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

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

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Rubi [A]  time = 0.07, antiderivative size = 130, 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 {c^{3/2} (5 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}}-\frac {5 b B-7 A c}{3 b^3 x^{3/2}}+\frac {5 b B-7 A c}{5 b^2 c x^{5/2}}+\frac {c (5 b B-7 A c)}{b^4 \sqrt {x}}-\frac {b B-A c}{b c x^{5/2} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.02, size = 64, normalized size = 0.49 \begin {gather*} \frac {(b+c x) (5 b B-7 A c) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-\frac {c x}{b}\right )+5 b (A c-b B)}{5 b^2 c x^{5/2} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*(-(b*B) + A*c) + (5*b*B - 7*A*c)*(b + c*x)*Hypergeometric2F1[-5/2, 1, -3/2, -((c*x)/b)])/(5*b^2*c*x^(5/2)
*(b + c*x))

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IntegrateAlgebraic [A]  time = 0.15, size = 122, normalized size = 0.94 \begin {gather*} \frac {\left (5 b B c^{3/2}-7 A c^{5/2}\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}}+\frac {-6 A b^3+14 A b^2 c x-70 A b c^2 x^2-105 A c^3 x^3-10 b^3 B x+50 b^2 B c x^2+75 b B c^2 x^3}{15 b^4 x^{5/2} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 319, normalized size = 2.45 \begin {gather*} \left [-\frac {15 \, {\left ({\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} + {\left (5 \, B b^{2} c - 7 \, A b 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 (6 \, A b^{3} - 15 \, {\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} - 10 \, {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 2 \, {\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x\right )} \sqrt {x}}{30 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}, -\frac {15 \, {\left ({\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} + {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) + {\left (6 \, A b^{3} - 15 \, {\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} - 10 \, {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 2 \, {\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x\right )} \sqrt {x}}{15 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 110, normalized size = 0.85 \begin {gather*} \frac {{\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{4}} + \frac {B b c^{2} \sqrt {x} - A c^{3} \sqrt {x}}{{\left (c x + b\right )} b^{4}} + \frac {2 \, {\left (30 \, B b c x^{2} - 45 \, A c^{2} x^{2} - 5 \, B b^{2} x + 10 \, A b c x - 3 \, A b^{2}\right )}}{15 \, b^{4} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*B*b*c^2 - 7*A*c^3)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^4) + (B*b*c^2*sqrt(x) - A*c^3*sqrt(x))/((c*x +
b)*b^4) + 2/15*(30*B*b*c*x^2 - 45*A*c^2*x^2 - 5*B*b^2*x + 10*A*b*c*x - 3*A*b^2)/(b^4*x^(5/2))

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maple [A]  time = 0.06, size = 139, normalized size = 1.07 \begin {gather*} -\frac {7 A \,c^{3} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{4}}+\frac {5 B \,c^{2} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{3}}-\frac {A \,c^{3} \sqrt {x}}{\left (c x +b \right ) b^{4}}+\frac {B \,c^{2} \sqrt {x}}{\left (c x +b \right ) b^{3}}-\frac {6 A \,c^{2}}{b^{4} \sqrt {x}}+\frac {4 B c}{b^{3} \sqrt {x}}+\frac {4 A c}{3 b^{3} x^{\frac {3}{2}}}-\frac {2 B}{3 b^{2} x^{\frac {3}{2}}}-\frac {2 A}{5 b^{2} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.17, size = 118, normalized size = 0.91 \begin {gather*} -\frac {6 \, A b^{3} - 15 \, {\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{3} - 10 \, {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2} + 2 \, {\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x}{15 \, {\left (b^{4} c x^{\frac {7}{2}} + b^{5} x^{\frac {5}{2}}\right )}} + \frac {{\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(6*A*b^3 - 15*(5*B*b*c^2 - 7*A*c^3)*x^3 - 10*(5*B*b^2*c - 7*A*b*c^2)*x^2 + 2*(5*B*b^3 - 7*A*b^2*c)*x)/(b
^4*c*x^(7/2) + b^5*x^(5/2)) + (5*B*b*c^2 - 7*A*c^3)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^4)

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mupad [B]  time = 1.11, size = 103, normalized size = 0.79 \begin {gather*} -\frac {\frac {2\,A}{5\,b}-\frac {2\,x\,\left (7\,A\,c-5\,B\,b\right )}{15\,b^2}+\frac {c^2\,x^3\,\left (7\,A\,c-5\,B\,b\right )}{b^4}+\frac {2\,c\,x^2\,\left (7\,A\,c-5\,B\,b\right )}{3\,b^3}}{b\,x^{5/2}+c\,x^{7/2}}-\frac {c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (7\,A\,c-5\,B\,b\right )}{b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 81.93, size = 1127, normalized size = 8.67

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(3/2)/(c*x**2+b*x)**2,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**2, Eq(b, 0)), ((-2*A/(5*x**(5/2)) - 2*B/(3*x**(3/2)))/b**2, Eq(c, 0)), (-12*I*A*b**(7/2)*sqrt(1/c)/(
30*I*b**(11/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) + 28*I*A*b**(5/2)*c*x*sqrt(1/c)/(30*I*
b**(11/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) - 140*I*A*b**(3/2)*c**2*x**2*sqrt(1/c)/(30*
I*b**(11/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) - 210*I*A*sqrt(b)*c**3*x**3*sqrt(1/c)/(30
*I*b**(11/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) - 105*A*b*c**2*x**(5/2)*log(-I*sqrt(b)*s
qrt(1/c) + sqrt(x))/(30*I*b**(11/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) + 105*A*b*c**2*x*
*(5/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(30*I*b**(11/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1
/c)) - 105*A*c**3*x**(7/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(30*I*b**(11/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9
/2)*c*x**(7/2)*sqrt(1/c)) + 105*A*c**3*x**(7/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(30*I*b**(11/2)*x**(5/2)*sq
rt(1/c) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) - 20*I*B*b**(7/2)*x*sqrt(1/c)/(30*I*b**(11/2)*x**(5/2)*sqrt(1/c)
+ 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) + 100*I*B*b**(5/2)*c*x**2*sqrt(1/c)/(30*I*b**(11/2)*x**(5/2)*sqrt(1/c)
+ 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) + 150*I*B*b**(3/2)*c**2*x**3*sqrt(1/c)/(30*I*b**(11/2)*x**(5/2)*sqrt(1/c
) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) + 75*B*b**2*c*x**(5/2)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(30*I*b**(1
1/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) - 75*B*b**2*c*x**(5/2)*log(I*sqrt(b)*sqrt(1/c) +
sqrt(x))/(30*I*b**(11/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) + 75*B*b*c**2*x**(7/2)*log(
-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(30*I*b**(11/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9/2)*c*x**(7/2)*sqrt(1/c)) - 75*
B*b*c**2*x**(7/2)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(30*I*b**(11/2)*x**(5/2)*sqrt(1/c) + 30*I*b**(9/2)*c*x**(
7/2)*sqrt(1/c)), True))

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