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

Optimal. Leaf size=189 \[ -\frac{2 \left (x \left (24 a^2 B c^2+8 a A b c^2-22 a b^2 B c+3 b^4 B\right )+a \left (16 a A c^2-20 a b B c+3 b^3 B\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{B \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{5/2}} \]

[Out]

(-2*x^2*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x))/(3*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) - (2*(a*(
3*b^3*B - 20*a*b*B*c + 16*a*A*c^2) + (3*b^4*B - 22*a*b^2*B*c + 8*a*A*b*c^2 + 24*a^2*B*c^2)*x))/(3*c^2*(b^2 - 4
*a*c)^2*Sqrt[a + b*x + c*x^2]) + (B*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(5/2)

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Rubi [A]  time = 0.13251, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {818, 777, 621, 206} \[ -\frac{2 \left (x \left (24 a^2 B c^2+8 a A b c^2-22 a b^2 B c+3 b^4 B\right )+a \left (16 a A c^2-20 a b B c+3 b^3 B\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{B \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x))/(3*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) - (2*(a*(
3*b^3*B - 20*a*b*B*c + 16*a*A*c^2) + (3*b^4*B - 22*a*b^2*B*c + 8*a*A*b*c^2 + 24*a^2*B*c^2)*x))/(3*c^2*(b^2 - 4
*a*c)^2*Sqrt[a + b*x + c*x^2]) + (B*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(5/2)

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 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 621

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

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{x^3 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 x^2 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{2 \int \frac{x \left (2 a (b B-2 A c)+\frac{3}{2} B \left (b^2-4 a c\right ) x\right )}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{3 c \left (b^2-4 a c\right )}\\ &=-\frac{2 x^2 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac{2 \left (a \left (3 b^3 B-20 a b B c+16 a A c^2\right )+\left (3 b^4 B-22 a b^2 B c+8 a A b c^2+24 a^2 B c^2\right ) x\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}+\frac{B \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{c^2}\\ &=-\frac{2 x^2 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac{2 \left (a \left (3 b^3 B-20 a b B c+16 a A c^2\right )+\left (3 b^4 B-22 a b^2 B c+8 a A b c^2+24 a^2 B c^2\right ) x\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}+\frac{(2 B) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{c^2}\\ &=-\frac{2 x^2 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac{2 \left (a \left (3 b^3 B-20 a b B c+16 a A c^2\right )+\left (3 b^4 B-22 a b^2 B c+8 a A b c^2+24 a^2 B c^2\right ) x\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}+\frac{B \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.412834, size = 201, normalized size = 1.06 \[ \frac{B \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{c^{5/2}}-\frac{2 \left (a^2 \left (24 A b c^2 x+8 c^3 x^2 (3 A+4 B x)-42 b^2 B c x+3 b^3 B\right )+4 a^3 c (4 A c-5 b B+6 B c x)+2 a b x \left (b c^2 x (3 A-14 B x)+6 A c^3 x^2-9 b^2 B c x+3 b^3 B\right )+b^3 x^2 \left (-A c^2 x+3 b^2 B+4 b B c x\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(4*a^3*c*(-5*b*B + 4*A*c + 6*B*c*x) + b^3*x^2*(3*b^2*B + 4*b*B*c*x - A*c^2*x) + 2*a*b*x*(3*b^3*B - 9*b^2*B
*c*x + 6*A*c^3*x^2 + b*c^2*x*(3*A - 14*B*x)) + a^2*(3*b^3*B - 42*b^2*B*c*x + 24*A*b*c^2*x + 8*c^3*x^2*(3*A + 4
*B*x))))/(3*c^2*(b^2 - 4*a*c)^2*(a + x*(b + c*x))^(3/2)) + (B*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c
*x)])])/c^(5/2)

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Maple [B]  time = 0.01, size = 860, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*A*b/c^2*x/(c*x^2+b*x+a)^(3/2)+1/24*A*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+1/2*B*b^2/c^2*a/(4*a*c-b^2)/
(c*x^2+b*x+a)^(3/2)*x+4*B*b^2/c*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-A*b/c*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*
x+2/3*A*b^3/c/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-1/2*A*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)-8*A*b*a/(4*a
*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-4*A*b^2/c*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+1/12*A*b^3/c^2/(4*a*c-b^2)/(c*x^
2+b*x+a)^(3/2)*x+1/4*B*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+2*B*b^3/c^2*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/
2)+B/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-1/24*B*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x-1/3*B*b^4/c^2/
(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-1/48*B*b^3/c^4/(c*x^2+b*x+a)^(3/2)-A*x^2/c/(c*x^2+b*x+a)^(3/2)+1/24*A*b^2/
c^3/(c*x^2+b*x+a)^(3/2)-2/3*A*a/c^2/(c*x^2+b*x+a)^(3/2)-B/c^2*x/(c*x^2+b*x+a)^(1/2)+1/2*B/c^3*b/(c*x^2+b*x+a)^
(1/2)-1/3*B*x^3/c/(c*x^2+b*x+a)^(3/2)+B/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/6*B*b^5/c^3/(4*a
*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+1/3*B*b/c^3*a/(c*x^2+b*x+a)^(3/2)+1/2*B/c^3*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+
1/2*B*b/c^2*x^2/(c*x^2+b*x+a)^(3/2)+1/8*B*b^2/c^3*x/(c*x^2+b*x+a)^(3/2)-1/48*B*b^5/c^4/(4*a*c-b^2)/(c*x^2+b*x+
a)^(3/2)+1/3*A*b^4/c^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(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(x^3*(B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 8.98737, size = 2261, normalized size = 11.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(B*a^2*b^4 - 8*B*a^3*b^2*c + 16*B*a^4*c^2 + (B*b^4*c^2 - 8*B*a*b^2*c^3 + 16*B*a^2*c^4)*x^4 + 2*(B*b^5*
c - 8*B*a*b^3*c^2 + 16*B*a^2*b*c^3)*x^3 + (B*b^6 - 6*B*a*b^4*c + 32*B*a^3*c^3)*x^2 + 2*(B*a*b^5 - 8*B*a^2*b^3*
c + 16*B*a^3*b*c^2)*x)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) -
4*a*c) - 4*(3*B*a^2*b^3*c - 20*B*a^3*b*c^2 + 16*A*a^3*c^3 + (4*B*b^4*c^2 + 4*(8*B*a^2 + 3*A*a*b)*c^4 - (28*B*a
*b^2 + A*b^3)*c^3)*x^3 + 3*(B*b^5*c - 6*B*a*b^3*c^2 + 2*A*a*b^2*c^3 + 8*A*a^2*c^4)*x^2 + 6*(B*a*b^4*c - 7*B*a^
2*b^2*c^2 + 4*(B*a^3 + A*a^2*b)*c^3)*x)*sqrt(c*x^2 + b*x + a))/(a^2*b^4*c^3 - 8*a^3*b^2*c^4 + 16*a^4*c^5 + (b^
4*c^5 - 8*a*b^2*c^6 + 16*a^2*c^7)*x^4 + 2*(b^5*c^4 - 8*a*b^3*c^5 + 16*a^2*b*c^6)*x^3 + (b^6*c^3 - 6*a*b^4*c^4
+ 32*a^3*c^6)*x^2 + 2*(a*b^5*c^3 - 8*a^2*b^3*c^4 + 16*a^3*b*c^5)*x), -1/3*(3*(B*a^2*b^4 - 8*B*a^3*b^2*c + 16*B
*a^4*c^2 + (B*b^4*c^2 - 8*B*a*b^2*c^3 + 16*B*a^2*c^4)*x^4 + 2*(B*b^5*c - 8*B*a*b^3*c^2 + 16*B*a^2*b*c^3)*x^3 +
(B*b^6 - 6*B*a*b^4*c + 32*B*a^3*c^3)*x^2 + 2*(B*a*b^5 - 8*B*a^2*b^3*c + 16*B*a^3*b*c^2)*x)*sqrt(-c)*arctan(1/
2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(3*B*a^2*b^3*c - 20*B*a^3*b*c^2 + 16
*A*a^3*c^3 + (4*B*b^4*c^2 + 4*(8*B*a^2 + 3*A*a*b)*c^4 - (28*B*a*b^2 + A*b^3)*c^3)*x^3 + 3*(B*b^5*c - 6*B*a*b^3
*c^2 + 2*A*a*b^2*c^3 + 8*A*a^2*c^4)*x^2 + 6*(B*a*b^4*c - 7*B*a^2*b^2*c^2 + 4*(B*a^3 + A*a^2*b)*c^3)*x)*sqrt(c*
x^2 + b*x + a))/(a^2*b^4*c^3 - 8*a^3*b^2*c^4 + 16*a^4*c^5 + (b^4*c^5 - 8*a*b^2*c^6 + 16*a^2*c^7)*x^4 + 2*(b^5*
c^4 - 8*a*b^3*c^5 + 16*a^2*b*c^6)*x^3 + (b^6*c^3 - 6*a*b^4*c^4 + 32*a^3*c^6)*x^2 + 2*(a*b^5*c^3 - 8*a^2*b^3*c^
4 + 16*a^3*b*c^5)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.21137, size = 424, normalized size = 2.24 \begin{align*} -\frac{2 \,{\left ({\left ({\left (\frac{{\left (4 \, B b^{4} c - 28 \, B a b^{2} c^{2} - A b^{3} c^{2} + 32 \, B a^{2} c^{3} + 12 \, A a b c^{3}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{3 \,{\left (B b^{5} - 6 \, B a b^{3} c + 2 \, A a b^{2} c^{2} + 8 \, A a^{2} c^{3}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{6 \,{\left (B a b^{4} - 7 \, B a^{2} b^{2} c + 4 \, B a^{3} c^{2} + 4 \, A a^{2} b c^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{3 \, B a^{2} b^{3} - 20 \, B a^{3} b c + 16 \, A a^{3} c^{2}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )}}{3 \,{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}} - \frac{B \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((((4*B*b^4*c - 28*B*a*b^2*c^2 - A*b^3*c^2 + 32*B*a^2*c^3 + 12*A*a*b*c^3)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a
^2*c^4) + 3*(B*b^5 - 6*B*a*b^3*c + 2*A*a*b^2*c^2 + 8*A*a^2*c^3)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + 6*(B
*a*b^4 - 7*B*a^2*b^2*c + 4*B*a^3*c^2 + 4*A*a^2*b*c^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + (3*B*a^2*b^3 -
20*B*a^3*b*c + 16*A*a^3*c^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2) - B*log(abs(-2*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)