3.4.45 \(\int \frac {x^{5/2} (A+B x)}{a+b x} \, dx\) [345]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {81, 52, 65, 211} \begin {gather*} -\frac {2 a^{5/2} (A b-a B) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{9/2}}+\frac {2 a^2 \sqrt {x} (A b-a B)}{b^4}-\frac {2 a x^{3/2} (A b-a B)}{3 b^3}+\frac {2 x^{5/2} (A b-a B)}{5 b^2}+\frac {2 B x^{7/2}}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x))/(a + b*x),x]

[Out]

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

Rule 52

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 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 81

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 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]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 101, normalized size = 0.89 \begin {gather*} \frac {2 \sqrt {x} \left (-105 a^3 B+35 a^2 b (3 A+B x)-7 a b^2 x (5 A+3 B x)+3 b^3 x^2 (7 A+5 B x)\right )}{105 b^4}+\frac {2 a^{5/2} (-A b+a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x))/(a + b*x),x]

[Out]

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

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Maple [A]
time = 0.06, size = 106, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {\frac {2 b^{3} B \,x^{\frac {7}{2}}}{7}+\frac {2 A \,b^{3} x^{\frac {5}{2}}}{5}-\frac {2 B a \,b^{2} x^{\frac {5}{2}}}{5}-\frac {2 A a \,b^{2} x^{\frac {3}{2}}}{3}+\frac {2 B \,a^{2} b \,x^{\frac {3}{2}}}{3}+2 a^{2} b A \sqrt {x}-2 a^{3} B \sqrt {x}}{b^{4}}-\frac {2 a^{3} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{4} \sqrt {a b}}\)	\(106\)
default	\(\frac {\frac {2 b^{3} B \,x^{\frac {7}{2}}}{7}+\frac {2 A \,b^{3} x^{\frac {5}{2}}}{5}-\frac {2 B a \,b^{2} x^{\frac {5}{2}}}{5}-\frac {2 A a \,b^{2} x^{\frac {3}{2}}}{3}+\frac {2 B \,a^{2} b \,x^{\frac {3}{2}}}{3}+2 a^{2} b A \sqrt {x}-2 a^{3} B \sqrt {x}}{b^{4}}-\frac {2 a^{3} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{4} \sqrt {a b}}\)	\(106\)
risch	\(\frac {2 \left (15 b^{3} B \,x^{3}+21 A \,b^{3} x^{2}-21 B a \,b^{2} x^{2}-35 a \,b^{2} A x +35 a^{2} b B x +105 a^{2} b A -105 a^{3} B \right ) \sqrt {x}}{105 b^{4}}-\frac {2 a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) A}{b^{3} \sqrt {a b}}+\frac {2 a^{4} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) B}{b^{4} \sqrt {a b}}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 105, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (B a^{4} - A a^{3} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {2 \, {\left (15 \, B b^{3} x^{\frac {7}{2}} - 21 \, {\left (B a b^{2} - A b^{3}\right )} x^{\frac {5}{2}} + 35 \, {\left (B a^{2} b - A a b^{2}\right )} x^{\frac {3}{2}} - 105 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {x}\right )}}{105 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.21, size = 229, normalized size = 2.03 \begin {gather*} \left [-\frac {105 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (15 \, B b^{3} x^{3} - 105 \, B a^{3} + 105 \, A a^{2} b - 21 \, {\left (B a b^{2} - A b^{3}\right )} x^{2} + 35 \, {\left (B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {x}}{105 \, b^{4}}, \frac {2 \, {\left (105 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (15 \, B b^{3} x^{3} - 105 \, B a^{3} + 105 \, A a^{2} b - 21 \, {\left (B a b^{2} - A b^{3}\right )} x^{2} + 35 \, {\left (B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {x}\right )}}{105 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (109) = 218\).
time = 6.18, size = 294, normalized size = 2.60 \begin {gather*} \begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {7}{2}}}{7}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {7}{2}}}{7}}{b} & \text {for}\: a = 0 \\\frac {\frac {2 A x^{\frac {7}{2}}}{7} + \frac {2 B x^{\frac {9}{2}}}{9}}{a} & \text {for}\: b = 0 \\- \frac {A a^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{4} \sqrt {- \frac {a}{b}}} + \frac {A a^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{4} \sqrt {- \frac {a}{b}}} + \frac {2 A a^{2} \sqrt {x}}{b^{3}} - \frac {2 A a x^{\frac {3}{2}}}{3 b^{2}} + \frac {2 A x^{\frac {5}{2}}}{5 b} + \frac {B a^{4} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{5} \sqrt {- \frac {a}{b}}} - \frac {B a^{4} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{5} \sqrt {- \frac {a}{b}}} - \frac {2 B a^{3} \sqrt {x}}{b^{4}} + \frac {2 B a^{2} x^{\frac {3}{2}}}{3 b^{3}} - \frac {2 B a x^{\frac {5}{2}}}{5 b^{2}} + \frac {2 B x^{\frac {7}{2}}}{7 b} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x+A)/(b*x+a),x)

[Out]

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

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Giac [A]
time = 1.31, size = 115, normalized size = 1.02 \begin {gather*} \frac {2 \, {\left (B a^{4} - A a^{3} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {2 \, {\left (15 \, B b^{6} x^{\frac {7}{2}} - 21 \, B a b^{5} x^{\frac {5}{2}} + 21 \, A b^{6} x^{\frac {5}{2}} + 35 \, B a^{2} b^{4} x^{\frac {3}{2}} - 35 \, A a b^{5} x^{\frac {3}{2}} - 105 \, B a^{3} b^{3} \sqrt {x} + 105 \, A a^{2} b^{4} \sqrt {x}\right )}}{105 \, b^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.36, size = 125, normalized size = 1.11 \begin {gather*} x^{5/2}\,\left (\frac {2\,A}{5\,b}-\frac {2\,B\,a}{5\,b^2}\right )+\frac {2\,B\,x^{7/2}}{7\,b}+\frac {a^2\,\sqrt {x}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{b^2}+\frac {2\,a^{5/2}\,\mathrm {atan}\left (\frac {a^{5/2}\,\sqrt {b}\,\sqrt {x}\,\left (A\,b-B\,a\right )}{B\,a^4-A\,a^3\,b}\right )\,\left (A\,b-B\,a\right )}{b^{9/2}}-\frac {a\,x^{3/2}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{3\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(5/2)*(A + B*x))/(a + b*x),x)

[Out]

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

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