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

Optimal. Leaf size=92 \[ -\frac {8 (x (2 c d-b e)+b d) (A b e-2 A c d+b B d)}{3 b^4 \sqrt {b x+c x^2}}-\frac {2 (d+e x)^2 (A b-x (b B-2 A c))}{3 b^2 \left (b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {804, 636} \begin {gather*} -\frac {2 (d+e x)^2 (A b-x (b B-2 A c))}{3 b^2 \left (b x+c x^2\right )^{3/2}}-\frac {8 (x (2 c d-b e)+b d) (A b e-2 A c d+b B d)}{3 b^4 \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 636

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-2*(b*d - 2*a*e + (2*c*
d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

Rule 804

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 149, normalized size = 1.62 \begin {gather*} \frac {2 \left (A \left (-\left (b^3 \left (d^2+6 d e x-3 e^2 x^2\right )\right )+2 b^2 c x \left (3 d^2-12 d e x+e^2 x^2\right )+8 b c^2 d x^2 (3 d-2 e x)+16 c^3 d^2 x^3\right )+b B x \left (b^2 \left (-3 d^2+6 d e x+e^2 x^2\right )+4 b c d x (e x-3 d)-8 c^2 d^2 x^2\right )\right )}{3 b^4 (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 0.57, size = 204, normalized size = 2.22 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-A b^3 d^2-6 A b^3 d e x+3 A b^3 e^2 x^2+6 A b^2 c d^2 x-24 A b^2 c d e x^2+2 A b^2 c e^2 x^3+24 A b c^2 d^2 x^2-16 A b c^2 d e x^3+16 A c^3 d^2 x^3-3 b^3 B d^2 x+6 b^3 B d e x^2+b^3 B e^2 x^3-12 b^2 B c d^2 x^2+4 b^2 B c d e x^3-8 b B c^2 d^2 x^3\right )}{3 b^4 x^2 (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 189, normalized size = 2.05 \begin {gather*} -\frac {2 \, {\left (A b^{3} d^{2} + {\left (8 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 4 \, {\left (B b^{2} c - 4 \, A b c^{2}\right )} d e - {\left (B b^{3} + 2 \, A b^{2} c\right )} e^{2}\right )} x^{3} - 3 \, {\left (A b^{3} e^{2} - 4 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )} d^{2} + 2 \, {\left (B b^{3} - 4 \, A b^{2} c\right )} d e\right )} x^{2} + 3 \, {\left (2 \, A b^{3} d e + {\left (B b^{3} - 2 \, A b^{2} c\right )} d^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.29, size = 176, normalized size = 1.91 \begin {gather*} -\frac {2 \, {\left (\frac {A d^{2}}{b} + {\left (x {\left (\frac {{\left (8 \, B b c^{2} d^{2} - 16 \, A c^{3} d^{2} - 4 \, B b^{2} c d e + 16 \, A b c^{2} d e - B b^{3} e^{2} - 2 \, A b^{2} c e^{2}\right )} x}{b^{4}} + \frac {3 \, {\left (4 \, B b^{2} c d^{2} - 8 \, A b c^{2} d^{2} - 2 \, B b^{3} d e + 8 \, A b^{2} c d e - A b^{3} e^{2}\right )}}{b^{4}}\right )} + \frac {3 \, {\left (B b^{3} d^{2} - 2 \, A b^{2} c d^{2} + 2 \, A b^{3} d e\right )}}{b^{4}}\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)^2/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

-2/3*(A*d^2/b + (x*((8*B*b*c^2*d^2 - 16*A*c^3*d^2 - 4*B*b^2*c*d*e + 16*A*b*c^2*d*e - B*b^3*e^2 - 2*A*b^2*c*e^2
)*x/b^4 + 3*(4*B*b^2*c*d^2 - 8*A*b*c^2*d^2 - 2*B*b^3*d*e + 8*A*b^2*c*d*e - A*b^3*e^2)/b^4) + 3*(B*b^3*d^2 - 2*
A*b^2*c*d^2 + 2*A*b^3*d*e)/b^4)*x)/(c*x^2 + b*x)^(3/2)

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maple [B]  time = 0.05, size = 197, normalized size = 2.14 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-2 A \,b^{2} c \,e^{2} x^{3}+16 A b \,c^{2} d e \,x^{3}-16 A \,c^{3} d^{2} x^{3}-B \,b^{3} e^{2} x^{3}-4 B \,b^{2} c d e \,x^{3}+8 B b \,c^{2} d^{2} x^{3}-3 A \,b^{3} e^{2} x^{2}+24 A \,b^{2} c d e \,x^{2}-24 A b \,c^{2} d^{2} x^{2}-6 B \,b^{3} d e \,x^{2}+12 B \,b^{2} c \,d^{2} x^{2}+6 A \,b^{3} d e x -6 A \,b^{2} c \,d^{2} x +3 B \,b^{3} d^{2} x +A \,d^{2} b^{3}\right ) x}{3 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.52, size = 347, normalized size = 3.77 \begin {gather*} -\frac {B e^{2} x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {4 \, A c d^{2} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} + \frac {32 \, A c^{2} d^{2} x}{3 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {B b e^{2} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} + \frac {2 \, B e^{2} x}{3 \, \sqrt {c x^{2} + b x} b c} - \frac {2 \, A d^{2}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} + \frac {16 \, A c d^{2}}{3 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {B e^{2}}{3 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {4 \, {\left (2 \, B d e + A e^{2}\right )} x}{3 \, \sqrt {c x^{2} + b x} b^{2}} + \frac {2 \, {\left (B d^{2} + 2 \, A d e\right )} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} - \frac {2 \, {\left (2 \, B d e + A e^{2}\right )} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {16 \, {\left (B d^{2} + 2 \, A d e\right )} c x}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {8 \, {\left (B d^{2} + 2 \, A d e\right )}}{3 \, \sqrt {c x^{2} + b x} b^{2}} + \frac {2 \, {\left (2 \, B d e + A e^{2}\right )}}{3 \, \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)^2/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.91, size = 190, normalized size = 2.07 \begin {gather*} \frac {2\,\left (-3\,B\,b^3\,d^2\,x-A\,b^3\,d^2+6\,B\,b^3\,d\,e\,x^2-6\,A\,b^3\,d\,e\,x+B\,b^3\,e^2\,x^3+3\,A\,b^3\,e^2\,x^2-12\,B\,b^2\,c\,d^2\,x^2+6\,A\,b^2\,c\,d^2\,x+4\,B\,b^2\,c\,d\,e\,x^3-24\,A\,b^2\,c\,d\,e\,x^2+2\,A\,b^2\,c\,e^2\,x^3-8\,B\,b\,c^2\,d^2\,x^3+24\,A\,b\,c^2\,d^2\,x^2-16\,A\,b\,c^2\,d\,e\,x^3+16\,A\,c^3\,d^2\,x^3\right )}{3\,b^4\,{\left (c\,x^2+b\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )^{2}}{\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)**2/(c*x**2+b*x)**(5/2),x)

[Out]

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

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