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

Optimal. Leaf size=184 \[ \frac{2 \left (c x \left (16 a^2 B c-20 a A b c+3 A b^3\right )+A \left (24 a^2 c^2-22 a b^2 c+3 b^4\right )+8 a^2 b B c\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{A \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{a^{5/2}}+\frac{2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.128345, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {822, 12, 724, 206} \[ \frac{2 \left (c x \left (16 a^2 B c-20 a A b c+3 A b^3\right )+A \left (24 a^2 c^2-22 a b^2 c+3 b^4\right )+8 a^2 b B c\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{A \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{a^{5/2}}+\frac{2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*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] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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}{x \left (a+b x+c x^2\right )^{5/2}} \, dx &=\frac{2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac{2 \int \frac{-\frac{3}{2} A \left (b^2-4 a c\right )-2 (A b-2 a B) c x}{x \left (a+b x+c x^2\right )^{3/2}} \, dx}{3 a \left (b^2-4 a c\right )}\\ &=\frac{2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{2 \left (8 a^2 b B c+A \left (3 b^4-22 a b^2 c+24 a^2 c^2\right )+c \left (3 A b^3-20 a A b c+16 a^2 B c\right ) x\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}+\frac{4 \int \frac{3 A \left (b^2-4 a c\right )^2}{4 x \sqrt{a+b x+c x^2}} \, dx}{3 a^2 \left (b^2-4 a c\right )^2}\\ &=\frac{2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{2 \left (8 a^2 b B c+A \left (3 b^4-22 a b^2 c+24 a^2 c^2\right )+c \left (3 A b^3-20 a A b c+16 a^2 B c\right ) x\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}+\frac{A \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx}{a^2}\\ &=\frac{2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{2 \left (8 a^2 b B c+A \left (3 b^4-22 a b^2 c+24 a^2 c^2\right )+c \left (3 A b^3-20 a A b c+16 a^2 B c\right ) x\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{(2 A) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x}{\sqrt{a+b x+c x^2}}\right )}{a^2}\\ &=\frac{2 \left (A b^2-a b B-2 a A c+(A b-2 a B) c x\right )}{3 a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{2 \left (8 a^2 b B c+A \left (3 b^4-22 a b^2 c+24 a^2 c^2\right )+c \left (3 A b^3-20 a A b c+16 a^2 B c\right ) x\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{A \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.142383, size = 179, normalized size = 0.97 \[ \frac{A \left (48 a^2 c^2-44 a b^2 c-40 a b c^2 x+6 b^3 c x+6 b^4\right )+16 a^2 B c (b+2 c x)}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)}}-\frac{A \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{a^{5/2}}+\frac{2 a B (b+2 c x)-2 A \left (-2 a c+b^2+b c x\right )}{3 a \left (4 a c-b^2\right ) (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*B*(b + 2*c*x) - 2*A*(b^2 - 2*a*c + b*c*x))/(3*a*(-b^2 + 4*a*c)*(a + x*(b + c*x))^(3/2)) + (16*a^2*B*c*(b
+ 2*c*x) + A*(6*b^4 - 44*a*b^2*c + 48*a^2*c^2 + 6*b^3*c*x - 40*a*b*c^2*x))/(3*a^2*(b^2 - 4*a*c)^2*Sqrt[a + x*(
b + c*x)]) - (A*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/a^(5/2)

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Maple [B]  time = 0.008, size = 390, normalized size = 2.1 \begin{align*}{\frac{4\,Bcx}{12\,ac-3\,{b}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,bB}{12\,ac-3\,{b}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}}+{\frac{32\,B{c}^{2}x}{3\, \left ( 4\,ac-{b}^{2} \right ) ^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{16\,Bcb}{3\, \left ( 4\,ac-{b}^{2} \right ) ^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{A}{3\,a} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,Abcx}{3\,a \left ( 4\,ac-{b}^{2} \right ) } \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}}-{\frac{A{b}^{2}}{3\,a \left ( 4\,ac-{b}^{2} \right ) } \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}}-{\frac{16\,Ab{c}^{2}x}{3\,a \left ( 4\,ac-{b}^{2} \right ) ^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}-{\frac{8\,A{b}^{2}c}{3\,a \left ( 4\,ac-{b}^{2} \right ) ^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{A}{{a}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}-2\,{\frac{Abcx}{{a}^{2} \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}-{\frac{A{b}^{2}}{{a}^{2} \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}-{A\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/x/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 18.5425, size = 2282, normalized size = 12.4 \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)/x/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.22176, size = 450, normalized size = 2.45 \begin{align*} \frac{{\left ({\left (\frac{{\left (3 \, A a^{5} b^{3} c^{2} + 16 \, B a^{7} c^{3} - 20 \, A a^{6} b c^{3}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{6 \,{\left (A a^{5} b^{4} c + 4 \, B a^{7} b c^{2} - 7 \, A a^{6} b^{2} c^{2} + 4 \, A a^{7} c^{3}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{3 \,{\left (A a^{5} b^{5} + 2 \, B a^{7} b^{2} c - 6 \, A a^{6} b^{3} c + 8 \, B a^{8} c^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac{B a^{7} b^{3} - 4 \, A a^{6} b^{4} - 12 \, B a^{8} b c + 28 \, A a^{7} b^{2} c - 32 \, A a^{8} c^{2}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}}{3 \,{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}} + \frac{2 \, A \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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