\(\int \frac {A+B x}{\sqrt {x} (b x+c x^2)^3} \, dx\) [192]

Optimal result

Integrand size = 22, antiderivative size = 169 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {7 c^{3/2} (5 b B-9 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}} \]

[Out]

7/20*(-9*A*c+5*B*b)/b^3/c/x^(5/2)-7/12*(-9*A*c+5*B*b)/b^4/x^(3/2)+1/2*(A*c-B*b)/b/c/x^(5/2)/(c*x+b)^2+1/4*(9*A
*c-5*B*b)/b^2/c/x^(5/2)/(c*x+b)+7/4*c^(3/2)*(-9*A*c+5*B*b)*arctan(c^(1/2)*x^(1/2)/b^(1/2))/b^(11/2)+7/4*c*(-9*
A*c+5*B*b)/b^5/x^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {795, 79, 44, 53, 65, 211} \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\frac {7 c^{3/2} (5 b B-9 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2} \]

[In]

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

[Out]

(7*(5*b*B - 9*A*c))/(20*b^3*c*x^(5/2)) - (7*(5*b*B - 9*A*c))/(12*b^4*x^(3/2)) + (7*c*(5*b*B - 9*A*c))/(4*b^5*S
qrt[x]) - (b*B - A*c)/(2*b*c*x^(5/2)*(b + c*x)^2) - (5*b*B - 9*A*c)/(4*b^2*c*x^(5/2)*(b + c*x)) + (7*c^(3/2)*(
5*b*B - 9*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(11/2))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 65

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 79

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] || L
tQ[p, n]))))

Rule 211

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

Rule 795

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*} \text {integral}& = \int \frac {A+B x}{x^{7/2} (b+c x)^3} \, dx \\ & = -\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {\left (\frac {5 b B}{2}-\frac {9 A c}{2}\right ) \int \frac {1}{x^{7/2} (b+c x)^2} \, dx}{2 b c} \\ & = -\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}-\frac {(7 (5 b B-9 A c)) \int \frac {1}{x^{7/2} (b+c x)} \, dx}{8 b^2 c} \\ & = \frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {(7 (5 b B-9 A c)) \int \frac {1}{x^{5/2} (b+c x)} \, dx}{8 b^3} \\ & = \frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}-\frac {(7 c (5 b B-9 A c)) \int \frac {1}{x^{3/2} (b+c x)} \, dx}{8 b^4} \\ & = \frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {\left (7 c^2 (5 b B-9 A c)\right ) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{8 b^5} \\ & = \frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {\left (7 c^2 (5 b B-9 A c)\right ) \text {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{4 b^5} \\ & = \frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {7 c^{3/2} (5 b B-9 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\frac {5 b B x \left (-8 b^3+56 b^2 c x+175 b c^2 x^2+105 c^3 x^3\right )-3 A \left (8 b^4-24 b^3 c x+168 b^2 c^2 x^2+525 b c^3 x^3+315 c^4 x^4\right )}{60 b^5 x^{5/2} (b+c x)^2}+\frac {7 c^{3/2} (5 b B-9 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}} \]

[In]

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

[Out]

(5*b*B*x*(-8*b^3 + 56*b^2*c*x + 175*b*c^2*x^2 + 105*c^3*x^3) - 3*A*(8*b^4 - 24*b^3*c*x + 168*b^2*c^2*x^2 + 525
*b*c^3*x^3 + 315*c^4*x^4))/(60*b^5*x^(5/2)*(b + c*x)^2) + (7*c^(3/2)*(5*b*B - 9*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/
Sqrt[b]])/(4*b^(11/2))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(-\frac {2 c^{2} \left (\frac {\left (\frac {15}{8} A \,c^{2}-\frac {11}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (17 A c -13 B b \right ) \sqrt {x}}{8}}{\left (c x +b \right )^{2}}+\frac {7 \left (9 A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{5}}-\frac {2 A}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 \left (-3 A c +B b \right )}{3 b^{4} x^{\frac {3}{2}}}-\frac {6 c \left (2 A c -B b \right )}{b^{5} \sqrt {x}}\)	\(121\)
default	\(-\frac {2 c^{2} \left (\frac {\left (\frac {15}{8} A \,c^{2}-\frac {11}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (17 A c -13 B b \right ) \sqrt {x}}{8}}{\left (c x +b \right )^{2}}+\frac {7 \left (9 A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{5}}-\frac {2 A}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 \left (-3 A c +B b \right )}{3 b^{4} x^{\frac {3}{2}}}-\frac {6 c \left (2 A c -B b \right )}{b^{5} \sqrt {x}}\)	\(121\)
risch	\(-\frac {2 \left (90 A \,c^{2} x^{2}-45 B b c \,x^{2}-15 A b c x +5 b^{2} B x +3 A \,b^{2}\right )}{15 b^{5} x^{\frac {5}{2}}}-\frac {c^{2} \left (\frac {2 \left (\frac {15}{8} A \,c^{2}-\frac {11}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (17 A c -13 B b \right ) \sqrt {x}}{4}}{\left (c x +b \right )^{2}}+\frac {7 \left (9 A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}}\right )}{b^{5}}\)	\(124\)

[In]

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

[Out]

-2/b^5*c^2*(((15/8*A*c^2-11/8*B*b*c)*x^(3/2)+1/8*b*(17*A*c-13*B*b)*x^(1/2))/(c*x+b)^2+7/8*(9*A*c-5*B*b)/(b*c)^
(1/2)*arctan(c*x^(1/2)/(b*c)^(1/2)))-2/5*A/b^3/x^(5/2)-2/3*(-3*A*c+B*b)/b^4/x^(3/2)-6*c*(2*A*c-B*b)/b^5/x^(1/2
)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.59 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\left [-\frac {105 \, {\left ({\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{5} + 2 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{4} + {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x - 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (24 \, A b^{4} - 105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} - 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x\right )} \sqrt {x}}{120 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}, -\frac {105 \, {\left ({\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{5} + 2 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{4} + {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) + {\left (24 \, A b^{4} - 105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} - 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x\right )} \sqrt {x}}{60 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}\right ] \]

[In]

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

[Out]

[-1/120*(105*((5*B*b*c^3 - 9*A*c^4)*x^5 + 2*(5*B*b^2*c^2 - 9*A*b*c^3)*x^4 + (5*B*b^3*c - 9*A*b^2*c^2)*x^3)*sqr
t(-c/b)*log((c*x - 2*b*sqrt(x)*sqrt(-c/b) - b)/(c*x + b)) + 2*(24*A*b^4 - 105*(5*B*b*c^3 - 9*A*c^4)*x^4 - 175*
(5*B*b^2*c^2 - 9*A*b*c^3)*x^3 - 56*(5*B*b^3*c - 9*A*b^2*c^2)*x^2 + 8*(5*B*b^4 - 9*A*b^3*c)*x)*sqrt(x))/(b^5*c^
2*x^5 + 2*b^6*c*x^4 + b^7*x^3), -1/60*(105*((5*B*b*c^3 - 9*A*c^4)*x^5 + 2*(5*B*b^2*c^2 - 9*A*b*c^3)*x^4 + (5*B
*b^3*c - 9*A*b^2*c^2)*x^3)*sqrt(c/b)*arctan(b*sqrt(c/b)/(c*sqrt(x))) + (24*A*b^4 - 105*(5*B*b*c^3 - 9*A*c^4)*x
^4 - 175*(5*B*b^2*c^2 - 9*A*b*c^3)*x^3 - 56*(5*B*b^3*c - 9*A*b^2*c^2)*x^2 + 8*(5*B*b^4 - 9*A*b^3*c)*x)*sqrt(x)
)/(b^5*c^2*x^5 + 2*b^6*c*x^4 + b^7*x^3)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1882 vs. \(2 (163) = 326\).

Time = 122.82 (sec) , antiderivative size = 1882, normalized size of antiderivative = 11.14 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((zoo*(-2*A/(11*x**(11/2)) - 2*B/(9*x**(9/2))), Eq(b, 0) & Eq(c, 0)), ((-2*A/(5*x**(5/2)) - 2*B/(3*x*
*(3/2)))/b**3, Eq(c, 0)), ((-2*A/(11*x**(11/2)) - 2*B/(9*x**(9/2)))/c**3, Eq(b, 0)), (-48*A*b**4*sqrt(-b/c)/(1
20*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) + 144*A*b**3
*c*x*sqrt(-b/c)/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-
b/c)) - 945*A*b**2*c**2*x**(5/2)*log(sqrt(x) - sqrt(-b/c))/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)
*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) + 945*A*b**2*c**2*x**(5/2)*log(sqrt(x) + sqrt(-b/c))/(120*b**
7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) - 1008*A*b**2*c**2
*x**2*sqrt(-b/c)/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(
-b/c)) - 1890*A*b*c**3*x**(7/2)*log(sqrt(x) - sqrt(-b/c))/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*
sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) + 1890*A*b*c**3*x**(7/2)*log(sqrt(x) + sqrt(-b/c))/(120*b**7*x
**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) - 3150*A*b*c**3*x**3*
sqrt(-b/c)/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c))
- 945*A*c**4*x**(9/2)*log(sqrt(x) - sqrt(-b/c))/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c
) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) + 945*A*c**4*x**(9/2)*log(sqrt(x) + sqrt(-b/c))/(120*b**7*x**(5/2)*sqrt
(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) - 1890*A*c**4*x**4*sqrt(-b/c)/(12
0*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) - 80*B*b**4*x
*sqrt(-b/c)/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)
) + 525*B*b**3*c*x**(5/2)*log(sqrt(x) - sqrt(-b/c))/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-
b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) - 525*B*b**3*c*x**(5/2)*log(sqrt(x) + sqrt(-b/c))/(120*b**7*x**(5/2)
*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) + 560*B*b**3*c*x**2*sqrt(-b/
c)/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) + 1050*
B*b**2*c**2*x**(7/2)*log(sqrt(x) - sqrt(-b/c))/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c)
+ 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) - 1050*B*b**2*c**2*x**(7/2)*log(sqrt(x) + sqrt(-b/c))/(120*b**7*x**(5/2)*
sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) + 1750*B*b**2*c**2*x**3*sqrt(
-b/c)/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) + 52
5*B*b*c**3*x**(9/2)*log(sqrt(x) - sqrt(-b/c))/(120*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) +
120*b**5*c**2*x**(9/2)*sqrt(-b/c)) - 525*B*b*c**3*x**(9/2)*log(sqrt(x) + sqrt(-b/c))/(120*b**7*x**(5/2)*sqrt(
-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)) + 1050*B*b*c**3*x**4*sqrt(-b/c)/(1
20*b**7*x**(5/2)*sqrt(-b/c) + 240*b**6*c*x**(7/2)*sqrt(-b/c) + 120*b**5*c**2*x**(9/2)*sqrt(-b/c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=-\frac {24 \, A b^{4} - 105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} - 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x}{60 \, {\left (b^{5} c^{2} x^{\frac {9}{2}} + 2 \, b^{6} c x^{\frac {7}{2}} + b^{7} x^{\frac {5}{2}}\right )}} + \frac {7 \, {\left (5 \, B b c^{2} - 9 \, A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{5}} \]

[In]

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

[Out]

-1/60*(24*A*b^4 - 105*(5*B*b*c^3 - 9*A*c^4)*x^4 - 175*(5*B*b^2*c^2 - 9*A*b*c^3)*x^3 - 56*(5*B*b^3*c - 9*A*b^2*
c^2)*x^2 + 8*(5*B*b^4 - 9*A*b^3*c)*x)/(b^5*c^2*x^(9/2) + 2*b^6*c*x^(7/2) + b^7*x^(5/2)) + 7/4*(5*B*b*c^2 - 9*A
*c^3)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\frac {7 \, {\left (5 \, B b c^{2} - 9 \, A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{5}} + \frac {11 \, B b c^{3} x^{\frac {3}{2}} - 15 \, A c^{4} x^{\frac {3}{2}} + 13 \, B b^{2} c^{2} \sqrt {x} - 17 \, A b c^{3} \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} b^{5}} + \frac {2 \, {\left (45 \, B b c x^{2} - 90 \, A c^{2} x^{2} - 5 \, B b^{2} x + 15 \, A b c x - 3 \, A b^{2}\right )}}{15 \, b^{5} x^{\frac {5}{2}}} \]

[In]

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

[Out]

7/4*(5*B*b*c^2 - 9*A*c^3)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^5) + 1/4*(11*B*b*c^3*x^(3/2) - 15*A*c^4*x^(
3/2) + 13*B*b^2*c^2*sqrt(x) - 17*A*b*c^3*sqrt(x))/((c*x + b)^2*b^5) + 2/15*(45*B*b*c*x^2 - 90*A*c^2*x^2 - 5*B*
b^2*x + 15*A*b*c*x - 3*A*b^2)/(b^5*x^(5/2))

Mupad [B] (verification not implemented)

Time = 10.29 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=-\frac {\frac {2\,A}{5\,b}-\frac {2\,x\,\left (9\,A\,c-5\,B\,b\right )}{15\,b^2}+\frac {35\,c^2\,x^3\,\left (9\,A\,c-5\,B\,b\right )}{12\,b^4}+\frac {7\,c^3\,x^4\,\left (9\,A\,c-5\,B\,b\right )}{4\,b^5}+\frac {14\,c\,x^2\,\left (9\,A\,c-5\,B\,b\right )}{15\,b^3}}{b^2\,x^{5/2}+c^2\,x^{9/2}+2\,b\,c\,x^{7/2}}-\frac {7\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (9\,A\,c-5\,B\,b\right )}{4\,b^{11/2}} \]

[In]

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

[Out]

- ((2*A)/(5*b) - (2*x*(9*A*c - 5*B*b))/(15*b^2) + (35*c^2*x^3*(9*A*c - 5*B*b))/(12*b^4) + (7*c^3*x^4*(9*A*c -
5*B*b))/(4*b^5) + (14*c*x^2*(9*A*c - 5*B*b))/(15*b^3))/(b^2*x^(5/2) + c^2*x^(9/2) + 2*b*c*x^(7/2)) - (7*c^(3/2
)*atan((c^(1/2)*x^(1/2))/b^(1/2))*(9*A*c - 5*B*b))/(4*b^(11/2))