3.418 \(\int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx\)

Optimal. Leaf size=139 \[ \frac {5 b^2 \sqrt {a+b x} (6 a B+A b)}{8 a}-\frac {5 b^2 (6 a B+A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 \sqrt {a}}-\frac {(a+b x)^{5/2} (6 a B+A b)}{12 a x^2}-\frac {5 b (a+b x)^{3/2} (6 a B+A b)}{24 a x}-\frac {A (a+b x)^{7/2}}{3 a x^3} \]

[Out]

-5/24*b*(A*b+6*B*a)*(b*x+a)^(3/2)/a/x-1/12*(A*b+6*B*a)*(b*x+a)^(5/2)/a/x^2-1/3*A*(b*x+a)^(7/2)/x^3/a-5/8*b^2*(
A*b+6*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(1/2)+5/8*b^2*(A*b+6*B*a)*(b*x+a)^(1/2)/a

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Rubi [A]  time = 0.06, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {78, 47, 50, 63, 208} \[ \frac {5 b^2 \sqrt {a+b x} (6 a B+A b)}{8 a}-\frac {5 b^2 (6 a B+A b) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 \sqrt {a}}-\frac {(a+b x)^{5/2} (6 a B+A b)}{12 a x^2}-\frac {5 b (a+b x)^{3/2} (6 a B+A b)}{24 a x}-\frac {A (a+b x)^{7/2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^4,x]

[Out]

(5*b^2*(A*b + 6*a*B)*Sqrt[a + b*x])/(8*a) - (5*b*(A*b + 6*a*B)*(a + b*x)^(3/2))/(24*a*x) - ((A*b + 6*a*B)*(a +
b*x)^(5/2))/(12*a*x^2) - (A*(a + b*x)^(7/2))/(3*a*x^3) - (5*b^2*(A*b + 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])
/(8*Sqrt[a])

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{5/2} (A+B x)}{x^4} \, dx &=-\frac {A (a+b x)^{7/2}}{3 a x^3}+\frac {\left (\frac {A b}{2}+3 a B\right ) \int \frac {(a+b x)^{5/2}}{x^3} \, dx}{3 a}\\ &=-\frac {(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac {A (a+b x)^{7/2}}{3 a x^3}+\frac {(5 b (A b+6 a B)) \int \frac {(a+b x)^{3/2}}{x^2} \, dx}{24 a}\\ &=-\frac {5 b (A b+6 a B) (a+b x)^{3/2}}{24 a x}-\frac {(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac {A (a+b x)^{7/2}}{3 a x^3}+\frac {\left (5 b^2 (A b+6 a B)\right ) \int \frac {\sqrt {a+b x}}{x} \, dx}{16 a}\\ &=\frac {5 b^2 (A b+6 a B) \sqrt {a+b x}}{8 a}-\frac {5 b (A b+6 a B) (a+b x)^{3/2}}{24 a x}-\frac {(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac {A (a+b x)^{7/2}}{3 a x^3}+\frac {1}{16} \left (5 b^2 (A b+6 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx\\ &=\frac {5 b^2 (A b+6 a B) \sqrt {a+b x}}{8 a}-\frac {5 b (A b+6 a B) (a+b x)^{3/2}}{24 a x}-\frac {(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac {A (a+b x)^{7/2}}{3 a x^3}+\frac {1}{8} (5 b (A b+6 a B)) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )\\ &=\frac {5 b^2 (A b+6 a B) \sqrt {a+b x}}{8 a}-\frac {5 b (A b+6 a B) (a+b x)^{3/2}}{24 a x}-\frac {(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac {A (a+b x)^{7/2}}{3 a x^3}-\frac {5 b^2 (A b+6 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 \sqrt {a}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 57, normalized size = 0.41 \[ -\frac {(a+b x)^{7/2} \left (7 a^3 A+b^2 x^3 (6 a B+A b) \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};\frac {b x}{a}+1\right )\right )}{21 a^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^4,x]

[Out]

-1/21*((a + b*x)^(7/2)*(7*a^3*A + b^2*(A*b + 6*a*B)*x^3*Hypergeometric2F1[3, 7/2, 9/2, 1 + (b*x)/a]))/(a^4*x^3
)

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fricas [A]  time = 0.68, size = 229, normalized size = 1.65 \[ \left [\frac {15 \, {\left (6 \, B a b^{2} + A b^{3}\right )} \sqrt {a} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (48 \, B a b^{2} x^{3} - 8 \, A a^{3} - 3 \, {\left (18 \, B a^{2} b + 11 \, A a b^{2}\right )} x^{2} - 2 \, {\left (6 \, B a^{3} + 13 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{48 \, a x^{3}}, \frac {15 \, {\left (6 \, B a b^{2} + A b^{3}\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (48 \, B a b^{2} x^{3} - 8 \, A a^{3} - 3 \, {\left (18 \, B a^{2} b + 11 \, A a b^{2}\right )} x^{2} - 2 \, {\left (6 \, B a^{3} + 13 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{24 \, a x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^4,x, algorithm="fricas")

[Out]

[1/48*(15*(6*B*a*b^2 + A*b^3)*sqrt(a)*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(48*B*a*b^2*x^3 - 8
*A*a^3 - 3*(18*B*a^2*b + 11*A*a*b^2)*x^2 - 2*(6*B*a^3 + 13*A*a^2*b)*x)*sqrt(b*x + a))/(a*x^3), 1/24*(15*(6*B*a
*b^2 + A*b^3)*sqrt(-a)*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (48*B*a*b^2*x^3 - 8*A*a^3 - 3*(18*B*a^2*b + 11*A
*a*b^2)*x^2 - 2*(6*B*a^3 + 13*A*a^2*b)*x)*sqrt(b*x + a))/(a*x^3)]

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giac [A]  time = 1.55, size = 151, normalized size = 1.09 \[ \frac {48 \, \sqrt {b x + a} B b^{3} + \frac {15 \, {\left (6 \, B a b^{3} + A b^{4}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {54 \, {\left (b x + a\right )}^{\frac {5}{2}} B a b^{3} - 96 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{2} b^{3} + 42 \, \sqrt {b x + a} B a^{3} b^{3} + 33 \, {\left (b x + a\right )}^{\frac {5}{2}} A b^{4} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b^{4} + 15 \, \sqrt {b x + a} A a^{2} b^{4}}{b^{3} x^{3}}}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^4,x, algorithm="giac")

[Out]

1/24*(48*sqrt(b*x + a)*B*b^3 + 15*(6*B*a*b^3 + A*b^4)*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - (54*(b*x + a)^
(5/2)*B*a*b^3 - 96*(b*x + a)^(3/2)*B*a^2*b^3 + 42*sqrt(b*x + a)*B*a^3*b^3 + 33*(b*x + a)^(5/2)*A*b^4 - 40*(b*x
+ a)^(3/2)*A*a*b^4 + 15*sqrt(b*x + a)*A*a^2*b^4)/(b^3*x^3))/b

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maple [A]  time = 0.02, size = 108, normalized size = 0.78 \[ 2 \left (\sqrt {b x +a}\, B -\frac {5 \left (A b +6 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}+\frac {\left (-\frac {11 A b}{16}-\frac {9 B a}{8}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (\frac {5}{6} A a b +2 B \,a^{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (-\frac {5}{16} A \,a^{2} b -\frac {7}{8} B \,a^{3}\right ) \sqrt {b x +a}}{b^{3} x^{3}}\right ) b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^4,x)

[Out]

2*b^2*((b*x+a)^(1/2)*B+((-11/16*A*b-9/8*B*a)*(b*x+a)^(5/2)+(5/6*A*a*b+2*B*a^2)*(b*x+a)^(3/2)+(-7/8*B*a^3-5/16*
A*a^2*b)*(b*x+a)^(1/2))/x^3/b^3-5/16*(A*b+6*B*a)/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))

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maxima [A]  time = 2.07, size = 168, normalized size = 1.21 \[ \frac {1}{48} \, b^{3} {\left (\frac {96 \, \sqrt {b x + a} B}{b} + \frac {15 \, {\left (6 \, B a + A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{\sqrt {a} b} - \frac {2 \, {\left (3 \, {\left (18 \, B a + 11 \, A b\right )} {\left (b x + a\right )}^{\frac {5}{2}} - 8 \, {\left (12 \, B a^{2} + 5 \, A a b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 3 \, {\left (14 \, B a^{3} + 5 \, A a^{2} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{3} b - 3 \, {\left (b x + a\right )}^{2} a b + 3 \, {\left (b x + a\right )} a^{2} b - a^{3} b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^4,x, algorithm="maxima")

[Out]

1/48*b^3*(96*sqrt(b*x + a)*B/b + 15*(6*B*a + A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(sq
rt(a)*b) - 2*(3*(18*B*a + 11*A*b)*(b*x + a)^(5/2) - 8*(12*B*a^2 + 5*A*a*b)*(b*x + a)^(3/2) + 3*(14*B*a^3 + 5*A
*a^2*b)*sqrt(b*x + a))/((b*x + a)^3*b - 3*(b*x + a)^2*a*b + 3*(b*x + a)*a^2*b - a^3*b))

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mupad [B]  time = 0.44, size = 182, normalized size = 1.31 \[ \frac {\left (\frac {11\,A\,b^3}{8}+\frac {9\,B\,a\,b^2}{4}\right )\,{\left (a+b\,x\right )}^{5/2}+\left (\frac {7\,B\,a^3\,b^2}{4}+\frac {5\,A\,a^2\,b^3}{8}\right )\,\sqrt {a+b\,x}-\left (4\,B\,a^2\,b^2+\frac {5\,A\,a\,b^3}{3}\right )\,{\left (a+b\,x\right )}^{3/2}}{3\,a\,{\left (a+b\,x\right )}^2-3\,a^2\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^3+a^3}+2\,B\,b^2\,\sqrt {a+b\,x}-\frac {5\,b^2\,\mathrm {atanh}\left (\frac {5\,b^2\,\left (A\,b+6\,B\,a\right )\,\sqrt {a+b\,x}}{4\,\sqrt {a}\,\left (\frac {5\,A\,b^3}{4}+\frac {15\,B\,a\,b^2}{2}\right )}\right )\,\left (A\,b+6\,B\,a\right )}{8\,\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(5/2))/x^4,x)

[Out]

(((11*A*b^3)/8 + (9*B*a*b^2)/4)*(a + b*x)^(5/2) + ((5*A*a^2*b^3)/8 + (7*B*a^3*b^2)/4)*(a + b*x)^(1/2) - (4*B*a
^2*b^2 + (5*A*a*b^3)/3)*(a + b*x)^(3/2))/(3*a*(a + b*x)^2 - 3*a^2*(a + b*x) - (a + b*x)^3 + a^3) + 2*B*b^2*(a
+ b*x)^(1/2) - (5*b^2*atanh((5*b^2*(A*b + 6*B*a)*(a + b*x)^(1/2))/(4*a^(1/2)*((5*A*b^3)/4 + (15*B*a*b^2)/2)))*
(A*b + 6*B*a))/(8*a^(1/2))

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sympy [B]  time = 132.74, size = 877, normalized size = 6.31 \[ - \frac {66 A a^{5} b^{3} \sqrt {a + b x}}{96 a^{6} + 144 a^{5} b x - 144 a^{4} \left (a + b x\right )^{2} + 48 a^{3} \left (a + b x\right )^{3}} + \frac {80 A a^{4} b^{3} \left (a + b x\right )^{\frac {3}{2}}}{96 a^{6} + 144 a^{5} b x - 144 a^{4} \left (a + b x\right )^{2} + 48 a^{3} \left (a + b x\right )^{3}} - \frac {30 A a^{3} b^{3} \left (a + b x\right )^{\frac {5}{2}}}{96 a^{6} + 144 a^{5} b x - 144 a^{4} \left (a + b x\right )^{2} + 48 a^{3} \left (a + b x\right )^{3}} - \frac {30 A a^{3} b^{3} \sqrt {a + b x}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} - \frac {5 A a^{3} b^{3} \sqrt {\frac {1}{a^{7}}} \log {\left (- a^{4} \sqrt {\frac {1}{a^{7}}} + \sqrt {a + b x} \right )}}{16} + \frac {5 A a^{3} b^{3} \sqrt {\frac {1}{a^{7}}} \log {\left (a^{4} \sqrt {\frac {1}{a^{7}}} + \sqrt {a + b x} \right )}}{16} + \frac {18 A a^{2} b^{3} \left (a + b x\right )^{\frac {3}{2}}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac {9 A a^{2} b^{3} \sqrt {\frac {1}{a^{5}}} \log {\left (- a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {9 A a^{2} b^{3} \sqrt {\frac {1}{a^{5}}} \log {\left (a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {3 A a b^{3} \sqrt {\frac {1}{a^{3}}} \log {\left (- a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {3 A a b^{3} \sqrt {\frac {1}{a^{3}}} \log {\left (a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {2 A b^{3} \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - \frac {3 A b^{2} \sqrt {a + b x}}{x} - \frac {10 B a^{4} b^{2} \sqrt {a + b x}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac {6 B a^{3} b^{2} \left (a + b x\right )^{\frac {3}{2}}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac {3 B a^{3} b^{2} \sqrt {\frac {1}{a^{5}}} \log {\left (- a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {3 B a^{3} b^{2} \sqrt {\frac {1}{a^{5}}} \log {\left (a^{3} \sqrt {\frac {1}{a^{5}}} + \sqrt {a + b x} \right )}}{8} - \frac {3 B a^{2} b^{2} \sqrt {\frac {1}{a^{3}}} \log {\left (- a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {3 B a^{2} b^{2} \sqrt {\frac {1}{a^{3}}} \log {\left (a^{2} \sqrt {\frac {1}{a^{3}}} + \sqrt {a + b x} \right )}}{2} + \frac {6 B a b^{2} \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - \frac {3 B a b \sqrt {a + b x}}{x} + 2 B b^{2} \sqrt {a + b x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(B*x+A)/x**4,x)

[Out]

-66*A*a**5*b**3*sqrt(a + b*x)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) + 80*A*a
**4*b**3*(a + b*x)**(3/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) - 30*A*a**3*
b**3*(a + b*x)**(5/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) - 30*A*a**3*b**3
*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) - 5*A*a**3*b**3*sqrt(a**(-7))*log(-a**4*sqrt(a**(
-7)) + sqrt(a + b*x))/16 + 5*A*a**3*b**3*sqrt(a**(-7))*log(a**4*sqrt(a**(-7)) + sqrt(a + b*x))/16 + 18*A*a**2*
b**3*(a + b*x)**(3/2)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 9*A*a**2*b**3*sqrt(a**(-5))*log(-a**3*sq
rt(a**(-5)) + sqrt(a + b*x))/8 - 9*A*a**2*b**3*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*A*a
*b**3*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + 3*A*a*b**3*sqrt(a**(-3))*log(a**2*sqrt(a**(-3
)) + sqrt(a + b*x))/2 + 2*A*b**3*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - 3*A*b**2*sqrt(a + b*x)/x - 10*B*a**4*
b**2*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 6*B*a**3*b**2*(a + b*x)**(3/2)/(-8*a**4 - 1
6*a**3*b*x + 8*a**2*(a + b*x)**2) + 3*B*a**3*b**2*sqrt(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3
*B*a**3*b**2*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*B*a**2*b**2*sqrt(a**(-3))*log(-a**2*s
qrt(a**(-3)) + sqrt(a + b*x))/2 + 3*B*a**2*b**2*sqrt(a**(-3))*log(a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + 6*B*
a*b**2*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - 3*B*a*b*sqrt(a + b*x)/x + 2*B*b**2*sqrt(a + b*x)

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