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

Optimal. Leaf size=69 \[ -2 a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {2}{3} A (a+b x)^{3/2}+2 a A \sqrt {a+b x}+\frac {2 B (a+b x)^{5/2}}{5 b} \]

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Rubi [A]  time = 0.02, antiderivative size = 69, 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, 208} \begin {gather*} -2 a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {2}{3} A (a+b x)^{3/2}+2 a A \sqrt {a+b x}+\frac {2 B (a+b x)^{5/2}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/x,x]

[Out]

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

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 71, normalized size = 1.03 \begin {gather*} A \left (\frac {2}{3} (a+b x)^{3/2}+a \left (2 \sqrt {a+b x}-2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )\right )+\frac {2 B (a+b x)^{5/2}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.04, size = 72, normalized size = 1.04 \begin {gather*} \frac {2 \left (5 A b (a+b x)^{3/2}+15 a A b \sqrt {a+b x}+3 B (a+b x)^{5/2}\right )}{15 b}-2 a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.07, size = 158, normalized size = 2.29 \begin {gather*} \left [\frac {15 \, A a^{\frac {3}{2}} b \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, B b^{2} x^{2} + 3 \, B a^{2} + 20 \, A a b + {\left (6 \, B a b + 5 \, A b^{2}\right )} x\right )} \sqrt {b x + a}}{15 \, b}, \frac {2 \, {\left (15 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, B b^{2} x^{2} + 3 \, B a^{2} + 20 \, A a b + {\left (6 \, B a b + 5 \, A b^{2}\right )} x\right )} \sqrt {b x + a}\right )}}{15 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*A*a^(3/2)*b*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(3*B*b^2*x^2 + 3*B*a^2 + 20*A*a*b + (6*
B*a*b + 5*A*b^2)*x)*sqrt(b*x + a))/b, 2/15*(15*A*sqrt(-a)*a*b*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (3*B*b^2*x^2
+ 3*B*a^2 + 20*A*a*b + (6*B*a*b + 5*A*b^2)*x)*sqrt(b*x + a))/b]

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giac [A]  time = 1.20, size = 72, normalized size = 1.04 \begin {gather*} \frac {2 \, A a^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B b^{4} + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{5} + 15 \, \sqrt {b x + a} A a b^{5}\right )}}{15 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 58, normalized size = 0.84 \begin {gather*} \frac {-2 A \,a^{\frac {3}{2}} b \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 \sqrt {b x +a}\, A a b +\frac {2 \left (b x +a \right )^{\frac {3}{2}} A b}{3}+\frac {2 \left (b x +a \right )^{\frac {5}{2}} B}{5}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.96, size = 73, normalized size = 1.06 \begin {gather*} A a^{\frac {3}{2}} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} A b + 15 \, \sqrt {b x + a} A a b\right )}}{15 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.39, size = 95, normalized size = 1.38 \begin {gather*} \left (\frac {2\,A\,b-2\,B\,a}{3\,b}+\frac {2\,B\,a}{3\,b}\right )\,{\left (a+b\,x\right )}^{3/2}+\frac {2\,B\,{\left (a+b\,x\right )}^{5/2}}{5\,b}+a\,\left (\frac {2\,A\,b-2\,B\,a}{b}+\frac {2\,B\,a}{b}\right )\,\sqrt {a+b\,x}+A\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^(3/2))/x,x)

[Out]

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

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sympy [A]  time = 30.34, size = 71, normalized size = 1.03 \begin {gather*} \frac {2 A a^{2} \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A a \sqrt {a + b x} + \frac {2 A \left (a + b x\right )^{\frac {3}{2}}}{3} + \frac {2 B \left (a + b x\right )^{\frac {5}{2}}}{5 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*A*a*sqrt(a + b*x) + 2*A*(a + b*x)**(3/2)/3 + 2*B*(a + b*x)*
*(5/2)/(5*b)

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