\(\int \frac {(a+b x)^{3/2} (A+B x)}{x^{3/2}} \, dx\) [494]

Optimal result

Integrand size = 20, antiderivative size = 115 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{3/2}} \, dx=\frac {3}{4} (4 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {(4 A b+a B) \sqrt {x} (a+b x)^{3/2}}{2 a}-\frac {2 A (a+b x)^{5/2}}{a \sqrt {x}}+\frac {3 a (4 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 \sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 52, 65, 223, 212} \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{3/2}} \, dx=\frac {3 a (a B+4 A b) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 \sqrt {b}}+\frac {\sqrt {x} (a+b x)^{3/2} (a B+4 A b)}{2 a}+\frac {3}{4} \sqrt {x} \sqrt {a+b x} (a B+4 A b)-\frac {2 A (a+b x)^{5/2}}{a \sqrt {x}} \]

[In]

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

[Out]

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

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 79

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A (a+b x)^{5/2}}{a \sqrt {x}}+\frac {\left (2 \left (2 A b+\frac {a B}{2}\right )\right ) \int \frac {(a+b x)^{3/2}}{\sqrt {x}} \, dx}{a} \\ & = \frac {(4 A b+a B) \sqrt {x} (a+b x)^{3/2}}{2 a}-\frac {2 A (a+b x)^{5/2}}{a \sqrt {x}}+\frac {1}{4} (3 (4 A b+a B)) \int \frac {\sqrt {a+b x}}{\sqrt {x}} \, dx \\ & = \frac {3}{4} (4 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {(4 A b+a B) \sqrt {x} (a+b x)^{3/2}}{2 a}-\frac {2 A (a+b x)^{5/2}}{a \sqrt {x}}+\frac {1}{8} (3 a (4 A b+a B)) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx \\ & = \frac {3}{4} (4 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {(4 A b+a B) \sqrt {x} (a+b x)^{3/2}}{2 a}-\frac {2 A (a+b x)^{5/2}}{a \sqrt {x}}+\frac {1}{4} (3 a (4 A b+a B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {3}{4} (4 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {(4 A b+a B) \sqrt {x} (a+b x)^{3/2}}{2 a}-\frac {2 A (a+b x)^{5/2}}{a \sqrt {x}}+\frac {1}{4} (3 a (4 A b+a B)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right ) \\ & = \frac {3}{4} (4 A b+a B) \sqrt {x} \sqrt {a+b x}+\frac {(4 A b+a B) \sqrt {x} (a+b x)^{3/2}}{2 a}-\frac {2 A (a+b x)^{5/2}}{a \sqrt {x}}+\frac {3 a (4 A b+a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 \sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{3/2}} \, dx=\frac {\sqrt {a+b x} \left (-8 a A+4 A b x+5 a B x+2 b B x^2\right )}{4 \sqrt {x}}+\frac {3 a (4 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{2 \sqrt {b}} \]

[In]

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

[Out]

(Sqrt[a + b*x]*(-8*a*A + 4*A*b*x + 5*a*B*x + 2*b*B*x^2))/(4*Sqrt[x]) + (3*a*(4*A*b + a*B)*ArcTanh[(Sqrt[b]*Sqr
t[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(2*Sqrt[b])

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.81

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-2 b B \,x^{2}-4 A b x -5 B a x +8 A a \right )}{4 \sqrt {x}}+\frac {3 a \left (4 A b +B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{8 \sqrt {b}\, \sqrt {x}\, \sqrt {b x +a}}\)	\(93\)
default	\(\frac {\sqrt {b x +a}\, \left (4 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, x^{2}+12 A b \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a x +8 A \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}+3 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} x +10 B a x \sqrt {x \left (b x +a \right )}\, \sqrt {b}-16 A a \sqrt {x \left (b x +a \right )}\, \sqrt {b}\right )}{8 \sqrt {x}\, \sqrt {x \left (b x +a \right )}\, \sqrt {b}}\)	\(158\)

[In]

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

[Out]

-1/4*(b*x+a)^(1/2)*(-2*B*b*x^2-4*A*b*x-5*B*a*x+8*A*a)/x^(1/2)+3/8*a*(4*A*b+B*a)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+
a*x)^(1/2))/b^(1/2)*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.56 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{3/2}} \, dx=\left [\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, B b^{2} x^{2} - 8 \, A a b + {\left (5 \, B a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{8 \, b x}, -\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, B b^{2} x^{2} - 8 \, A a b + {\left (5 \, B a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{4 \, b x}\right ] \]

[In]

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

[Out]

[1/8*(3*(B*a^2 + 4*A*a*b)*sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*B*b^2*x^2 - 8*A*a*
b + (5*B*a*b + 4*A*b^2)*x)*sqrt(b*x + a)*sqrt(x))/(b*x), -1/4*(3*(B*a^2 + 4*A*a*b)*sqrt(-b)*x*arctan(sqrt(b*x
+ a)*sqrt(-b)/(b*sqrt(x))) - (2*B*b^2*x^2 - 8*A*a*b + (5*B*a*b + 4*A*b^2)*x)*sqrt(b*x + a)*sqrt(x))/(b*x)]

Sympy [A] (verification not implemented)

Time = 3.75 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.22 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{3/2}} \, dx=- \frac {2 A a^{\frac {3}{2}}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + A \sqrt {a} b \sqrt {x} \sqrt {1 + \frac {b x}{a}} - \frac {2 A \sqrt {a} b \sqrt {x}}{\sqrt {1 + \frac {b x}{a}}} + 3 A a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} + B a^{\frac {3}{2}} \sqrt {x} \sqrt {1 + \frac {b x}{a}} + \frac {B a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{\sqrt {b}} + 2 B b \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a \sqrt {x} \sqrt {a + b x}}{8 b} + \frac {x^{\frac {3}{2}} \sqrt {a + b x}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

-2*A*a**(3/2)/(sqrt(x)*sqrt(1 + b*x/a)) + A*sqrt(a)*b*sqrt(x)*sqrt(1 + b*x/a) - 2*A*sqrt(a)*b*sqrt(x)/sqrt(1 +
b*x/a) + 3*A*a*sqrt(b)*asinh(sqrt(b)*sqrt(x)/sqrt(a)) + B*a**(3/2)*sqrt(x)*sqrt(1 + b*x/a) + B*a**2*asinh(sqr
t(b)*sqrt(x)/sqrt(a))/sqrt(b) + 2*B*b*Piecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/s
qrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(8*b) + a*sqrt(x)*sqrt(a + b*x)/(8*b) + x**(3/2)*sq
rt(a + b*x)/4, Ne(b, 0)), (sqrt(a)*x**(3/2)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{3/2}} \, dx=\frac {3 \, B a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, \sqrt {b}} + \frac {3}{2} \, A a \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + \frac {3}{4} \, \sqrt {b x^{2} + a x} B a + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{2 \, x} - \frac {3 \, \sqrt {b x^{2} + a x} A a}{x} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{x^{2}} \]

[In]

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

[Out]

3/8*B*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/sqrt(b) + 3/2*A*a*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2
+ a*x)*sqrt(b)) + 3/4*sqrt(b*x^2 + a*x)*B*a + 1/2*(b*x^2 + a*x)^(3/2)*B/x - 3*sqrt(b*x^2 + a*x)*A*a/x + (b*x^2
+ a*x)^(3/2)*A/x^2

Giac [A] (verification not implemented)

none

Time = 75.63 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{3/2}} \, dx=\frac {{\left (\frac {{\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (b x + a\right )} B}{b} + \frac {B a b + 4 \, A b^{2}}{b^{2}}\right )} - \frac {3 \, {\left (B a^{2} b + 4 \, A a b^{2}\right )}}{b^{2}}\right )} \sqrt {b x + a}}{\sqrt {{\left (b x + a\right )} b - a b}} - \frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}}\right )} b^{2}}{4 \, {\left | b \right |}} \]

[In]

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

[Out]

1/4*(((b*x + a)*(2*(b*x + a)*B/b + (B*a*b + 4*A*b^2)/b^2) - 3*(B*a^2*b + 4*A*a*b^2)/b^2)*sqrt(b*x + a)/sqrt((b
*x + a)*b - a*b) - 3*(B*a^2 + 4*A*a*b)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b*x + a)*b - a*b)))/b^(3/2))*b^2
/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2}}{x^{3/2}} \,d x \]

[In]

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

[Out]

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