3.2.75 \(\int \frac {A+B x}{x^{7/2} (b x+c x^2)} \, dx\)

Optimal. Leaf size=113 \[ -\frac {2 c^{5/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}}-\frac {2 c^2 (b B-A c)}{b^4 \sqrt {x}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}-\frac {2 (b B-A c)}{5 b^2 x^{5/2}}-\frac {2 A}{7 b x^{7/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 113, 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 {2 c^2 (b B-A c)}{b^4 \sqrt {x}}-\frac {2 c^{5/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}-\frac {2 (b B-A c)}{5 b^2 x^{5/2}}-\frac {2 A}{7 b x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(7/2)*(b*x + c*x^2)),x]

[Out]

(-2*A)/(7*b*x^(7/2)) - (2*(b*B - A*c))/(5*b^2*x^(5/2)) + (2*c*(b*B - A*c))/(3*b^3*x^(3/2)) - (2*c^2*(b*B - A*c
))/(b^4*Sqrt[x]) - (2*c^(5/2)*(b*B - 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^{7/2} \left (b x+c x^2\right )} \, dx &=\int \frac {A+B x}{x^{9/2} (b+c x)} \, dx\\ &=-\frac {2 A}{7 b x^{7/2}}+\frac {\left (2 \left (\frac {7 b B}{2}-\frac {7 A c}{2}\right )\right ) \int \frac {1}{x^{7/2} (b+c x)} \, dx}{7 b}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{5 b^2 x^{5/2}}-\frac {(c (b B-A c)) \int \frac {1}{x^{5/2} (b+c x)} \, dx}{b^2}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{5 b^2 x^{5/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}+\frac {\left (c^2 (b B-A c)\right ) \int \frac {1}{x^{3/2} (b+c x)} \, dx}{b^3}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{5 b^2 x^{5/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}-\frac {2 c^2 (b B-A c)}{b^4 \sqrt {x}}-\frac {\left (c^3 (b B-A c)\right ) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{b^4}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{5 b^2 x^{5/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}-\frac {2 c^2 (b B-A c)}{b^4 \sqrt {x}}-\frac {\left (2 c^3 (b B-A c)\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{b^4}\\ &=-\frac {2 A}{7 b x^{7/2}}-\frac {2 (b B-A c)}{5 b^2 x^{5/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}-\frac {2 c^2 (b B-A c)}{b^4 \sqrt {x}}-\frac {2 c^{5/2} (b B-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 = 0.01, size = 44, normalized size = 0.39 \begin {gather*} \frac {2 \left (\, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-\frac {c x}{b}\right ) (7 A c x-7 b B x)-5 A b\right )}{35 b^2 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(7/2)*(b*x + c*x^2)),x]

[Out]

(2*(-5*A*b + (-7*b*B*x + 7*A*c*x)*Hypergeometric2F1[-5/2, 1, -3/2, -((c*x)/b)]))/(35*b^2*x^(7/2))

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IntegrateAlgebraic [A]  time = 0.13, size = 115, normalized size = 1.02 \begin {gather*} -\frac {2 \left (b B c^{5/2}-A c^{7/2}\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{9/2}}-\frac {2 \left (15 A b^3-21 A b^2 c x+35 A b c^2 x^2-105 A c^3 x^3+21 b^3 B x-35 b^2 B c x^2+105 b B c^2 x^3\right )}{105 b^4 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(7/2)*(b*x + c*x^2)),x]

[Out]

(-2*(15*A*b^3 + 21*b^3*B*x - 21*A*b^2*c*x - 35*b^2*B*c*x^2 + 35*A*b*c^2*x^2 + 105*b*B*c^2*x^3 - 105*A*c^3*x^3)
)/(105*b^4*x^(7/2)) - (2*(b*B*c^(5/2) - A*c^(7/2))*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/b^(9/2)

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fricas [A]  time = 0.44, size = 246, normalized size = 2.18 \begin {gather*} \left [-\frac {105 \, {\left (B b c^{2} - A c^{3}\right )} x^{4} \sqrt {-\frac {c}{b}} \log \left (\frac {c x + 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (15 \, A b^{3} + 105 \, {\left (B b c^{2} - A c^{3}\right )} x^{3} - 35 \, {\left (B b^{2} c - A b c^{2}\right )} x^{2} + 21 \, {\left (B b^{3} - A b^{2} c\right )} x\right )} \sqrt {x}}{105 \, b^{4} x^{4}}, \frac {2 \, {\left (105 \, {\left (B b c^{2} - A c^{3}\right )} x^{4} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) - {\left (15 \, A b^{3} + 105 \, {\left (B b c^{2} - A c^{3}\right )} x^{3} - 35 \, {\left (B b^{2} c - A b c^{2}\right )} x^{2} + 21 \, {\left (B b^{3} - A b^{2} c\right )} x\right )} \sqrt {x}\right )}}{105 \, b^{4} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

[-1/105*(105*(B*b*c^2 - A*c^3)*x^4*sqrt(-c/b)*log((c*x + 2*b*sqrt(x)*sqrt(-c/b) - b)/(c*x + b)) + 2*(15*A*b^3
+ 105*(B*b*c^2 - A*c^3)*x^3 - 35*(B*b^2*c - A*b*c^2)*x^2 + 21*(B*b^3 - A*b^2*c)*x)*sqrt(x))/(b^4*x^4), 2/105*(
105*(B*b*c^2 - A*c^3)*x^4*sqrt(c/b)*arctan(b*sqrt(c/b)/(c*sqrt(x))) - (15*A*b^3 + 105*(B*b*c^2 - A*c^3)*x^3 -
35*(B*b^2*c - A*b*c^2)*x^2 + 21*(B*b^3 - A*b^2*c)*x)*sqrt(x))/(b^4*x^4)]

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giac [A]  time = 0.16, size = 104, normalized size = 0.92 \begin {gather*} -\frac {2 \, {\left (B b c^{3} - A c^{4}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{4}} - \frac {2 \, {\left (105 \, B b c^{2} x^{3} - 105 \, A c^{3} x^{3} - 35 \, B b^{2} c x^{2} + 35 \, A b c^{2} x^{2} + 21 \, B b^{3} x - 21 \, A b^{2} c x + 15 \, A b^{3}\right )}}{105 \, b^{4} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

-2*(B*b*c^3 - A*c^4)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^4) - 2/105*(105*B*b*c^2*x^3 - 105*A*c^3*x^3 - 35
*B*b^2*c*x^2 + 35*A*b*c^2*x^2 + 21*B*b^3*x - 21*A*b^2*c*x + 15*A*b^3)/(b^4*x^(7/2))

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maple [A]  time = 0.06, size = 126, normalized size = 1.12 \begin {gather*} \frac {2 A \,c^{4} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{4}}-\frac {2 B \,c^{3} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, b^{3}}+\frac {2 A \,c^{3}}{b^{4} \sqrt {x}}-\frac {2 B \,c^{2}}{b^{3} \sqrt {x}}-\frac {2 A \,c^{2}}{3 b^{3} x^{\frac {3}{2}}}+\frac {2 B c}{3 b^{2} x^{\frac {3}{2}}}+\frac {2 A c}{5 b^{2} x^{\frac {5}{2}}}-\frac {2 B}{5 b \,x^{\frac {5}{2}}}-\frac {2 A}{7 b \,x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(7/2)/(c*x^2+b*x),x)

[Out]

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

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maxima [A]  time = 1.32, size = 103, normalized size = 0.91 \begin {gather*} -\frac {2 \, {\left (B b c^{3} - A c^{4}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{4}} - \frac {2 \, {\left (15 \, A b^{3} + 105 \, {\left (B b c^{2} - A c^{3}\right )} x^{3} - 35 \, {\left (B b^{2} c - A b c^{2}\right )} x^{2} + 21 \, {\left (B b^{3} - A b^{2} c\right )} x\right )}}{105 \, b^{4} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x),x, algorithm="maxima")

[Out]

-2*(B*b*c^3 - A*c^4)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*b^4) - 2/105*(15*A*b^3 + 105*(B*b*c^2 - A*c^3)*x^3
- 35*(B*b^2*c - A*b*c^2)*x^2 + 21*(B*b^3 - A*b^2*c)*x)/(b^4*x^(7/2))

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mupad [B]  time = 1.09, size = 90, normalized size = 0.80 \begin {gather*} \frac {2\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (A\,c-B\,b\right )}{b^{9/2}}-\frac {\frac {2\,A}{7\,b}-\frac {2\,x\,\left (A\,c-B\,b\right )}{5\,b^2}-\frac {2\,c^2\,x^3\,\left (A\,c-B\,b\right )}{b^4}+\frac {2\,c\,x^2\,\left (A\,c-B\,b\right )}{3\,b^3}}{x^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(7/2)*(b*x + c*x^2)),x)

[Out]

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

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sympy [A]  time = 52.05, size = 326, normalized size = 2.88 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{7 x^{\frac {7}{2}}} - \frac {2 B}{5 x^{\frac {5}{2}}}}{b} & \text {for}\: c = 0 \\\frac {- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}}{c} & \text {for}\: b = 0 \\- \frac {2 A}{7 b x^{\frac {7}{2}}} + \frac {2 A c}{5 b^{2} x^{\frac {5}{2}}} - \frac {2 A c^{2}}{3 b^{3} x^{\frac {3}{2}}} + \frac {2 A c^{3}}{b^{4} \sqrt {x}} - \frac {i A c^{3} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {9}{2}} \sqrt {\frac {1}{c}}} + \frac {i A c^{3} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {9}{2}} \sqrt {\frac {1}{c}}} - \frac {2 B}{5 b x^{\frac {5}{2}}} + \frac {2 B c}{3 b^{2} x^{\frac {3}{2}}} - \frac {2 B c^{2}}{b^{3} \sqrt {x}} + \frac {i B c^{2} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {7}{2}} \sqrt {\frac {1}{c}}} - \frac {i B c^{2} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{b^{\frac {7}{2}} \sqrt {\frac {1}{c}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(7/2)/(c*x**2+b*x),x)

[Out]

Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(b, 0) & Eq(c, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(5*x**(
5/2)))/b, Eq(c, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2)))/c, Eq(b, 0)), (-2*A/(7*b*x**(7/2)) + 2*A*c/(5*b**
2*x**(5/2)) - 2*A*c**2/(3*b**3*x**(3/2)) + 2*A*c**3/(b**4*sqrt(x)) - I*A*c**3*log(-I*sqrt(b)*sqrt(1/c) + sqrt(
x))/(b**(9/2)*sqrt(1/c)) + I*A*c**3*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(9/2)*sqrt(1/c)) - 2*B/(5*b*x**(5/2
)) + 2*B*c/(3*b**2*x**(3/2)) - 2*B*c**2/(b**3*sqrt(x)) + I*B*c**2*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(7/2
)*sqrt(1/c)) - I*B*c**2*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(b**(7/2)*sqrt(1/c)), True))

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