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

Optimal. Leaf size=108 \[ -\frac {\sqrt {a} (3 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}+\frac {\sqrt {x} (3 A b-5 a B)}{b^3}-\frac {x^{3/2} (3 A b-5 a B)}{3 a b^2}+\frac {x^{5/2} (A b-a B)}{a b (a+b x)} \]

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Rubi [A]  time = 0.05, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {27, 78, 50, 63, 205} \begin {gather*} -\frac {x^{3/2} (3 A b-5 a B)}{3 a b^2}+\frac {\sqrt {x} (3 A b-5 a B)}{b^3}-\frac {\sqrt {a} (3 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}+\frac {x^{5/2} (A b-a B)}{a b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

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Mathematica [A]  time = 0.06, size = 88, normalized size = 0.81 \begin {gather*} \frac {\sqrt {x} \left (-15 a^2 B+a b (9 A-10 B x)+2 b^2 x (3 A+B x)\right )}{3 b^3 (a+b x)}+\frac {\sqrt {a} (5 a B-3 A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-15*a^2*B + a*b*(9*A - 10*B*x) + 2*b^2*x*(3*A + B*x)))/(3*b^3*(a + b*x)) + (Sqrt[a]*(-3*A*b + 5*a*B)
*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(7/2)

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IntegrateAlgebraic [A]  time = 0.12, size = 95, normalized size = 0.88 \begin {gather*} \frac {\left (5 a^{3/2} B-3 \sqrt {a} A b\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}+\frac {\sqrt {x} \left (-15 a^2 B+9 a A b-10 a b B x+6 A b^2 x+2 b^2 B x^2\right )}{3 b^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 231, normalized size = 2.14 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B a^{2} - 3 \, A a b + {\left (5 \, B a b - 3 \, 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 (2 \, B b^{2} x^{2} - 15 \, B a^{2} + 9 \, A a b - 2 \, {\left (5 \, B a b - 3 \, A b^{2}\right )} x\right )} \sqrt {x}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {3 \, {\left (5 \, B a^{2} - 3 \, A a b + {\left (5 \, B a b - 3 \, A b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (2 \, B b^{2} x^{2} - 15 \, B a^{2} + 9 \, A a b - 2 \, {\left (5 \, B a b - 3 \, A b^{2}\right )} x\right )} \sqrt {x}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*(5*B*a^2 - 3*A*a*b + (5*B*a*b - 3*A*b^2)*x)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x +
a)) - 2*(2*B*b^2*x^2 - 15*B*a^2 + 9*A*a*b - 2*(5*B*a*b - 3*A*b^2)*x)*sqrt(x))/(b^4*x + a*b^3), 1/3*(3*(5*B*a^2
- 3*A*a*b + (5*B*a*b - 3*A*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (2*B*b^2*x^2 - 15*B*a^2 + 9*A*a*
b - 2*(5*B*a*b - 3*A*b^2)*x)*sqrt(x))/(b^4*x + a*b^3)]

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giac [A]  time = 0.19, size = 95, normalized size = 0.88 \begin {gather*} \frac {{\left (5 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {B a^{2} \sqrt {x} - A a b \sqrt {x}}{{\left (b x + a\right )} b^{3}} + \frac {2 \, {\left (B b^{4} x^{\frac {3}{2}} - 6 \, B a b^{3} \sqrt {x} + 3 \, A b^{4} \sqrt {x}\right )}}{3 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.13, size = 113, normalized size = 1.05 \begin {gather*} -\frac {3 A a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{2}}+\frac {5 B \,a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{3}}+\frac {A a \sqrt {x}}{\left (b x +a \right ) b^{2}}-\frac {B \,a^{2} \sqrt {x}}{\left (b x +a \right ) b^{3}}+\frac {2 B \,x^{\frac {3}{2}}}{3 b^{2}}+\frac {2 A \sqrt {x}}{b^{2}}-\frac {4 B a \sqrt {x}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.35, size = 88, normalized size = 0.81 \begin {gather*} -\frac {{\left (B a^{2} - A a b\right )} \sqrt {x}}{b^{4} x + a b^{3}} + \frac {{\left (5 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (B b x^{\frac {3}{2}} - 3 \, {\left (2 \, B a - A b\right )} \sqrt {x}\right )}}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 107, normalized size = 0.99 \begin {gather*} \sqrt {x}\,\left (\frac {2\,A}{b^2}-\frac {4\,B\,a}{b^3}\right )-\frac {\sqrt {x}\,\left (B\,a^2-A\,a\,b\right )}{x\,b^4+a\,b^3}+\frac {2\,B\,x^{3/2}}{3\,b^2}+\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\sqrt {x}\,\left (3\,A\,b-5\,B\,a\right )}{5\,B\,a^2-3\,A\,a\,b}\right )\,\left (3\,A\,b-5\,B\,a\right )}{b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 14.34, size = 932, normalized size = 8.63 \begin {gather*} \begin {cases} \tilde {\infty } \left (2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {7}{2}}}{7}}{a^{2}} & \text {for}\: b = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}}{b^{2}} & \text {for}\: a = 0 \\\frac {18 i A a^{\frac {3}{2}} b^{2} \sqrt {x} \sqrt {\frac {1}{b}}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} + \frac {12 i A \sqrt {a} b^{3} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} - \frac {9 A a^{2} b \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} + \frac {9 A a^{2} b \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} - \frac {9 A a b^{2} x \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} + \frac {9 A a b^{2} x \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} - \frac {30 i B a^{\frac {5}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} - \frac {20 i B a^{\frac {3}{2}} b^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} + \frac {4 i B \sqrt {a} b^{3} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} + \frac {15 B a^{3} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} - \frac {15 B a^{3} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} + \frac {15 B a^{2} b x \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} - \frac {15 B a^{2} b x \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 i \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(3/2)/3), Eq(a, 0) & Eq(b, 0)), ((2*A*x**(5/2)/5 + 2*B*x**(7/2)/7)/a**2,
Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3)/b**2, Eq(a, 0)), (18*I*A*a**(3/2)*b**2*sqrt(x)*sqrt(1/b)/(6*I*a**(3
/2)*b**4*sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) + 12*I*A*sqrt(a)*b**3*x**(3/2)*sqrt(1/b)/(6*I*a**(3/2)*b**4
*sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) - 9*A*a**2*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4
*sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) + 9*A*a**2*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4*
sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) - 9*A*a*b**2*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**
4*sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) + 9*A*a*b**2*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b*
*4*sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) - 30*I*B*a**(5/2)*b*sqrt(x)*sqrt(1/b)/(6*I*a**(3/2)*b**4*sqrt(1/b
) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) - 20*I*B*a**(3/2)*b**2*x**(3/2)*sqrt(1/b)/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I
*sqrt(a)*b**5*x*sqrt(1/b)) + 4*I*B*sqrt(a)*b**3*x**(5/2)*sqrt(1/b)/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I*sqrt(a)*
b**5*x*sqrt(1/b)) + 15*B*a**3*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I*sqrt(a)*b
**5*x*sqrt(1/b)) - 15*B*a**3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I*sqrt(a)*b**
5*x*sqrt(1/b)) + 15*B*a**2*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I*sqrt(a)*
b**5*x*sqrt(1/b)) - 15*B*a**2*b*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I*sqrt(a
)*b**5*x*sqrt(1/b)), True))

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