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

Optimal. Leaf size=130 \[ \frac {a^{3/2} (5 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{9/2}}-\frac {a \sqrt {x} (5 A b-7 a B)}{b^4}+\frac {x^{3/2} (5 A b-7 a B)}{3 b^3}-\frac {x^{5/2} (5 A b-7 a B)}{5 a b^2}+\frac {x^{7/2} (A b-a B)}{a b (a+b x)} \]

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Rubi [A]  time = 0.06, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {78, 50, 63, 205} \begin {gather*} \frac {a^{3/2} (5 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{9/2}}-\frac {x^{5/2} (5 A b-7 a B)}{5 a b^2}+\frac {x^{3/2} (5 A b-7 a B)}{3 b^3}-\frac {a \sqrt {x} (5 A b-7 a B)}{b^4}+\frac {x^{7/2} (A b-a B)}{a b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(5*A*b - 7*a*B)*Sqrt[x])/b^4) + ((5*A*b - 7*a*B)*x^(3/2))/(3*b^3) - ((5*A*b - 7*a*B)*x^(5/2))/(5*a*b^2) +
((A*b - a*B)*x^(7/2))/(a*b*(a + b*x)) + (a^(3/2)*(5*A*b - 7*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(9/2)

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 205

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

Rubi steps

\begin {align*} \int \frac {x^{5/2} (A+B x)}{(a+b x)^2} \, dx &=\frac {(A b-a B) x^{7/2}}{a b (a+b x)}-\frac {\left (\frac {5 A b}{2}-\frac {7 a B}{2}\right ) \int \frac {x^{5/2}}{a+b x} \, dx}{a b}\\ &=-\frac {(5 A b-7 a B) x^{5/2}}{5 a b^2}+\frac {(A b-a B) x^{7/2}}{a b (a+b x)}+\frac {(5 A b-7 a B) \int \frac {x^{3/2}}{a+b x} \, dx}{2 b^2}\\ &=\frac {(5 A b-7 a B) x^{3/2}}{3 b^3}-\frac {(5 A b-7 a B) x^{5/2}}{5 a b^2}+\frac {(A b-a B) x^{7/2}}{a b (a+b x)}-\frac {(a (5 A b-7 a B)) \int \frac {\sqrt {x}}{a+b x} \, dx}{2 b^3}\\ &=-\frac {a (5 A b-7 a B) \sqrt {x}}{b^4}+\frac {(5 A b-7 a B) x^{3/2}}{3 b^3}-\frac {(5 A b-7 a B) x^{5/2}}{5 a b^2}+\frac {(A b-a B) x^{7/2}}{a b (a+b x)}+\frac {\left (a^2 (5 A b-7 a B)\right ) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{2 b^4}\\ &=-\frac {a (5 A b-7 a B) \sqrt {x}}{b^4}+\frac {(5 A b-7 a B) x^{3/2}}{3 b^3}-\frac {(5 A b-7 a B) x^{5/2}}{5 a b^2}+\frac {(A b-a B) x^{7/2}}{a b (a+b x)}+\frac {\left (a^2 (5 A b-7 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{b^4}\\ &=-\frac {a (5 A b-7 a B) \sqrt {x}}{b^4}+\frac {(5 A b-7 a B) x^{3/2}}{3 b^3}-\frac {(5 A b-7 a B) x^{5/2}}{5 a b^2}+\frac {(A b-a B) x^{7/2}}{a b (a+b x)}+\frac {a^{3/2} (5 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 110, normalized size = 0.85 \begin {gather*} \frac {\sqrt {x} \left (105 a^3 B+a^2 (70 b B x-75 A b)-2 a b^2 x (25 A+7 B x)+2 b^3 x^2 (5 A+3 B x)\right )}{15 b^4 (a+b x)}-\frac {a^{3/2} (7 a B-5 A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(105*a^3*B + 2*b^3*x^2*(5*A + 3*B*x) - 2*a*b^2*x*(25*A + 7*B*x) + a^2*(-75*A*b + 70*b*B*x)))/(15*b^4*
(a + b*x)) - (a^(3/2)*(-5*A*b + 7*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(9/2)

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IntegrateAlgebraic [A]  time = 0.14, size = 119, normalized size = 0.92 \begin {gather*} \frac {\left (5 a^{3/2} A b-7 a^{5/2} B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{9/2}}+\frac {\sqrt {x} \left (105 a^3 B-75 a^2 A b+70 a^2 b B x-50 a A b^2 x-14 a b^2 B x^2+10 A b^3 x^2+6 b^3 B x^3\right )}{15 b^4 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-75*a^2*A*b + 105*a^3*B - 50*a*A*b^2*x + 70*a^2*b*B*x + 10*A*b^3*x^2 - 14*a*b^2*B*x^2 + 6*b^3*B*x^3)
)/(15*b^4*(a + b*x)) + ((5*a^(3/2)*A*b - 7*a^(5/2)*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(9/2)

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fricas [A]  time = 0.83, size = 290, normalized size = 2.23 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b + {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (6 \, B b^{3} x^{3} + 105 \, B a^{3} - 75 \, A a^{2} b - 2 \, {\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} x^{2} + 10 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x\right )} \sqrt {x}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b + {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (6 \, B b^{3} x^{3} + 105 \, B a^{3} - 75 \, A a^{2} b - 2 \, {\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} x^{2} + 10 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x\right )} \sqrt {x}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/30*(15*(7*B*a^3 - 5*A*a^2*b + (7*B*a^2*b - 5*A*a*b^2)*x)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)
/(b*x + a)) - 2*(6*B*b^3*x^3 + 105*B*a^3 - 75*A*a^2*b - 2*(7*B*a*b^2 - 5*A*b^3)*x^2 + 10*(7*B*a^2*b - 5*A*a*b^
2)*x)*sqrt(x))/(b^5*x + a*b^4), -1/15*(15*(7*B*a^3 - 5*A*a^2*b + (7*B*a^2*b - 5*A*a*b^2)*x)*sqrt(a/b)*arctan(b
*sqrt(x)*sqrt(a/b)/a) - (6*B*b^3*x^3 + 105*B*a^3 - 75*A*a^2*b - 2*(7*B*a*b^2 - 5*A*b^3)*x^2 + 10*(7*B*a^2*b -
5*A*a*b^2)*x)*sqrt(x))/(b^5*x + a*b^4)]

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giac [A]  time = 1.30, size = 122, normalized size = 0.94 \begin {gather*} -\frac {{\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {B a^{3} \sqrt {x} - A a^{2} b \sqrt {x}}{{\left (b x + a\right )} b^{4}} + \frac {2 \, {\left (3 \, B b^{8} x^{\frac {5}{2}} - 10 \, B a b^{7} x^{\frac {3}{2}} + 5 \, A b^{8} x^{\frac {3}{2}} + 45 \, B a^{2} b^{6} \sqrt {x} - 30 \, A a b^{7} \sqrt {x}\right )}}{15 \, b^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(7*B*a^3 - 5*A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + (B*a^3*sqrt(x) - A*a^2*b*sqrt(x))/((b*x +
a)*b^4) + 2/15*(3*B*b^8*x^(5/2) - 10*B*a*b^7*x^(3/2) + 5*A*b^8*x^(3/2) + 45*B*a^2*b^6*sqrt(x) - 30*A*a*b^7*sq
rt(x))/b^10

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maple [A]  time = 0.02, size = 139, normalized size = 1.07 \begin {gather*} \frac {2 B \,x^{\frac {5}{2}}}{5 b^{2}}+\frac {5 A \,a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{3}}-\frac {7 B \,a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{4}}-\frac {A \,a^{2} \sqrt {x}}{\left (b x +a \right ) b^{3}}+\frac {2 A \,x^{\frac {3}{2}}}{3 b^{2}}+\frac {B \,a^{3} \sqrt {x}}{\left (b x +a \right ) b^{4}}-\frac {4 B a \,x^{\frac {3}{2}}}{3 b^{3}}-\frac {4 A a \sqrt {x}}{b^{3}}+\frac {6 B \,a^{2} \sqrt {x}}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/b^2*B*x^(5/2)+2/3/b^2*A*x^(3/2)-4/3/b^3*B*x^(3/2)*a-4/b^3*a*A*x^(1/2)+6/b^4*a^2*B*x^(1/2)-a^2/b^3*x^(1/2)/
(b*x+a)*A+a^3/b^4*x^(1/2)/(b*x+a)*B+5*a^2/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A-7*a^3/b^4/(a*b)^(1
/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*B

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maxima [A]  time = 2.03, size = 115, normalized size = 0.88 \begin {gather*} \frac {{\left (B a^{3} - A a^{2} b\right )} \sqrt {x}}{b^{5} x + a b^{4}} - \frac {{\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {2 \, {\left (3 \, B b^{2} x^{\frac {5}{2}} - 5 \, {\left (2 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} + 15 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} \sqrt {x}\right )}}{15 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*a^3 - A*a^2*b)*sqrt(x)/(b^5*x + a*b^4) - (7*B*a^3 - 5*A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4)
+ 2/15*(3*B*b^2*x^(5/2) - 5*(2*B*a*b - A*b^2)*x^(3/2) + 15*(3*B*a^2 - 2*A*a*b)*sqrt(x))/b^4

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mupad [B]  time = 0.06, size = 146, normalized size = 1.12 \begin {gather*} x^{3/2}\,\left (\frac {2\,A}{3\,b^2}-\frac {4\,B\,a}{3\,b^3}\right )-\sqrt {x}\,\left (\frac {2\,a\,\left (\frac {2\,A}{b^2}-\frac {4\,B\,a}{b^3}\right )}{b}+\frac {2\,B\,a^2}{b^4}\right )+\frac {2\,B\,x^{5/2}}{5\,b^2}+\frac {\sqrt {x}\,\left (B\,a^3-A\,a^2\,b\right )}{x\,b^5+a\,b^4}-\frac {a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,\sqrt {x}\,\left (5\,A\,b-7\,B\,a\right )}{7\,B\,a^3-5\,A\,a^2\,b}\right )\,\left (5\,A\,b-7\,B\,a\right )}{b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(3/2)*((2*A)/(3*b^2) - (4*B*a)/(3*b^3)) - x^(1/2)*((2*a*((2*A)/b^2 - (4*B*a)/b^3))/b + (2*B*a^2)/b^4) + (2*B
*x^(5/2))/(5*b^2) + (x^(1/2)*(B*a^3 - A*a^2*b))/(a*b^4 + b^5*x) - (a^(3/2)*atan((a^(3/2)*b^(1/2)*x^(1/2)*(5*A*
b - 7*B*a))/(7*B*a^3 - 5*A*a^2*b))*(5*A*b - 7*B*a))/b^(9/2)

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sympy [A]  time = 54.88, size = 1068, normalized size = 8.22 \begin {gather*} \begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{b^{2}} & \text {for}\: a = 0 \\\frac {\frac {2 A x^{\frac {7}{2}}}{7} + \frac {2 B x^{\frac {9}{2}}}{9}}{a^{2}} & \text {for}\: b = 0 \\- \frac {150 i A a^{\frac {5}{2}} b^{2} \sqrt {x} \sqrt {\frac {1}{b}}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} - \frac {100 i A a^{\frac {3}{2}} b^{3} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} + \frac {20 i A \sqrt {a} b^{4} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} + \frac {75 A a^{3} b \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} - \frac {75 A a^{3} b \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} + \frac {75 A a^{2} b^{2} x \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} - \frac {75 A a^{2} b^{2} x \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} + \frac {210 i B a^{\frac {7}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} + \frac {140 i B a^{\frac {5}{2}} b^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} - \frac {28 i B a^{\frac {3}{2}} b^{3} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} + \frac {12 i B \sqrt {a} b^{4} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} - \frac {105 B a^{4} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} + \frac {105 B a^{4} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} - \frac {105 B a^{3} b x \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} + \frac {105 B a^{3} b x \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{30 i a^{\frac {3}{2}} b^{5} \sqrt {\frac {1}{b}} + 30 i \sqrt {a} b^{6} x \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(2*A*x**(3/2)/3 + 2*B*x**(5/2)/5), Eq(a, 0) & Eq(b, 0)), ((2*A*x**(3/2)/3 + 2*B*x**(5/2)/5)/b**
2, Eq(a, 0)), ((2*A*x**(7/2)/7 + 2*B*x**(9/2)/9)/a**2, Eq(b, 0)), (-150*I*A*a**(5/2)*b**2*sqrt(x)*sqrt(1/b)/(3
0*I*a**(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) - 100*I*A*a**(3/2)*b**3*x**(3/2)*sqrt(1/b)/(30*I*
a**(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) + 20*I*A*sqrt(a)*b**4*x**(5/2)*sqrt(1/b)/(30*I*a**(3/
2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) + 75*A*a**3*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*a**
(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) - 75*A*a**3*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*a
**(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) + 75*A*a**2*b**2*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))
/(30*I*a**(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) - 75*A*a**2*b**2*x*log(I*sqrt(a)*sqrt(1/b) + s
qrt(x))/(30*I*a**(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) + 210*I*B*a**(7/2)*b*sqrt(x)*sqrt(1/b)/
(30*I*a**(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) + 140*I*B*a**(5/2)*b**2*x**(3/2)*sqrt(1/b)/(30*
I*a**(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) - 28*I*B*a**(3/2)*b**3*x**(5/2)*sqrt(1/b)/(30*I*a**
(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) + 12*I*B*sqrt(a)*b**4*x**(7/2)*sqrt(1/b)/(30*I*a**(3/2)*
b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) - 105*B*a**4*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*a**(3/2
)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) + 105*B*a**4*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*a**(3/
2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) - 105*B*a**3*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(30*I*
a**(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)) + 105*B*a**3*b*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(
30*I*a**(3/2)*b**5*sqrt(1/b) + 30*I*sqrt(a)*b**6*x*sqrt(1/b)), True))

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