3.11.62 \(\int \frac {(A+B x) (d+e x)^3}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=220 \[ -\frac {2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac {2 \left (b c d^2 \left (-4 b c (2 A e+B d)+8 A c^2 d+b^2 B e\right )+x \left (2 b^2 c^2 d e (4 A e+3 B d)-8 b c^3 d^2 (3 A e+B d)+16 A c^4 d^3-3 b^4 B e^3+2 b^3 B c d e^2\right )\right )}{3 b^4 c^2 \sqrt {b x+c x^2}}+\frac {2 B e^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \]

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Rubi [A]  time = 0.20, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {818, 777, 620, 206} \begin {gather*} \frac {2 \left (x \left (2 b^2 c^2 d e (4 A e+3 B d)-8 b c^3 d^2 (3 A e+B d)+16 A c^4 d^3+2 b^3 B c d e^2-3 b^4 B e^3\right )+b c d^2 \left (-4 b c (2 A e+B d)+8 A c^2 d+b^2 B e\right )\right )}{3 b^4 c^2 \sqrt {b x+c x^2}}-\frac {2 (d+e x)^2 \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 {2 B e^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^3)/(b*x + c*x^2)^(5/2),x]

[Out]

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

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 777

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((2
*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x)*(a + b*x + c*x^2)^
(p + 1))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p
+ 3))/(c*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N
eQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

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])

Rubi steps

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

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Mathematica [A]  time = 5.48, size = 199, normalized size = 0.90 \begin {gather*} \frac {x^{5/2} \left (\frac {2 B e^3 (b+c x)^{5/2} \log \left (\sqrt {c} \sqrt {b+c x}+c \sqrt {x}\right )}{c^{5/2}}-\frac {2 (b+c x) \left (x^2 (b+c x) (c d-b e)^2 \left (b c (5 B d-A e)-8 A c^2 d+4 b^2 B e\right )+c^2 d^2 x (b+c x)^2 (9 A b e-8 A c d+3 b B d)+b x^2 (b B-A c) (c d-b e)^3+A b c^2 d^3 (b+c x)^2\right )}{3 b^4 c^2 x^{3/2}}\right )}{(x (b+c x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^3)/(b*x + c*x^2)^(5/2),x]

[Out]

(x^(5/2)*((-2*(b + c*x)*(b*(b*B - A*c)*(c*d - b*e)^3*x^2 + (c*d - b*e)^2*(-8*A*c^2*d + 4*b^2*B*e + b*c*(5*B*d
- A*e))*x^2*(b + c*x) + A*b*c^2*d^3*(b + c*x)^2 + c^2*d^2*(3*b*B*d - 8*A*c*d + 9*A*b*e)*x*(b + c*x)^2))/(3*b^4
*c^2*x^(3/2)) + (2*B*e^3*(b + c*x)^(5/2)*Log[c*Sqrt[x] + Sqrt[c]*Sqrt[b + c*x]])/c^(5/2)))/(x*(b + c*x))^(5/2)

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IntegrateAlgebraic [A]  time = 0.96, size = 333, normalized size = 1.51 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (A b^3 c^2 d^3+9 A b^3 c^2 d^2 e x-9 A b^3 c^2 d e^2 x^2-A b^3 c^2 e^3 x^3-6 A b^2 c^3 d^3 x+36 A b^2 c^3 d^2 e x^2-6 A b^2 c^3 d e^2 x^3-24 A b c^4 d^3 x^2+24 A b c^4 d^2 e x^3-16 A c^5 d^3 x^3+3 b^5 B e^3 x^2+4 b^4 B c e^3 x^3+3 b^3 B c^2 d^3 x-9 b^3 B c^2 d^2 e x^2-3 b^3 B c^2 d e^2 x^3+12 b^2 B c^3 d^3 x^2-6 b^2 B c^3 d^2 e x^3+8 b B c^4 d^3 x^3\right )}{3 b^4 c^2 x^2 (b+c x)^2}-\frac {B e^3 \log \left (-2 c^{5/2} \sqrt {b x+c x^2}+b c^2+2 c^3 x\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^3)/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*Sqrt[b*x + c*x^2]*(A*b^3*c^2*d^3 + 3*b^3*B*c^2*d^3*x - 6*A*b^2*c^3*d^3*x + 9*A*b^3*c^2*d^2*e*x + 12*b^2*B*
c^3*d^3*x^2 - 24*A*b*c^4*d^3*x^2 - 9*b^3*B*c^2*d^2*e*x^2 + 36*A*b^2*c^3*d^2*e*x^2 - 9*A*b^3*c^2*d*e^2*x^2 + 3*
b^5*B*e^3*x^2 + 8*b*B*c^4*d^3*x^3 - 16*A*c^5*d^3*x^3 - 6*b^2*B*c^3*d^2*e*x^3 + 24*A*b*c^4*d^2*e*x^3 - 3*b^3*B*
c^2*d*e^2*x^3 - 6*A*b^2*c^3*d*e^2*x^3 + 4*b^4*B*c*e^3*x^3 - A*b^3*c^2*e^3*x^3))/(3*b^4*c^2*x^2*(b + c*x)^2) -
(B*e^3*Log[b*c^2 + 2*c^3*x - 2*c^(5/2)*Sqrt[b*x + c*x^2]])/c^(5/2)

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fricas [A]  time = 0.43, size = 671, normalized size = 3.05 \begin {gather*} \left [\frac {3 \, {\left (B b^{4} c^{2} e^{3} x^{4} + 2 \, B b^{5} c e^{3} x^{3} + B b^{6} e^{3} x^{2}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (A b^{3} c^{3} d^{3} + {\left (8 \, {\left (B b c^{5} - 2 \, A c^{6}\right )} d^{3} - 6 \, {\left (B b^{2} c^{4} - 4 \, A b c^{5}\right )} d^{2} e - 3 \, {\left (B b^{3} c^{3} + 2 \, A b^{2} c^{4}\right )} d e^{2} + {\left (4 \, B b^{4} c^{2} - A b^{3} c^{3}\right )} e^{3}\right )} x^{3} - 3 \, {\left (3 \, A b^{3} c^{3} d e^{2} - B b^{5} c e^{3} - 4 \, {\left (B b^{2} c^{4} - 2 \, A b c^{5}\right )} d^{3} + 3 \, {\left (B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} d^{2} e\right )} x^{2} + 3 \, {\left (3 \, A b^{3} c^{3} d^{2} e + {\left (B b^{3} c^{3} - 2 \, A b^{2} c^{4}\right )} d^{3}\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 b^{4} c^{2} e^{3} x^{4} + 2 \, B b^{5} c e^{3} x^{3} + B b^{6} e^{3} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (A b^{3} c^{3} d^{3} + {\left (8 \, {\left (B b c^{5} - 2 \, A c^{6}\right )} d^{3} - 6 \, {\left (B b^{2} c^{4} - 4 \, A b c^{5}\right )} d^{2} e - 3 \, {\left (B b^{3} c^{3} + 2 \, A b^{2} c^{4}\right )} d e^{2} + {\left (4 \, B b^{4} c^{2} - A b^{3} c^{3}\right )} e^{3}\right )} x^{3} - 3 \, {\left (3 \, A b^{3} c^{3} d e^{2} - B b^{5} c e^{3} - 4 \, {\left (B b^{2} c^{4} - 2 \, A b c^{5}\right )} d^{3} + 3 \, {\left (B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} d^{2} e\right )} x^{2} + 3 \, {\left (3 \, A b^{3} c^{3} d^{2} e + {\left (B b^{3} c^{3} - 2 \, A b^{2} c^{4}\right )} d^{3}\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*(B*b^4*c^2*e^3*x^4 + 2*B*b^5*c*e^3*x^3 + B*b^6*e^3*x^2)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sq
rt(c)) - 2*(A*b^3*c^3*d^3 + (8*(B*b*c^5 - 2*A*c^6)*d^3 - 6*(B*b^2*c^4 - 4*A*b*c^5)*d^2*e - 3*(B*b^3*c^3 + 2*A*
b^2*c^4)*d*e^2 + (4*B*b^4*c^2 - A*b^3*c^3)*e^3)*x^3 - 3*(3*A*b^3*c^3*d*e^2 - B*b^5*c*e^3 - 4*(B*b^2*c^4 - 2*A*
b*c^5)*d^3 + 3*(B*b^3*c^3 - 4*A*b^2*c^4)*d^2*e)*x^2 + 3*(3*A*b^3*c^3*d^2*e + (B*b^3*c^3 - 2*A*b^2*c^4)*d^3)*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*b^4*c^2*e^3*x^4 + 2*B*b^5*c*e^3*x^
3 + B*b^6*e^3*x^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (A*b^3*c^3*d^3 + (8*(B*b*c^5 - 2*A*c^6)
*d^3 - 6*(B*b^2*c^4 - 4*A*b*c^5)*d^2*e - 3*(B*b^3*c^3 + 2*A*b^2*c^4)*d*e^2 + (4*B*b^4*c^2 - A*b^3*c^3)*e^3)*x^
3 - 3*(3*A*b^3*c^3*d*e^2 - B*b^5*c*e^3 - 4*(B*b^2*c^4 - 2*A*b*c^5)*d^3 + 3*(B*b^3*c^3 - 4*A*b^2*c^4)*d^2*e)*x^
2 + 3*(3*A*b^3*c^3*d^2*e + (B*b^3*c^3 - 2*A*b^2*c^4)*d^3)*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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giac [A]  time = 0.31, size = 289, normalized size = 1.31 \begin {gather*} -\frac {B e^{3} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {5}{2}}} - \frac {2 \, {\left (\frac {A d^{3}}{b} + {\left (x {\left (\frac {{\left (8 \, B b c^{4} d^{3} - 16 \, A c^{5} d^{3} - 6 \, B b^{2} c^{3} d^{2} e + 24 \, A b c^{4} d^{2} e - 3 \, B b^{3} c^{2} d e^{2} - 6 \, A b^{2} c^{3} d e^{2} + 4 \, B b^{4} c e^{3} - A b^{3} c^{2} e^{3}\right )} x}{b^{4} c^{2}} + \frac {3 \, {\left (4 \, B b^{2} c^{3} d^{3} - 8 \, A b c^{4} d^{3} - 3 \, B b^{3} c^{2} d^{2} e + 12 \, A b^{2} c^{3} d^{2} e - 3 \, A b^{3} c^{2} d e^{2} + B b^{5} e^{3}\right )}}{b^{4} c^{2}}\right )} + \frac {3 \, {\left (B b^{3} c^{2} d^{3} - 2 \, A b^{2} c^{3} d^{3} + 3 \, A b^{3} c^{2} d^{2} e\right )}}{b^{4} c^{2}}\right )} x\right )}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

-B*e^3*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2) - 2/3*(A*d^3/b + (x*((8*B*b*c^4*d^3 -
16*A*c^5*d^3 - 6*B*b^2*c^3*d^2*e + 24*A*b*c^4*d^2*e - 3*B*b^3*c^2*d*e^2 - 6*A*b^2*c^3*d*e^2 + 4*B*b^4*c*e^3 -
A*b^3*c^2*e^3)*x/(b^4*c^2) + 3*(4*B*b^2*c^3*d^3 - 8*A*b*c^4*d^3 - 3*B*b^3*c^2*d^2*e + 12*A*b^2*c^3*d^2*e - 3*A
*b^3*c^2*d*e^2 + B*b^5*e^3)/(b^4*c^2)) + 3*(B*b^3*c^2*d^3 - 2*A*b^2*c^3*d^3 + 3*A*b^3*c^2*d^2*e)/(b^4*c^2))*x)
/(c*x^2 + b*x)^(3/2)

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maple [B]  time = 0.06, size = 680, normalized size = 3.09 \begin {gather*} -\frac {B \,e^{3} x^{3}}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}-\frac {A \,e^{3} x^{2}}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}+\frac {B b \,e^{3} x^{2}}{2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}-\frac {3 B d \,e^{2} x^{2}}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}-\frac {A b \,e^{3} x}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}+\frac {2 A \,d^{2} e x}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} b}-\frac {4 A c \,d^{3} x}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{2}}-\frac {2 A d \,e^{2} x}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}+\frac {B \,b^{2} e^{3} x}{6 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{3}}-\frac {B b d \,e^{2} x}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c^{2}}+\frac {2 B \,d^{3} x}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b}-\frac {2 B \,d^{2} e x}{\left (c \,x^{2}+b x \right )^{\frac {3}{2}} c}+\frac {2 A \,e^{3} x}{3 \sqrt {c \,x^{2}+b x}\, b c}-\frac {2 A \,d^{3}}{3 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b}+\frac {4 A d \,e^{2} x}{\sqrt {c \,x^{2}+b x}\, b^{2}}-\frac {16 A c \,d^{2} e x}{\sqrt {c \,x^{2}+b x}\, b^{3}}+\frac {32 A \,c^{2} d^{3} x}{3 \sqrt {c \,x^{2}+b x}\, b^{4}}+\frac {2 B d \,e^{2} x}{\sqrt {c \,x^{2}+b x}\, b c}+\frac {4 B \,d^{2} e x}{\sqrt {c \,x^{2}+b x}\, b^{2}}-\frac {16 B c \,d^{3} x}{3 \sqrt {c \,x^{2}+b x}\, b^{3}}-\frac {7 B \,e^{3} x}{3 \sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {B \,e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {5}{2}}}+\frac {2 A d \,e^{2}}{\sqrt {c \,x^{2}+b x}\, b c}-\frac {8 A \,d^{2} e}{\sqrt {c \,x^{2}+b x}\, b^{2}}+\frac {16 A c \,d^{3}}{3 \sqrt {c \,x^{2}+b x}\, b^{3}}+\frac {A \,e^{3}}{3 \sqrt {c \,x^{2}+b x}\, c^{2}}-\frac {B b \,e^{3}}{6 \sqrt {c \,x^{2}+b x}\, c^{3}}+\frac {2 B \,d^{2} e}{\sqrt {c \,x^{2}+b x}\, b c}-\frac {8 B \,d^{3}}{3 \sqrt {c \,x^{2}+b x}\, b^{2}}+\frac {B d \,e^{2}}{\sqrt {c \,x^{2}+b x}\, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^3/(c*x^2+b*x)^(5/2),x)

[Out]

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

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maxima [B]  time = 0.72, size = 550, normalized size = 2.50 \begin {gather*} -\frac {1}{3} \, B e^{3} x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {b x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} - \frac {2 \, x}{\sqrt {c x^{2} + b x} b c} - \frac {1}{\sqrt {c x^{2} + b x} c^{2}}\right )} - \frac {4 \, A c d^{3} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} + \frac {32 \, A c^{2} d^{3} x}{3 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {4 \, B e^{3} x}{3 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {B e^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {2 \, A d^{3}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} + \frac {16 \, A c d^{3}}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} B e^{3}}{3 \, b c^{2}} - \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {4 \, {\left (B d^{2} e + A d e^{2}\right )} x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {2 \, {\left (B d^{3} + 3 \, A d^{2} e\right )} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} - \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} b x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} - \frac {2 \, {\left (B d^{2} e + A d e^{2}\right )} x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {2 \, {\left (3 \, B d e^{2} + A e^{3}\right )} x}{3 \, \sqrt {c x^{2} + b x} b c} - \frac {16 \, {\left (B d^{3} + 3 \, A d^{2} e\right )} c x}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {8 \, {\left (B d^{3} + 3 \, A d^{2} e\right )}}{3 \, \sqrt {c x^{2} + b x} b^{2}} + \frac {3 \, B d e^{2} + A e^{3}}{3 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {2 \, {\left (B d^{2} e + A d e^{2}\right )}}{\sqrt {c x^{2} + b x} b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^3}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^3)/(b*x + c*x^2)^(5/2),x)

[Out]

int(((A + B*x)*(d + e*x)^3)/(b*x + c*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (d + e x\right )^{3}}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**3/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((A + B*x)*(d + e*x)**3/(x*(b + c*x))**(5/2), x)

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