\(\int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx\) [495]

Optimal result

Integrand size = 20, antiderivative size = 117 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\frac {b (2 A b+3 a B) \sqrt {x} \sqrt {a+b x}}{a}-\frac {2 (2 A b+3 a B) (a+b x)^{3/2}}{3 a \sqrt {x}}-\frac {2 A (a+b x)^{5/2}}{3 a x^{3/2}}+\sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {79, 49, 52, 65, 223, 212} \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\sqrt {b} (3 a B+2 A b) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )-\frac {2 (a+b x)^{3/2} (3 a B+2 A b)}{3 a \sqrt {x}}+\frac {b \sqrt {x} \sqrt {a+b x} (3 a B+2 A b)}{a}-\frac {2 A (a+b x)^{5/2}}{3 a x^{3/2}} \]

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/x^(5/2),x]

[Out]

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

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\frac {\sqrt {a+b x} \left (-2 a A-8 A b x-6 a B x+3 b B x^2\right )}{3 x^{3/2}}+2 \sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right ) \]

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/x^(5/2),x]

[Out]

(Sqrt[a + b*x]*(-2*a*A - 8*A*b*x - 6*a*B*x + 3*b*B*x^2))/(3*x^(3/2)) + 2*Sqrt[b]*(2*A*b + 3*a*B)*ArcTanh[(Sqrt
[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])]

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-3 b B \,x^{2}+8 A b x +6 B a x +2 A a \right )}{3 x^{\frac {3}{2}}}+\frac {\sqrt {b}\, \left (2 A b +3 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{2 \sqrt {x}\, \sqrt {b x +a}}\)	\(93\)
default	\(\frac {\sqrt {b x +a}\, \left (6 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) b^{2} x^{2}+9 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a b \,x^{2}+6 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, x^{2}-16 A \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}-12 B a x \sqrt {x \left (b x +a \right )}\, \sqrt {b}-4 A a \sqrt {x \left (b x +a \right )}\, \sqrt {b}\right )}{6 x^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, \sqrt {b}}\)	\(162\)

[In]

int((b*x+a)^(3/2)*(B*x+A)/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*(b*x+a)^(1/2)*(-3*B*b*x^2+8*A*b*x+6*B*a*x+2*A*a)/x^(3/2)+1/2*b^(1/2)*(2*A*b+3*B*a)*ln((1/2*a+b*x)/b^(1/2)
+(b*x^2+a*x)^(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 = 163, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\left [\frac {3 \, {\left (3 \, B a + 2 \, A b\right )} \sqrt {b} x^{2} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, B b x^{2} - 2 \, A a - 2 \, {\left (3 \, B a + 4 \, A b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{6 \, x^{2}}, -\frac {3 \, {\left (3 \, B a + 2 \, A b\right )} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (3 \, B b x^{2} - 2 \, A a - 2 \, {\left (3 \, B a + 4 \, A b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3 \, x^{2}}\right ] \]

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(3*B*a + 2*A*b)*sqrt(b)*x^2*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(3*B*b*x^2 - 2*A*a -
2*(3*B*a + 4*A*b)*x)*sqrt(b*x + a)*sqrt(x))/x^2, -1/3*(3*(3*B*a + 2*A*b)*sqrt(-b)*x^2*arctan(sqrt(b*x + a)*sqr
t(-b)/(b*sqrt(x))) - (3*B*b*x^2 - 2*A*a - 2*(3*B*a + 4*A*b)*x)*sqrt(b*x + a)*sqrt(x))/x^2]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (109) = 218\).

Time = 3.83 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.89 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=- \frac {2 A \sqrt {a} b}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} - \frac {2 A a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x} - \frac {2 A b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3} + 2 A b^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} - \frac {2 A b^{2} \sqrt {x}}{\sqrt {a} \sqrt {1 + \frac {b x}{a}}} - \frac {2 B a^{\frac {3}{2}}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + B \sqrt {a} b \sqrt {x} \sqrt {1 + \frac {b x}{a}} - \frac {2 B \sqrt {a} b \sqrt {x}}{\sqrt {1 + \frac {b x}{a}}} + 3 B a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} \]

[In]

integrate((b*x+a)**(3/2)*(B*x+A)/x**(5/2),x)

[Out]

-2*A*sqrt(a)*b/(sqrt(x)*sqrt(1 + b*x/a)) - 2*A*a*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 2*A*b**(3/2)*sqrt(a/(b*x) +
1)/3 + 2*A*b**(3/2)*asinh(sqrt(b)*sqrt(x)/sqrt(a)) - 2*A*b**2*sqrt(x)/(sqrt(a)*sqrt(1 + b*x/a)) - 2*B*a**(3/2
)/(sqrt(x)*sqrt(1 + b*x/a)) + B*sqrt(a)*b*sqrt(x)*sqrt(1 + b*x/a) - 2*B*sqrt(a)*b*sqrt(x)/sqrt(1 + b*x/a) + 3*
B*a*sqrt(b)*asinh(sqrt(b)*sqrt(x)/sqrt(a))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.25 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\frac {3}{2} \, B a \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + A b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {3 \, \sqrt {b x^{2} + a x} B a}{x} - \frac {7 \, \sqrt {b x^{2} + a x} A b}{3 \, x} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{x^{2}} - \frac {\sqrt {b x^{2} + a x} A a}{3 \, x^{2}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{3 \, x^{3}} \]

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^(5/2),x, algorithm="maxima")

[Out]

3/2*B*a*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + A*b^(3/2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*s
qrt(b)) - 3*sqrt(b*x^2 + a*x)*B*a/x - 7/3*sqrt(b*x^2 + a*x)*A*b/x + (b*x^2 + a*x)^(3/2)*B/x^2 - 1/3*sqrt(b*x^2
+ a*x)*A*a/x^2 - 1/3*(b*x^2 + a*x)^(3/2)*A/x^3

Giac [A] (verification not implemented)

none

Time = 75.34 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=-\frac {{\left (\frac {3 \, {\left (3 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{\sqrt {b}} - \frac {{\left ({\left (3 \, {\left (b x + a\right )} B b^{2} - \frac {4 \, {\left (3 \, B a^{2} b^{3} + 2 \, A a b^{4}\right )}}{a b}\right )} {\left (b x + a\right )} + \frac {3 \, {\left (3 \, B a^{3} b^{3} + 2 \, A a^{2} b^{4}\right )}}{a b}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b - a b\right )}^{\frac {3}{2}}}\right )} b}{3 \, {\left | b \right |}} \]

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^(5/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^{5/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2}}{x^{5/2}} \,d x \]

[In]

int(((A + B*x)*(a + b*x)^(3/2))/x^(5/2),x)

[Out]

int(((A + B*x)*(a + b*x)^(3/2))/x^(5/2), x)