\(\int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx\) [403]

Optimal result

Integrand size = 18, antiderivative size = 69 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=2 a A \sqrt {a+b x}+\frac {2}{3} A (a+b x)^{3/2}+\frac {2 B (a+b x)^{5/2}}{5 b}-2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {81, 52, 65, 214} \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=-2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {2}{3} A (a+b x)^{3/2}+2 a A \sqrt {a+b x}+\frac {2 B (a+b x)^{5/2}}{5 b} \]

[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 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 214

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

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=\frac {2 \sqrt {a+b x} \left (15 a A b+5 A b (a+b x)+3 B (a+b x)^2\right )}{15 b}-2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A a b \sqrt {b x +a}-2 A \,a^{\frac {3}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b}\)	\(58\)
default	\(\frac {\frac {2 B \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A a b \sqrt {b x +a}-2 A \,a^{\frac {3}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b}\)	\(58\)
pseudoelliptic	\(\frac {-2 A \,a^{\frac {3}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {8 \left (\frac {\left (\frac {3 B x}{5}+A \right ) x \,b^{2}}{4}+a \left (\frac {3 B x}{10}+A \right ) b +\frac {3 a^{2} B}{20}\right ) \sqrt {b x +a}}{3}}{b}\)	\(63\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.29 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=\left [\frac {15 \, A a^{\frac {3}{2}} b \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, B b^{2} x^{2} + 3 \, B a^{2} + 20 \, A a b + {\left (6 \, B a b + 5 \, A b^{2}\right )} x\right )} \sqrt {b x + a}}{15 \, b}, \frac {2 \, {\left (15 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, B b^{2} x^{2} + 3 \, B a^{2} + 20 \, A a b + {\left (6 \, B a b + 5 \, A b^{2}\right )} x\right )} \sqrt {b x + a}\right )}}{15 \, b}\right ] \]

[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]

Sympy [A] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=\begin {cases} \frac {2 A a^{2} \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A a \sqrt {a + b x} + \frac {2 A \left (a + b x\right )^{\frac {3}{2}}}{3} + \frac {2 B \left (a + b x\right )^{\frac {5}{2}}}{5 b} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A \log {\left (B x \right )} + B x\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((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), Ne(b, 0)), (a**(3/2)*(A*log(B*x) + B*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=A a^{\frac {3}{2}} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} A b + 15 \, \sqrt {b x + a} A a b\right )}}{15 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=\frac {2 \, A a^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} B b^{4} + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{5} + 15 \, \sqrt {b x + a} A a b^{5}\right )}}{15 \, b^{5}} \]

[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

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x} \, dx=\left (\frac {2\,A\,b-2\,B\,a}{3\,b}+\frac {2\,B\,a}{3\,b}\right )\,{\left (a+b\,x\right )}^{3/2}+\frac {2\,B\,{\left (a+b\,x\right )}^{5/2}}{5\,b}+a\,\left (\frac {2\,A\,b-2\,B\,a}{b}+\frac {2\,B\,a}{b}\right )\,\sqrt {a+b\,x}+A\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]

[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)