \, B b^{4} c^{2} d e^{3} - 2 \, A b^{3} c^{3} d e^{3} - 5 \, B b^{5} c e^{4} + 2 \, A b^{4} c^{2} e^{4}\right )}}{b^{4} c^{3}}\right )} x - \frac{3 \,{\left (8 \, B b^{2} c^{4} d^{4} - 16 \, A b c^{5} d^{4} - 8 \, B b^{3} c^{3} d^{3} e + 32 \, A b^{2} c^{4} d^{3} e - 12 \, A b^{3} c^{3} d^{2} e^{2} + 8 \, B b^{5} c d e^{3} - 5 \, B b^{6} e^{4} + 2 \, A b^{5} c e^{4}\right )}}{b^{4} c^{3}}\right )} x - \frac{6 \,{\left (B b^{3} c^{3} d^{4} - 2 \, A b^{2} c^{4} d^{4} + 4 \, A b^{3} c^{3} d^{3} e\right )}}{b^{4} c^{3}}\right )} x}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}}} - \frac{{\left (8 \, B c d e^{3} - 5 \, B b e^{4} + 2 \, A c e^{4}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{2 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*A*d^4/b - (((3*B*x*e^4/c - 4*(4*B*b*c^5*d^4 - 8*A*c^6*d^4 - 4*B*b^2*c^4*d^3*e + 16*A*b*c^5*d^3*e - 3*B
*b^3*c^3*d^2*e^2 - 6*A*b^2*c^4*d^2*e^2 + 8*B*b^4*c^2*d*e^3 - 2*A*b^3*c^3*d*e^3 - 5*B*b^5*c*e^4 + 2*A*b^4*c^2*e
^4)/(b^4*c^3))*x - 3*(8*B*b^2*c^4*d^4 - 16*A*b*c^5*d^4 - 8*B*b^3*c^3*d^3*e + 32*A*b^2*c^4*d^3*e - 12*A*b^3*c^3
*d^2*e^2 + 8*B*b^5*c*d*e^3 - 5*B*b^6*e^4 + 2*A*b^5*c*e^4)/(b^4*c^3))*x - 6*(B*b^3*c^3*d^4 - 2*A*b^2*c^4*d^4 +
4*A*b^3*c^3*d^3*e)/(b^4*c^3))*x)/(c*x^2 + b*x)^(3/2) - 1/2*(8*B*c*d*e^3 - 5*B*b*e^4 + 2*A*c*e^4)*log(abs(-2*(s
qrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)