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

Optimal. Leaf size=169 \[ \frac {7 c^{3/2} (5 b B-9 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2} \]

[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)

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Rubi [A]  time = 0.09, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {7 c^{3/2} (5 b B-9 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2} \]

Antiderivative was successfully verified.

[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 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 {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx &=\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 ) \operatorname {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*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.36 \[ \frac {\frac {5 b^2 (A c-b B)}{(b+c x)^2}+(5 b B-9 A c) \, _2F_1\left (-\frac {5}{2},2;-\frac {3}{2};-\frac {c x}{b}\right )}{10 b^3 c x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*b^2*(-(b*B) + A*c))/(b + c*x)^2 + (5*b*B - 9*A*c)*Hypergeometric2F1[-5/2, 2, -3/2, -((c*x)/b)])/(10*b^3*c*
x^(5/2))

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fricas [A]  time = 0.91, size = 437, normalized size = 2.59 \[ \left [-\frac {105 \, {\left ({\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{5} + 2 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{4} + {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x - 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (24 \, A b^{4} - 105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} - 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x\right )} \sqrt {x}}{120 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}, -\frac {105 \, {\left ({\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{5} + 2 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{4} + {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) + {\left (24 \, A b^{4} - 105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} - 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x\right )} \sqrt {x}}{60 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.17, size = 135, normalized size = 0.80 \[ \frac {7 \, {\left (5 \, B b c^{2} - 9 \, A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{5}} + \frac {11 \, B b c^{3} x^{\frac {3}{2}} - 15 \, A c^{4} x^{\frac {3}{2}} + 13 \, B b^{2} c^{2} \sqrt {x} - 17 \, A b c^{3} \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} b^{5}} + \frac {2 \, {\left (45 \, B b c x^{2} - 90 \, A c^{2} x^{2} - 5 \, B b^{2} x + 15 \, A b c x - 3 \, A b^{2}\right )}}{15 \, b^{5} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.08, size = 178, normalized size = 1.05 \[ -\frac {15 A \,c^{4} x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} b^{5}}+\frac {11 B \,c^{3} x^{\frac {3}{2}}}{4 \left (c x +b \right )^{2} b^{4}}-\frac {17 A \,c^{3} \sqrt {x}}{4 \left (c x +b \right )^{2} b^{4}}+\frac {13 B \,c^{2} \sqrt {x}}{4 \left (c x +b \right )^{2} b^{3}}-\frac {63 A \,c^{3} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, b^{5}}+\frac {35 B \,c^{2} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, b^{4}}-\frac {12 A \,c^{2}}{b^{5} \sqrt {x}}+\frac {6 B c}{b^{4} \sqrt {x}}+\frac {2 A c}{b^{4} x^{\frac {3}{2}}}-\frac {2 B}{3 b^{3} x^{\frac {3}{2}}}-\frac {2 A}{5 b^{3} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4/b^5*c^4/(c*x+b)^2*x^(3/2)*A+11/4/b^4*c^3/(c*x+b)^2*x^(3/2)*B-17/4/b^4*c^3/(c*x+b)^2*A*x^(1/2)+13/4/b^3*c
^2/(c*x+b)^2*B*x^(1/2)-63/4/b^5*c^3/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*A+35/4/b^4*c^2/(b*c)^(1/2)*arc
tan(1/(b*c)^(1/2)*c*x^(1/2))*B-2/5/b^3*A/x^(5/2)+2/b^4/x^(3/2)*A*c-2/3/b^3/x^(3/2)*B-12*c^2/b^5/x^(1/2)*A+6*c/
b^4/x^(1/2)*B

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maxima [A]  time = 1.56, size = 154, normalized size = 0.91 \[ -\frac {24 \, A b^{4} - 105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} - 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x}{60 \, {\left (b^{5} c^{2} x^{\frac {9}{2}} + 2 \, b^{6} c x^{\frac {7}{2}} + b^{7} x^{\frac {5}{2}}\right )}} + \frac {7 \, {\left (5 \, B b c^{2} - 9 \, A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.14, size = 135, normalized size = 0.80 \[ -\frac {\frac {2\,A}{5\,b}-\frac {2\,x\,\left (9\,A\,c-5\,B\,b\right )}{15\,b^2}+\frac {35\,c^2\,x^3\,\left (9\,A\,c-5\,B\,b\right )}{12\,b^4}+\frac {7\,c^3\,x^4\,\left (9\,A\,c-5\,B\,b\right )}{4\,b^5}+\frac {14\,c\,x^2\,\left (9\,A\,c-5\,B\,b\right )}{15\,b^3}}{b^2\,x^{5/2}+c^2\,x^{9/2}+2\,b\,c\,x^{7/2}}-\frac {7\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (9\,A\,c-5\,B\,b\right )}{4\,b^{11/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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