3.1187 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=574 \[ -\frac {\left (b x+c x^2\right )^{3/2} \left (6 c e x (-10 A c e-b B e+12 B c d)+10 A c e (8 c d-7 b e)-B \left (3 b^2 e^2-92 b c d e+96 c^2 d^2\right )\right )}{48 c e^4}-\frac {\sqrt {b x+c x^2} \left (-2 c e x \left (8 b c e (6 B d-5 A e) (2 c d-b e)-\left (-3 b^2 e^2-8 b c d e+16 c^2 d^2\right ) (-10 A c e-b B e+12 B c d)\right )+10 A c e \left (-b^3 e^3+48 b^2 c d e^2-112 b c^2 d^2 e+64 c^3 d^3\right )-B \left (-3 b^4 e^4-20 b^3 c d e^3+656 b^2 c^2 d^2 e^2-1408 b c^3 d^3 e+768 c^4 d^4\right )\right )}{128 c^2 e^6}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (10 A c e \left (-b^4 e^4-16 b^3 c d e^3+144 b^2 c^2 d^2 e^2-256 b c^3 d^3 e+128 c^4 d^4\right )-B \left (-3 b^5 e^5-20 b^4 c d e^4-240 b^3 c^2 d^2 e^3+1920 b^2 c^3 d^3 e^2-3200 b c^4 d^4 e+1536 c^5 d^5\right )\right )}{128 c^{5/2} e^7}+\frac {d^{3/2} (c d-b e)^{3/2} (B d (12 c d-7 b e)-5 A e (2 c d-b e)) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{2 e^7}+\frac {\left (b x+c x^2\right )^{5/2} (-5 A e+6 B d+B e x)}{5 e^2 (d+e x)} \]

[Out]

-1/48*(10*A*c*e*(-7*b*e+8*c*d)-B*(3*b^2*e^2-92*b*c*d*e+96*c^2*d^2)+6*c*e*(-10*A*c*e-B*b*e+12*B*c*d)*x)*(c*x^2+
b*x)^(3/2)/c/e^4+1/5*(B*e*x-5*A*e+6*B*d)*(c*x^2+b*x)^(5/2)/e^2/(e*x+d)+1/128*(10*A*c*e*(-b^4*e^4-16*b^3*c*d*e^
3+144*b^2*c^2*d^2*e^2-256*b*c^3*d^3*e+128*c^4*d^4)-B*(-3*b^5*e^5-20*b^4*c*d*e^4-240*b^3*c^2*d^2*e^3+1920*b^2*c
^3*d^3*e^2-3200*b*c^4*d^4*e+1536*c^5*d^5))*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(5/2)/e^7+1/2*d^(3/2)*(-b*e+
c*d)^(3/2)*(B*d*(-7*b*e+12*c*d)-5*A*e*(-b*e+2*c*d))*arctanh(1/2*(b*d+(-b*e+2*c*d)*x)/d^(1/2)/(-b*e+c*d)^(1/2)/
(c*x^2+b*x)^(1/2))/e^7-1/128*(10*A*c*e*(-b^3*e^3+48*b^2*c*d*e^2-112*b*c^2*d^2*e+64*c^3*d^3)-B*(-3*b^4*e^4-20*b
^3*c*d*e^3+656*b^2*c^2*d^2*e^2-1408*b*c^3*d^3*e+768*c^4*d^4)-2*c*e*(8*b*c*e*(-5*A*e+6*B*d)*(-b*e+2*c*d)-(-10*A
*c*e-B*b*e+12*B*c*d)*(-3*b^2*e^2-8*b*c*d*e+16*c^2*d^2))*x)*(c*x^2+b*x)^(1/2)/c^2/e^6

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Rubi [A]  time = 0.93, antiderivative size = 574, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {812, 814, 843, 620, 206, 724} \[ -\frac {\left (b x+c x^2\right )^{3/2} \left (6 c e x (-10 A c e-b B e+12 B c d)+10 A c e (8 c d-7 b e)-B \left (3 b^2 e^2-92 b c d e+96 c^2 d^2\right )\right )}{48 c e^4}-\frac {\sqrt {b x+c x^2} \left (-2 c e x \left (8 b c e (6 B d-5 A e) (2 c d-b e)-\left (-3 b^2 e^2-8 b c d e+16 c^2 d^2\right ) (-10 A c e-b B e+12 B c d)\right )+10 A c e \left (48 b^2 c d e^2-b^3 e^3-112 b c^2 d^2 e+64 c^3 d^3\right )-B \left (656 b^2 c^2 d^2 e^2-20 b^3 c d e^3-3 b^4 e^4-1408 b c^3 d^3 e+768 c^4 d^4\right )\right )}{128 c^2 e^6}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (10 A c e \left (144 b^2 c^2 d^2 e^2-16 b^3 c d e^3-b^4 e^4-256 b c^3 d^3 e+128 c^4 d^4\right )-B \left (1920 b^2 c^3 d^3 e^2-240 b^3 c^2 d^2 e^3-20 b^4 c d e^4-3 b^5 e^5-3200 b c^4 d^4 e+1536 c^5 d^5\right )\right )}{128 c^{5/2} e^7}+\frac {d^{3/2} (c d-b e)^{3/2} (B d (12 c d-7 b e)-5 A e (2 c d-b e)) \tanh ^{-1}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{2 e^7}+\frac {\left (b x+c x^2\right )^{5/2} (-5 A e+6 B d+B e x)}{5 e^2 (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((10*A*c*e*(64*c^3*d^3 - 112*b*c^2*d^2*e + 48*b^2*c*d*e^2 - b^3*e^3) - B*(768*c^4*d^4 - 1408*b*c^3*d^3*e + 65
6*b^2*c^2*d^2*e^2 - 20*b^3*c*d*e^3 - 3*b^4*e^4) - 2*c*e*(8*b*c*e*(6*B*d - 5*A*e)*(2*c*d - b*e) - (12*B*c*d - b
*B*e - 10*A*c*e)*(16*c^2*d^2 - 8*b*c*d*e - 3*b^2*e^2))*x)*Sqrt[b*x + c*x^2])/(128*c^2*e^6) - ((10*A*c*e*(8*c*d
- 7*b*e) - B*(96*c^2*d^2 - 92*b*c*d*e + 3*b^2*e^2) + 6*c*e*(12*B*c*d - b*B*e - 10*A*c*e)*x)*(b*x + c*x^2)^(3/
2))/(48*c*e^4) + ((6*B*d - 5*A*e + B*e*x)*(b*x + c*x^2)^(5/2))/(5*e^2*(d + e*x)) + ((10*A*c*e*(128*c^4*d^4 - 2
56*b*c^3*d^3*e + 144*b^2*c^2*d^2*e^2 - 16*b^3*c*d*e^3 - b^4*e^4) - B*(1536*c^5*d^5 - 3200*b*c^4*d^4*e + 1920*b
^2*c^3*d^3*e^2 - 240*b^3*c^2*d^2*e^3 - 20*b^4*c*d*e^4 - 3*b^5*e^5))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(1
28*c^(5/2)*e^7) + (d^(3/2)*(c*d - b*e)^(3/2)*(B*d*(12*c*d - 7*b*e) - 5*A*e*(2*c*d - b*e))*ArcTanh[(b*d + (2*c*
d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(2*e^7)

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])

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 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

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]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{(d+e x)^2} \, dx &=\frac {(6 B d-5 A e+B e x) \left (b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}-\frac {\int \frac {(b (6 B d-5 A e)+(12 B c d-b B e-10 A c e) x) \left (b x+c x^2\right )^{3/2}}{d+e x} \, dx}{2 e^2}\\ &=-\frac {\left (10 A c e (8 c d-7 b e)-B \left (96 c^2 d^2-92 b c d e+3 b^2 e^2\right )+6 c e (12 B c d-b B e-10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{48 c e^4}+\frac {(6 B d-5 A e+B e x) \left (b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}+\frac {\int \frac {\left (\frac {1}{2} b d \left (10 A c e (8 c d-7 b e)-2 B \left (48 c^2 d^2-46 b c d e+\frac {3 b^2 e^2}{2}\right )\right )+\frac {1}{2} \left (8 b c e (6 B d-5 A e) (2 c d-b e)-2 (12 B c d-b B e-10 A c e) \left (8 c^2 d^2-4 b c d e-\frac {3 b^2 e^2}{2}\right )\right ) x\right ) \sqrt {b x+c x^2}}{d+e x} \, dx}{16 c e^4}\\ &=-\frac {\left (10 A c e \left (64 c^3 d^3-112 b c^2 d^2 e+48 b^2 c d e^2-b^3 e^3\right )-B \left (768 c^4 d^4-1408 b c^3 d^3 e+656 b^2 c^2 d^2 e^2-20 b^3 c d e^3-3 b^4 e^4\right )-2 c e \left (8 b c e (6 B d-5 A e) (2 c d-b e)-(12 B c d-b B e-10 A c e) \left (16 c^2 d^2-8 b c d e-3 b^2 e^2\right )\right ) x\right ) \sqrt {b x+c x^2}}{128 c^2 e^6}-\frac {\left (10 A c e (8 c d-7 b e)-B \left (96 c^2 d^2-92 b c d e+3 b^2 e^2\right )+6 c e (12 B c d-b B e-10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{48 c e^4}+\frac {(6 B d-5 A e+B e x) \left (b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}-\frac {\int \frac {-\frac {1}{4} b d \left (10 A c e \left (64 c^3 d^3-112 b c^2 d^2 e+48 b^2 c d e^2-b^3 e^3\right )-B \left (768 c^4 d^4-1408 b c^3 d^3 e+656 b^2 c^2 d^2 e^2-20 b^3 c d e^3-3 b^4 e^4\right )\right )-\frac {1}{4} \left (10 A c e \left (128 c^4 d^4-256 b c^3 d^3 e+144 b^2 c^2 d^2 e^2-16 b^3 c d e^3-b^4 e^4\right )-B \left (1536 c^5 d^5-3200 b c^4 d^4 e+1920 b^2 c^3 d^3 e^2-240 b^3 c^2 d^2 e^3-20 b^4 c d e^4-3 b^5 e^5\right )\right ) x}{(d+e x) \sqrt {b x+c x^2}} \, dx}{64 c^2 e^6}\\ &=-\frac {\left (10 A c e \left (64 c^3 d^3-112 b c^2 d^2 e+48 b^2 c d e^2-b^3 e^3\right )-B \left (768 c^4 d^4-1408 b c^3 d^3 e+656 b^2 c^2 d^2 e^2-20 b^3 c d e^3-3 b^4 e^4\right )-2 c e \left (8 b c e (6 B d-5 A e) (2 c d-b e)-(12 B c d-b B e-10 A c e) \left (16 c^2 d^2-8 b c d e-3 b^2 e^2\right )\right ) x\right ) \sqrt {b x+c x^2}}{128 c^2 e^6}-\frac {\left (10 A c e (8 c d-7 b e)-B \left (96 c^2 d^2-92 b c d e+3 b^2 e^2\right )+6 c e (12 B c d-b B e-10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{48 c e^4}+\frac {(6 B d-5 A e+B e x) \left (b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}+\frac {\left (d^2 (c d-b e)^2 (B d (12 c d-7 b e)-5 A e (2 c d-b e))\right ) \int \frac {1}{(d+e x) \sqrt {b x+c x^2}} \, dx}{2 e^7}+\frac {\left (10 A c e \left (128 c^4 d^4-256 b c^3 d^3 e+144 b^2 c^2 d^2 e^2-16 b^3 c d e^3-b^4 e^4\right )-B \left (1536 c^5 d^5-3200 b c^4 d^4 e+1920 b^2 c^3 d^3 e^2-240 b^3 c^2 d^2 e^3-20 b^4 c d e^4-3 b^5 e^5\right )\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{256 c^2 e^7}\\ &=-\frac {\left (10 A c e \left (64 c^3 d^3-112 b c^2 d^2 e+48 b^2 c d e^2-b^3 e^3\right )-B \left (768 c^4 d^4-1408 b c^3 d^3 e+656 b^2 c^2 d^2 e^2-20 b^3 c d e^3-3 b^4 e^4\right )-2 c e \left (8 b c e (6 B d-5 A e) (2 c d-b e)-(12 B c d-b B e-10 A c e) \left (16 c^2 d^2-8 b c d e-3 b^2 e^2\right )\right ) x\right ) \sqrt {b x+c x^2}}{128 c^2 e^6}-\frac {\left (10 A c e (8 c d-7 b e)-B \left (96 c^2 d^2-92 b c d e+3 b^2 e^2\right )+6 c e (12 B c d-b B e-10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{48 c e^4}+\frac {(6 B d-5 A e+B e x) \left (b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}-\frac {\left (d^2 (c d-b e)^2 (B d (12 c d-7 b e)-5 A e (2 c d-b e))\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d^2-4 b d e-x^2} \, dx,x,\frac {-b d-(2 c d-b e) x}{\sqrt {b x+c x^2}}\right )}{e^7}+\frac {\left (10 A c e \left (128 c^4 d^4-256 b c^3 d^3 e+144 b^2 c^2 d^2 e^2-16 b^3 c d e^3-b^4 e^4\right )-B \left (1536 c^5 d^5-3200 b c^4 d^4 e+1920 b^2 c^3 d^3 e^2-240 b^3 c^2 d^2 e^3-20 b^4 c d e^4-3 b^5 e^5\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{128 c^2 e^7}\\ &=-\frac {\left (10 A c e \left (64 c^3 d^3-112 b c^2 d^2 e+48 b^2 c d e^2-b^3 e^3\right )-B \left (768 c^4 d^4-1408 b c^3 d^3 e+656 b^2 c^2 d^2 e^2-20 b^3 c d e^3-3 b^4 e^4\right )-2 c e \left (8 b c e (6 B d-5 A e) (2 c d-b e)-(12 B c d-b B e-10 A c e) \left (16 c^2 d^2-8 b c d e-3 b^2 e^2\right )\right ) x\right ) \sqrt {b x+c x^2}}{128 c^2 e^6}-\frac {\left (10 A c e (8 c d-7 b e)-B \left (96 c^2 d^2-92 b c d e+3 b^2 e^2\right )+6 c e (12 B c d-b B e-10 A c e) x\right ) \left (b x+c x^2\right )^{3/2}}{48 c e^4}+\frac {(6 B d-5 A e+B e x) \left (b x+c x^2\right )^{5/2}}{5 e^2 (d+e x)}+\frac {\left (10 A c e \left (128 c^4 d^4-256 b c^3 d^3 e+144 b^2 c^2 d^2 e^2-16 b^3 c d e^3-b^4 e^4\right )-B \left (1536 c^5 d^5-3200 b c^4 d^4 e+1920 b^2 c^3 d^3 e^2-240 b^3 c^2 d^2 e^3-20 b^4 c d e^4-3 b^5 e^5\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{5/2} e^7}+\frac {d^{3/2} (c d-b e)^{3/2} (B d (12 c d-7 b e)-5 A e (2 c d-b e)) \tanh ^{-1}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{2 e^7}\\ \end {align*}

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Mathematica [A]  time = 3.20, size = 618, normalized size = 1.08 \[ \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (\frac {e \sqrt {x} \left (10 A c e \left (15 b^3 e^3 (d+e x)+2 b^2 c e^2 \left (-360 d^2-205 d e x+59 e^2 x^2\right )+8 b c^2 e \left (210 d^3+110 d^2 e x-35 d e^2 x^2+17 e^3 x^3\right )-16 c^3 \left (60 d^4+30 d^3 e x-10 d^2 e^2 x^2+5 d e^3 x^3-3 e^4 x^4\right )\right )+B \left (-45 b^4 e^4 (d+e x)+30 b^3 c e^3 \left (-10 d^2-9 d e x+e^2 x^2\right )+8 b^2 c^2 e^2 \left (1230 d^3+695 d^2 e x-202 d e^2 x^2+93 e^3 x^3\right )+16 b c^3 e \left (-1320 d^4-690 d^3 e x+220 d^2 e^2 x^2-107 d e^3 x^3+63 e^4 x^4\right )+192 c^4 \left (60 d^5+30 d^4 e x-10 d^3 e^2 x^2+5 d^2 e^3 x^3-3 d e^4 x^4+2 e^5 x^5\right )\right )\right )}{d+e x}+\frac {1920 c^2 d^{3/2} (c d-b e)^{3/2} (5 A e (b e-2 c d)+B d (12 c d-7 b e)) \tanh ^{-1}\left (\frac {\sqrt {x} \sqrt {c d-b e}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {b+c x}}\right )+\frac {15 \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right ) \left (B \left (3 b^5 e^5+20 b^4 c d e^4+240 b^3 c^2 d^2 e^3-1920 b^2 c^3 d^3 e^2+3200 b c^4 d^4 e-1536 c^5 d^5\right )-10 A c e \left (b^4 e^4+16 b^3 c d e^3-144 b^2 c^2 d^2 e^2+256 b c^3 d^3 e-128 c^4 d^4\right )\right )}{\sqrt {b} \sqrt {\frac {c x}{b}+1}}\right )}{1920 c^{5/2} e^7 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*((15*(-10*A*c*e*(-128*c^4*d^4 + 256*b*c^3*d^3*e - 144*b^2*c^2*d^2*e^2 + 16*b^3*c*d*e^3 + b^
4*e^4) + B*(-1536*c^5*d^5 + 3200*b*c^4*d^4*e - 1920*b^2*c^3*d^3*e^2 + 240*b^3*c^2*d^2*e^3 + 20*b^4*c*d*e^4 + 3
*b^5*e^5))*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[b]*Sqrt[1 + (c*x)/b]) + Sqrt[c]*((e*Sqrt[x]*(10*A*c*e*(15
*b^3*e^3*(d + e*x) + 2*b^2*c*e^2*(-360*d^2 - 205*d*e*x + 59*e^2*x^2) + 8*b*c^2*e*(210*d^3 + 110*d^2*e*x - 35*d
*e^2*x^2 + 17*e^3*x^3) - 16*c^3*(60*d^4 + 30*d^3*e*x - 10*d^2*e^2*x^2 + 5*d*e^3*x^3 - 3*e^4*x^4)) + B*(-45*b^4
*e^4*(d + e*x) + 30*b^3*c*e^3*(-10*d^2 - 9*d*e*x + e^2*x^2) + 8*b^2*c^2*e^2*(1230*d^3 + 695*d^2*e*x - 202*d*e^
2*x^2 + 93*e^3*x^3) + 16*b*c^3*e*(-1320*d^4 - 690*d^3*e*x + 220*d^2*e^2*x^2 - 107*d*e^3*x^3 + 63*e^4*x^4) + 19
2*c^4*(60*d^5 + 30*d^4*e*x - 10*d^3*e^2*x^2 + 5*d^2*e^3*x^3 - 3*d*e^4*x^4 + 2*e^5*x^5))))/(d + e*x) + (1920*c^
2*d^(3/2)*(c*d - b*e)^(3/2)*(B*d*(12*c*d - 7*b*e) + 5*A*e*(-2*c*d + b*e))*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(S
qrt[d]*Sqrt[b + c*x])])/Sqrt[b + c*x])))/(1920*c^(5/2)*e^7*Sqrt[x])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.06, size = 7095, normalized size = 12.36 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{{\left (d+e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{\left (d + e x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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