3.2.91 \(\int \frac {\sqrt {x} (A+B x)}{(b x+c x^2)^3} \, dx\)

Optimal. Leaf size=147 \[ -\frac {5 \sqrt {c} (3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{9/2}}-\frac {5 (3 b B-7 A c)}{4 b^4 \sqrt {x}}+\frac {5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac {3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}-\frac {b B-A c}{2 b c x^{3/2} (b+c x)^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {781, 78, 51, 63, 205} \begin {gather*} -\frac {3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}+\frac {5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac {5 (3 b B-7 A c)}{4 b^4 \sqrt {x}}-\frac {5 \sqrt {c} (3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{9/2}}-\frac {b B-A c}{2 b c x^{3/2} (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(3*b*B - 7*A*c))/(12*b^3*c*x^(3/2)) - (5*(3*b*B - 7*A*c))/(4*b^4*Sqrt[x]) - (b*B - A*c)/(2*b*c*x^(3/2)*(b +
c*x)^2) - (3*b*B - 7*A*c)/(4*b^2*c*x^(3/2)*(b + c*x)) - (5*Sqrt[c]*(3*b*B - 7*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/S
qrt[b]])/(4*b^(9/2))

Rule 51

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

Rule 781

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e
*x)^(m + p)*(f + g*x)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^3} \, dx &=\int \frac {A+B x}{x^{5/2} (b+c x)^3} \, dx\\ &=-\frac {b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac {\left (\frac {3 b B}{2}-\frac {7 A c}{2}\right ) \int \frac {1}{x^{5/2} (b+c x)^2} \, dx}{2 b c}\\ &=-\frac {b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac {3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}-\frac {(5 (3 b B-7 A c)) \int \frac {1}{x^{5/2} (b+c x)} \, dx}{8 b^2 c}\\ &=\frac {5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac {b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac {3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}+\frac {(5 (3 b B-7 A c)) \int \frac {1}{x^{3/2} (b+c x)} \, dx}{8 b^3}\\ &=\frac {5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac {5 (3 b B-7 A c)}{4 b^4 \sqrt {x}}-\frac {b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac {3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}-\frac {(5 c (3 b B-7 A c)) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{8 b^4}\\ &=\frac {5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac {5 (3 b B-7 A c)}{4 b^4 \sqrt {x}}-\frac {b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac {3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}-\frac {(5 c (3 b B-7 A c)) \operatorname {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{4 b^4}\\ &=\frac {5 (3 b B-7 A c)}{12 b^3 c x^{3/2}}-\frac {5 (3 b B-7 A c)}{4 b^4 \sqrt {x}}-\frac {b B-A c}{2 b c x^{3/2} (b+c x)^2}-\frac {3 b B-7 A c}{4 b^2 c x^{3/2} (b+c x)}-\frac {5 \sqrt {c} (3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.41 \begin {gather*} \frac {\frac {3 b^2 (A c-b B)}{(b+c x)^2}+(3 b B-7 A c) \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};-\frac {c x}{b}\right )}{6 b^3 c x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.19, size = 125, normalized size = 0.85 \begin {gather*} \frac {-8 A b^3+56 A b^2 c x+175 A b c^2 x^2+105 A c^3 x^3-24 b^3 B x-75 b^2 B c x^2-45 b B c^2 x^3}{12 b^4 x^{3/2} (b+c x)^2}-\frac {5 \left (3 b B \sqrt {c}-7 A c^{3/2}\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*A*b^3 - 24*b^3*B*x + 56*A*b^2*c*x - 75*b^2*B*c*x^2 + 175*A*b*c^2*x^2 - 45*b*B*c^2*x^3 + 105*A*c^3*x^3)/(12
*b^4*x^(3/2)*(b + c*x)^2) - (5*(3*b*B*Sqrt[c] - 7*A*c^(3/2))*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(9/2))

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fricas [A]  time = 0.43, size = 380, normalized size = 2.59 \begin {gather*} \left [-\frac {15 \, {\left ({\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} + 2 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{3} + {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x + 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (8 \, A b^{3} + 15 \, {\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} + 25 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 8 \, {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x\right )} \sqrt {x}}{24 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}}, \frac {15 \, {\left ({\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} + 2 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{3} + {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2}\right )} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) - {\left (8 \, A b^{3} + 15 \, {\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} + 25 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 8 \, {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x\right )} \sqrt {x}}{12 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

[-1/24*(15*((3*B*b*c^2 - 7*A*c^3)*x^4 + 2*(3*B*b^2*c - 7*A*b*c^2)*x^3 + (3*B*b^3 - 7*A*b^2*c)*x^2)*sqrt(-c/b)*
log((c*x + 2*b*sqrt(x)*sqrt(-c/b) - b)/(c*x + b)) + 2*(8*A*b^3 + 15*(3*B*b*c^2 - 7*A*c^3)*x^3 + 25*(3*B*b^2*c
- 7*A*b*c^2)*x^2 + 8*(3*B*b^3 - 7*A*b^2*c)*x)*sqrt(x))/(b^4*c^2*x^4 + 2*b^5*c*x^3 + b^6*x^2), 1/12*(15*((3*B*b
*c^2 - 7*A*c^3)*x^4 + 2*(3*B*b^2*c - 7*A*b*c^2)*x^3 + (3*B*b^3 - 7*A*b^2*c)*x^2)*sqrt(c/b)*arctan(b*sqrt(c/b)/
(c*sqrt(x))) - (8*A*b^3 + 15*(3*B*b*c^2 - 7*A*c^3)*x^3 + 25*(3*B*b^2*c - 7*A*b*c^2)*x^2 + 8*(3*B*b^3 - 7*A*b^2
*c)*x)*sqrt(x))/(b^4*c^2*x^4 + 2*b^5*c*x^3 + b^6*x^2)]

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giac [A]  time = 0.16, size = 108, normalized size = 0.73 \begin {gather*} -\frac {5 \, {\left (3 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{4}} - \frac {2 \, {\left (3 \, B b x - 9 \, A c x + A b\right )}}{3 \, b^{4} x^{\frac {3}{2}}} - \frac {7 \, B b c^{2} x^{\frac {3}{2}} - 11 \, A c^{3} x^{\frac {3}{2}} + 9 \, B b^{2} c \sqrt {x} - 13 \, A b c^{2} \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

-5/4*(3*B*b*c - 7*A*c^2)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^4) - 2/3*(3*B*b*x - 9*A*c*x + A*b)/(b^4*x^(3
/2)) - 1/4*(7*B*b*c^2*x^(3/2) - 11*A*c^3*x^(3/2) + 9*B*b^2*c*sqrt(x) - 13*A*b*c^2*sqrt(x))/((c*x + b)^2*b^4)

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maple [A]  time = 0.09, size = 152, normalized size = 1.03 \begin {gather*} \frac {11 A \,c^{3} x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} b^{4}}-\frac {7 B \,c^{2} x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} b^{3}}+\frac {13 A \,c^{2} \sqrt {x}}{4 \left (c x +b \right )^{2} b^{3}}-\frac {9 B c \sqrt {x}}{4 \left (c x +b \right )^{2} b^{2}}+\frac {35 A \,c^{2} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, b^{4}}-\frac {15 B c \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, b^{3}}+\frac {6 A c}{b^{4} \sqrt {x}}-\frac {2 B}{b^{3} \sqrt {x}}-\frac {2 A}{3 b^{3} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*x^(1/2)/(c*x^2+b*x)^3,x)

[Out]

11/4/b^4*c^3/(c*x+b)^2*x^(3/2)*A-7/4/b^3*c^2/(c*x+b)^2*x^(3/2)*B+13/4/b^3*c^2/(c*x+b)^2*A*x^(1/2)-9/4/b^2*c/(c
*x+b)^2*B*x^(1/2)+35/4/b^4*c^2/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*A-15/4/b^3*c/(b*c)^(1/2)*arctan(1/(
b*c)^(1/2)*c*x^(1/2))*B-2/3/b^3*A/x^(3/2)+6/b^4/x^(1/2)*A*c-2/b^3/x^(1/2)*B

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maxima [A]  time = 1.29, size = 128, normalized size = 0.87 \begin {gather*} -\frac {8 \, A b^{3} + 15 \, {\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} + 25 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 8 \, {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x}{12 \, {\left (b^{4} c^{2} x^{\frac {7}{2}} + 2 \, b^{5} c x^{\frac {5}{2}} + b^{6} x^{\frac {3}{2}}\right )}} - \frac {5 \, {\left (3 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

-1/12*(8*A*b^3 + 15*(3*B*b*c^2 - 7*A*c^3)*x^3 + 25*(3*B*b^2*c - 7*A*b*c^2)*x^2 + 8*(3*B*b^3 - 7*A*b^2*c)*x)/(b
^4*c^2*x^(7/2) + 2*b^5*c*x^(5/2) + b^6*x^(3/2)) - 5/4*(3*B*b*c - 7*A*c^2)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*
c)*b^4)

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mupad [B]  time = 1.11, size = 114, normalized size = 0.78 \begin {gather*} \frac {\frac {2\,x\,\left (7\,A\,c-3\,B\,b\right )}{3\,b^2}-\frac {2\,A}{3\,b}+\frac {5\,c^2\,x^3\,\left (7\,A\,c-3\,B\,b\right )}{4\,b^4}+\frac {25\,c\,x^2\,\left (7\,A\,c-3\,B\,b\right )}{12\,b^3}}{b^2\,x^{3/2}+c^2\,x^{7/2}+2\,b\,c\,x^{5/2}}+\frac {5\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (7\,A\,c-3\,B\,b\right )}{4\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/2)*(A + B*x))/(b*x + c*x^2)^3,x)

[Out]

((2*x*(7*A*c - 3*B*b))/(3*b^2) - (2*A)/(3*b) + (5*c^2*x^3*(7*A*c - 3*B*b))/(4*b^4) + (25*c*x^2*(7*A*c - 3*B*b)
)/(12*b^3))/(b^2*x^(3/2) + c^2*x^(7/2) + 2*b*c*x^(5/2)) + (5*c^(1/2)*atan((c^(1/2)*x^(1/2))/b^(1/2))*(7*A*c -
3*B*b))/(4*b^(9/2))

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sympy [A]  time = 165.44, size = 1880, normalized size = 12.79

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x**(1/2)/(c*x**2+b*x)**3,x)

[Out]

Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(b, 0) & Eq(c, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(7*x**(
7/2)))/c**3, Eq(b, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/b**3, Eq(c, 0)), (-16*I*A*b**(7/2)*sqrt(1/c)/(24*I*
b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) +
112*I*A*b**(5/2)*c*x*sqrt(1/c)/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I
*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) + 350*I*A*b**(3/2)*c**2*x**2*sqrt(1/c)/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) +
48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) + 210*I*A*sqrt(b)*c**3*x**3*sqrt
(1/c)/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*s
qrt(1/c)) + 105*A*b**2*c*x**(3/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*
I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) - 105*A*b**2*c*x**(3/2)*log(I*sqrt(b
)*sqrt(1/c) + sqrt(x))/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2
)*c**2*x**(7/2)*sqrt(1/c)) + 210*A*b*c**2*x**(5/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(24*I*b**(13/2)*x**(3/2
)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) - 210*A*b*c**2*x**(
5/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/
c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) + 105*A*c**3*x**(7/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(24*I*b*
*(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) - 10
5*A*c**3*x**(7/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**
(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) - 48*I*B*b**(7/2)*x*sqrt(1/c)/(24*I*b**(13/2)*x**(3/2
)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) - 150*I*B*b**(5/2)*
c*x**2*sqrt(1/c)/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2
*x**(7/2)*sqrt(1/c)) - 90*I*B*b**(3/2)*c**2*x**3*sqrt(1/c)/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)
*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) - 45*B*b**3*x**(3/2)*log(-I*sqrt(b)*sqrt(1/c) +
sqrt(x))/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/
2)*sqrt(1/c)) + 45*B*b**3*x**(3/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*
I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) - 90*B*b**2*c*x**(5/2)*log(-I*sqrt(b
)*sqrt(1/c) + sqrt(x))/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2
)*c**2*x**(7/2)*sqrt(1/c)) + 90*B*b**2*c*x**(5/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(24*I*b**(13/2)*x**(3/2)*
sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) - 45*B*b*c**2*x**(7/2
)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(24*I*b**(13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c)
+ 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)) + 45*B*b*c**2*x**(7/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(24*I*b**(
13/2)*x**(3/2)*sqrt(1/c) + 48*I*b**(11/2)*c*x**(5/2)*sqrt(1/c) + 24*I*b**(9/2)*c**2*x**(7/2)*sqrt(1/c)), True)
)

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