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

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

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Rubi [A]  time = 0.13, antiderivative size = 184, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 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])

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*}

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Mathematica [A]  time = 0.13, size = 179, normalized size = 0.97 \begin {gather*} -\frac {A \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{a^{5/2}}+\frac {A \left (48 a^2 c^2-44 a b^2 c-40 a b c^2 x+6 b^4+6 b^3 c x\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 {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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 1.47, size = 241, normalized size = 1.31 \begin {gather*} \frac {2 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2 \left (32 a^3 A c^2+12 a^3 b B c+24 a^3 B c^2 x-28 a^2 A b^2 c+24 a^2 A c^3 x^2-a^2 b^3 B+6 a^2 b^2 B c x+24 a^2 b B c^2 x^2+16 a^2 B c^3 x^3+4 a A b^4-18 a A b^3 c x-42 a A b^2 c^2 x^2-20 a A b c^3 x^3+3 A b^5 x+6 A b^4 c x^2+3 A b^3 c^2 x^3\right )}{3 a^2 \left (4 a c-b^2\right )^2 \left (a+b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(4*a*A*b^4 - a^2*b^3*B - 28*a^2*A*b^2*c + 12*a^3*b*B*c + 32*a^3*A*c^2 + 3*A*b^5*x - 18*a*A*b^3*c*x + 6*a^2*
b^2*B*c*x + 24*a^3*B*c^2*x + 6*A*b^4*c*x^2 - 42*a*A*b^2*c^2*x^2 + 24*a^2*b*B*c^2*x^2 + 24*a^2*A*c^3*x^2 + 3*A*
b^3*c^2*x^3 - 20*a*A*b*c^3*x^3 + 16*a^2*B*c^3*x^3))/(3*a^2*(-b^2 + 4*a*c)^2*(a + b*x + c*x^2)^(3/2)) + (2*A*Ar
cTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + b*x + c*x^2]/Sqrt[a]])/a^(5/2)

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fricas [B]  time = 2.37, size = 1077, normalized size = 5.85 \begin {gather*} \left [\frac {3 \, {\left (A a^{2} b^{4} - 8 \, A a^{3} b^{2} c + 16 \, A a^{4} c^{2} + {\left (A b^{4} c^{2} - 8 \, A a b^{2} c^{3} + 16 \, A a^{2} c^{4}\right )} x^{4} + 2 \, {\left (A b^{5} c - 8 \, A a b^{3} c^{2} + 16 \, A a^{2} b c^{3}\right )} x^{3} + {\left (A b^{6} - 6 \, A a b^{4} c + 32 \, A a^{3} c^{3}\right )} x^{2} + 2 \, {\left (A a b^{5} - 8 \, A a^{2} b^{3} c + 16 \, A a^{3} b c^{2}\right )} x\right )} \sqrt {a} \log \left (-\frac {8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{2}}\right ) - 4 \, {\left (B a^{3} b^{3} - 4 \, A a^{2} b^{4} - 32 \, A a^{4} c^{2} - {\left (3 \, A a b^{3} c^{2} + 4 \, {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} c^{3}\right )} x^{3} - 6 \, {\left (A a b^{4} c + 4 \, A a^{3} c^{3} + {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} c^{2}\right )} x^{2} - 4 \, {\left (3 \, B a^{4} b - 7 \, A a^{3} b^{2}\right )} c - 3 \, {\left (A a b^{5} + 8 \, B a^{4} c^{2} + 2 \, {\left (B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} c\right )} x\right )} \sqrt {c x^{2} + b x + a}}{6 \, {\left (a^{5} b^{4} - 8 \, a^{6} b^{2} c + 16 \, a^{7} c^{2} + {\left (a^{3} b^{4} c^{2} - 8 \, a^{4} b^{2} c^{3} + 16 \, a^{5} c^{4}\right )} x^{4} + 2 \, {\left (a^{3} b^{5} c - 8 \, a^{4} b^{3} c^{2} + 16 \, a^{5} b c^{3}\right )} x^{3} + {\left (a^{3} b^{6} - 6 \, a^{4} b^{4} c + 32 \, a^{6} c^{3}\right )} x^{2} + 2 \, {\left (a^{4} b^{5} - 8 \, a^{5} b^{3} c + 16 \, a^{6} b c^{2}\right )} x\right )}}, \frac {3 \, {\left (A a^{2} b^{4} - 8 \, A a^{3} b^{2} c + 16 \, A a^{4} c^{2} + {\left (A b^{4} c^{2} - 8 \, A a b^{2} c^{3} + 16 \, A a^{2} c^{4}\right )} x^{4} + 2 \, {\left (A b^{5} c - 8 \, A a b^{3} c^{2} + 16 \, A a^{2} b c^{3}\right )} x^{3} + {\left (A b^{6} - 6 \, A a b^{4} c + 32 \, A a^{3} c^{3}\right )} x^{2} + 2 \, {\left (A a b^{5} - 8 \, A a^{2} b^{3} c + 16 \, A a^{3} b c^{2}\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{2} + a b x + a^{2}\right )}}\right ) - 2 \, {\left (B a^{3} b^{3} - 4 \, A a^{2} b^{4} - 32 \, A a^{4} c^{2} - {\left (3 \, A a b^{3} c^{2} + 4 \, {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} c^{3}\right )} x^{3} - 6 \, {\left (A a b^{4} c + 4 \, A a^{3} c^{3} + {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} c^{2}\right )} x^{2} - 4 \, {\left (3 \, B a^{4} b - 7 \, A a^{3} b^{2}\right )} c - 3 \, {\left (A a b^{5} + 8 \, B a^{4} c^{2} + 2 \, {\left (B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} c\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (a^{5} b^{4} - 8 \, a^{6} b^{2} c + 16 \, a^{7} c^{2} + {\left (a^{3} b^{4} c^{2} - 8 \, a^{4} b^{2} c^{3} + 16 \, a^{5} c^{4}\right )} x^{4} + 2 \, {\left (a^{3} b^{5} c - 8 \, a^{4} b^{3} c^{2} + 16 \, a^{5} b c^{3}\right )} x^{3} + {\left (a^{3} b^{6} - 6 \, a^{4} b^{4} c + 32 \, a^{6} c^{3}\right )} x^{2} + 2 \, {\left (a^{4} b^{5} - 8 \, a^{5} b^{3} c + 16 \, a^{6} b c^{2}\right )} x\right )}}\right ] \end {gather*}

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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giac [A]  time = 0.25, size = 333, normalized size = 1.81 \begin {gather*} \frac {2 \, {\left ({\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}{a^{7} b^{4} - 8 \, a^{8} b^{2} c + 16 \, a^{9} c^{2}} + \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 )}}{a^{7} b^{4} - 8 \, a^{8} b^{2} c + 16 \, a^{9} c^{2}}\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 )}}{a^{7} b^{4} - 8 \, a^{8} b^{2} c + 16 \, a^{9} c^{2}}\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}}{a^{7} b^{4} - 8 \, a^{8} b^{2} c + 16 \, a^{9} c^{2}}\right )}}{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 {gather*}

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]

2/3*((((3*A*a^5*b^3*c^2 + 16*B*a^7*c^3 - 20*A*a^6*b*c^3)*x/(a^7*b^4 - 8*a^8*b^2*c + 16*a^9*c^2) + 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)/(a^7*b^4 - 8*a^8*b^2*c + 16*a^9*c^2))*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)/(a^7*b^4 - 8*a^8*b^2*c + 16*a^9*c^2))*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)/(a^7*b^4 - 8*a^8*b^2*c + 16*a^9*c^2))/(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)

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maple [B]  time = 0.05, size = 390, normalized size = 2.12 \begin {gather*} -\frac {16 A b \,c^{2} x}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, a}+\frac {32 B \,c^{2} x}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}-\frac {8 A \,b^{2} c}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}\, a}-\frac {2 A b c x}{3 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a}+\frac {16 B b c}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}+\frac {4 B c x}{3 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {A \,b^{2}}{3 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a}-\frac {2 A b c x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2}}+\frac {2 B b}{3 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {A \,b^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2}}+\frac {A}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a}-\frac {A \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{a^{\frac {5}{2}}}+\frac {A}{\sqrt {c \,x^{2}+b x +a}\, a^{2}} \end {gather*}

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)*c*x+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)*c*x-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)*c*x-A/a^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-A/a^(5/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a
^(1/2))/x)

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

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{x\,{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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