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

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

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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, 80, 50, 63, 205} \begin {gather*} -\frac {2 b^2 \sqrt {x} (b B-A c)}{c^4}+\frac {2 b^{5/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{9/2}}-\frac {2 x^{5/2} (b B-A c)}{5 c^2}+\frac {2 b x^{3/2} (b B-A c)}{3 c^3}+\frac {2 B x^{7/2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

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Mathematica [A]  time = 0.06, size = 101, normalized size = 0.89 \begin {gather*} \frac {2 b^{5/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{9/2}}+\frac {2 \sqrt {x} \left (35 b^2 c (3 A+B x)-7 b c^2 x (5 A+3 B x)+3 c^3 x^2 (7 A+5 B x)-105 b^3 B\right )}{105 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.11, size = 131, normalized size = 1.16 \begin {gather*} \frac {2 \left (b^{7/2} B-A b^{5/2} c\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{9/2}}+\frac {2 \left (105 A b^2 c \sqrt {x}-35 A b c^2 x^{3/2}+21 A c^3 x^{5/2}-105 b^3 B \sqrt {x}+35 b^2 B c x^{3/2}-21 b B c^2 x^{5/2}+15 B c^3 x^{7/2}\right )}{105 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 229, normalized size = 2.03 \begin {gather*} \left [-\frac {105 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x - 2 \, c \sqrt {x} \sqrt {-\frac {b}{c}} - b}{c x + b}\right ) - 2 \, {\left (15 \, B c^{3} x^{3} - 105 \, B b^{3} + 105 \, A b^{2} c - 21 \, {\left (B b c^{2} - A c^{3}\right )} x^{2} + 35 \, {\left (B b^{2} c - A b c^{2}\right )} x\right )} \sqrt {x}}{105 \, c^{4}}, \frac {2 \, {\left (105 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) + {\left (15 \, B c^{3} x^{3} - 105 \, B b^{3} + 105 \, A b^{2} c - 21 \, {\left (B b c^{2} - A c^{3}\right )} x^{2} + 35 \, {\left (B b^{2} c - A b c^{2}\right )} x\right )} \sqrt {x}\right )}}{105 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 115, normalized size = 1.02 \begin {gather*} \frac {2 \, {\left (B b^{4} - A b^{3} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{4}} + \frac {2 \, {\left (15 \, B c^{6} x^{\frac {7}{2}} - 21 \, B b c^{5} x^{\frac {5}{2}} + 21 \, A c^{6} x^{\frac {5}{2}} + 35 \, B b^{2} c^{4} x^{\frac {3}{2}} - 35 \, A b c^{5} x^{\frac {3}{2}} - 105 \, B b^{3} c^{3} \sqrt {x} + 105 \, A b^{2} c^{4} \sqrt {x}\right )}}{105 \, c^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(B*b^4 - A*b^3*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^4) + 2/105*(15*B*c^6*x^(7/2) - 21*B*b*c^5*x^(5/2)
+ 21*A*c^6*x^(5/2) + 35*B*b^2*c^4*x^(3/2) - 35*A*b*c^5*x^(3/2) - 105*B*b^3*c^3*sqrt(x) + 105*A*b^2*c^4*sqrt(x
))/c^7

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maple [A]  time = 0.05, size = 126, normalized size = 1.12 \begin {gather*} \frac {2 B \,x^{\frac {7}{2}}}{7 c}+\frac {2 A \,x^{\frac {5}{2}}}{5 c}-\frac {2 B b \,x^{\frac {5}{2}}}{5 c^{2}}-\frac {2 A \,b^{3} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{3}}+\frac {2 B \,b^{4} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{4}}-\frac {2 A b \,x^{\frac {3}{2}}}{3 c^{2}}+\frac {2 B \,b^{2} x^{\frac {3}{2}}}{3 c^{3}}+\frac {2 A \,b^{2} \sqrt {x}}{c^{3}}-\frac {2 B \,b^{3} \sqrt {x}}{c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.28, size = 105, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (B b^{4} - A b^{3} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{4}} + \frac {2 \, {\left (15 \, B c^{3} x^{\frac {7}{2}} - 21 \, {\left (B b c^{2} - A c^{3}\right )} x^{\frac {5}{2}} + 35 \, {\left (B b^{2} c - A b c^{2}\right )} x^{\frac {3}{2}} - 105 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {x}\right )}}{105 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(B*b^4 - A*b^3*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^4) + 2/105*(15*B*c^3*x^(7/2) - 21*(B*b*c^2 - A*c^
3)*x^(5/2) + 35*(B*b^2*c - A*b*c^2)*x^(3/2) - 105*(B*b^3 - A*b^2*c)*sqrt(x))/c^4

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mupad [B]  time = 1.04, size = 125, normalized size = 1.11 \begin {gather*} x^{5/2}\,\left (\frac {2\,A}{5\,c}-\frac {2\,B\,b}{5\,c^2}\right )+\frac {2\,B\,x^{7/2}}{7\,c}+\frac {b^2\,\sqrt {x}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )}{c^2}+\frac {2\,b^{5/2}\,\mathrm {atan}\left (\frac {b^{5/2}\,\sqrt {c}\,\sqrt {x}\,\left (A\,c-B\,b\right )}{B\,b^4-A\,b^3\,c}\right )\,\left (A\,c-B\,b\right )}{c^{9/2}}-\frac {b\,x^{3/2}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )}{3\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 45.39, size = 279, normalized size = 2.47 \begin {gather*} \begin {cases} \frac {i A b^{\frac {5}{2}} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{4} \sqrt {\frac {1}{c}}} - \frac {i A b^{\frac {5}{2}} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{4} \sqrt {\frac {1}{c}}} + \frac {2 A b^{2} \sqrt {x}}{c^{3}} - \frac {2 A b x^{\frac {3}{2}}}{3 c^{2}} + \frac {2 A x^{\frac {5}{2}}}{5 c} - \frac {i B b^{\frac {7}{2}} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{5} \sqrt {\frac {1}{c}}} + \frac {i B b^{\frac {7}{2}} \log {\left (i \sqrt {b} \sqrt {\frac {1}{c}} + \sqrt {x} \right )}}{c^{5} \sqrt {\frac {1}{c}}} - \frac {2 B b^{3} \sqrt {x}}{c^{4}} + \frac {2 B b^{2} x^{\frac {3}{2}}}{3 c^{3}} - \frac {2 B b x^{\frac {5}{2}}}{5 c^{2}} + \frac {2 B x^{\frac {7}{2}}}{7 c} & \text {for}\: c \neq 0 \\\frac {\frac {2 A x^{\frac {7}{2}}}{7} + \frac {2 B x^{\frac {9}{2}}}{9}}{b} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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