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

Optimal. Leaf size=90 \[ \frac {2 a^{3/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}-\frac {2 a \sqrt {x} (A b-a B)}{b^3}+\frac {2 x^{3/2} (A b-a B)}{3 b^2}+\frac {2 B x^{5/2}}{5 b} \]

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Rubi [A]  time = 0.04, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {80, 50, 63, 205} \begin {gather*} \frac {2 a^{3/2} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}+\frac {2 x^{3/2} (A b-a B)}{3 b^2}-\frac {2 a \sqrt {x} (A b-a B)}{b^3}+\frac {2 B x^{5/2}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(3/2)*(A + B*x))/(a + b*x),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 81, normalized size = 0.90 \begin {gather*} \frac {2 \sqrt {x} \left (15 a^2 B-5 a b (3 A+B x)+b^2 x (5 A+3 B x)\right )}{15 b^3}-\frac {2 a^{3/2} (a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(15*a^2*B - 5*a*b*(3*A + B*x) + b^2*x*(5*A + 3*B*x)))/(15*b^3) - (2*a^(3/2)*(-(A*b) + a*B)*ArcTan[(
Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(7/2)

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IntegrateAlgebraic [A]  time = 0.10, size = 88, normalized size = 0.98 \begin {gather*} \frac {2 \sqrt {x} \left (15 a^2 B-15 a A b-5 a b B x+5 A b^2 x+3 b^2 B x^2\right )}{15 b^3}-\frac {2 \left (a^{5/2} B-a^{3/2} A b\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(-15*a*A*b + 15*a^2*B + 5*A*b^2*x - 5*a*b*B*x + 3*b^2*B*x^2))/(15*b^3) - (2*(-(a^(3/2)*A*b) + a^(5/
2)*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(7/2)

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fricas [A]  time = 0.79, size = 180, normalized size = 2.00 \begin {gather*} \left [-\frac {15 \, {\left (B a^{2} - A a 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 (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {x}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (B a^{2} - A a b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} \sqrt {x}\right )}}{15 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15*(15*(B*a^2 - A*a*b)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(3*B*b^2*x^2 + 15*
B*a^2 - 15*A*a*b - 5*(B*a*b - A*b^2)*x)*sqrt(x))/b^3, -2/15*(15*(B*a^2 - A*a*b)*sqrt(a/b)*arctan(b*sqrt(x)*sqr
t(a/b)/a) - (3*B*b^2*x^2 + 15*B*a^2 - 15*A*a*b - 5*(B*a*b - A*b^2)*x)*sqrt(x))/b^3]

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giac [A]  time = 1.21, size = 91, normalized size = 1.01 \begin {gather*} -\frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, B b^{4} x^{\frac {5}{2}} - 5 \, B a b^{3} x^{\frac {3}{2}} + 5 \, A b^{4} x^{\frac {3}{2}} + 15 \, B a^{2} b^{2} \sqrt {x} - 15 \, A a b^{3} \sqrt {x}\right )}}{15 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 102, normalized size = 1.13 \begin {gather*} \frac {2 B \,x^{\frac {5}{2}}}{5 b}+\frac {2 A \,a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{2}}-\frac {2 B \,a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{3}}+\frac {2 A \,x^{\frac {3}{2}}}{3 b}-\frac {2 B a \,x^{\frac {3}{2}}}{3 b^{2}}-\frac {2 A a \sqrt {x}}{b^{2}}+\frac {2 B \,a^{2} \sqrt {x}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.93, size = 82, normalized size = 0.91 \begin {gather*} -\frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, B b^{2} x^{\frac {5}{2}} - 5 \, {\left (B a b - A b^{2}\right )} x^{\frac {3}{2}} + 15 \, {\left (B a^{2} - A a b\right )} \sqrt {x}\right )}}{15 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*a^3 - A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/15*(3*B*b^2*x^(5/2) - 5*(B*a*b - A*b^2)*x
^(3/2) + 15*(B*a^2 - A*a*b)*sqrt(x))/b^3

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mupad [B]  time = 0.08, size = 101, normalized size = 1.12 \begin {gather*} x^{3/2}\,\left (\frac {2\,A}{3\,b}-\frac {2\,B\,a}{3\,b^2}\right )+\frac {2\,B\,x^{5/2}}{5\,b}-\frac {2\,a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,\sqrt {x}\,\left (A\,b-B\,a\right )}{B\,a^3-A\,a^2\,b}\right )\,\left (A\,b-B\,a\right )}{b^{7/2}}-\frac {a\,\sqrt {x}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(3/2)*(A + B*x))/(a + b*x),x)

[Out]

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

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sympy [A]  time = 4.82, size = 245, normalized size = 2.72 \begin {gather*} \begin {cases} - \frac {i A a^{\frac {3}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{b^{3} \sqrt {\frac {1}{b}}} + \frac {i A a^{\frac {3}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{b^{3} \sqrt {\frac {1}{b}}} - \frac {2 A a \sqrt {x}}{b^{2}} + \frac {2 A x^{\frac {3}{2}}}{3 b} + \frac {i B a^{\frac {5}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{b^{4} \sqrt {\frac {1}{b}}} - \frac {i B a^{\frac {5}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{b^{4} \sqrt {\frac {1}{b}}} + \frac {2 B a^{2} \sqrt {x}}{b^{3}} - \frac {2 B a x^{\frac {3}{2}}}{3 b^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 b} & \text {for}\: b \neq 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {7}{2}}}{7}}{a} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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