3.4.90 \(\int \frac {(a+b x)^{5/2} (A+B x)}{x} \, dx\)

Optimal. Leaf size=86 \[ -2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+2 a^2 A \sqrt {a+b x}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2}{3} a A (a+b x)^{3/2}+\frac {2 B (a+b x)^{7/2}}{7 b} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, 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^2 A \sqrt {a+b x}-2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {2}{5} A (a+b x)^{5/2}+\frac {2}{3} a A (a+b x)^{3/2}+\frac {2 B (a+b x)^{7/2}}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x,x]

[Out]

2*a^2*A*Sqrt[a + b*x] + (2*a*A*(a + b*x)^(3/2))/3 + (2*A*(a + b*x)^(5/2))/5 + (2*B*(a + b*x)^(7/2))/(7*b) - 2*
a^(5/2)*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)^{5/2} (A+B x)}{x} \, dx &=\frac {2 B (a+b x)^{7/2}}{7 b}+A \int \frac {(a+b x)^{5/2}}{x} \, dx\\ &=\frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}+(a A) \int \frac {(a+b x)^{3/2}}{x} \, dx\\ &=\frac {2}{3} a A (a+b x)^{3/2}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}+\left (a^2 A\right ) \int \frac {\sqrt {a+b x}}{x} \, dx\\ &=2 a^2 A \sqrt {a+b x}+\frac {2}{3} a A (a+b x)^{3/2}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}+\left (a^3 A\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx\\ &=2 a^2 A \sqrt {a+b x}+\frac {2}{3} a A (a+b x)^{3/2}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}+\frac {\left (2 a^3 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^2 A \sqrt {a+b x}+\frac {2}{3} a A (a+b x)^{3/2}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}-2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 78, normalized size = 0.91 \begin {gather*} -2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {2}{5} A (a+b x)^{5/2}+\frac {2}{3} a A (4 a+b x) \sqrt {a+b x}+\frac {2 B (a+b x)^{7/2}}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*(a + b*x)^(5/2))/5 + (2*B*(a + b*x)^(7/2))/(7*b) + (2*a*A*Sqrt[a + b*x]*(4*a + b*x))/3 - 2*a^(5/2)*A*ArcT
anh[Sqrt[a + b*x]/Sqrt[a]]

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.06, size = 88, normalized size = 1.02 \begin {gather*} \frac {2 \left (105 a^2 A b \sqrt {a+b x}+21 A b (a+b x)^{5/2}+35 a A b (a+b x)^{3/2}+15 B (a+b x)^{7/2}\right )}{105 b}-2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(105*a^2*A*b*Sqrt[a + b*x] + 35*a*A*b*(a + b*x)^(3/2) + 21*A*b*(a + b*x)^(5/2) + 15*B*(a + b*x)^(7/2)))/(10
5*b) - 2*a^(5/2)*A*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

________________________________________________________________________________________

fricas [A]  time = 1.50, size = 208, normalized size = 2.42 \begin {gather*} \left [\frac {105 \, A a^{\frac {5}{2}} b \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, B b^{3} x^{3} + 15 \, B a^{3} + 161 \, A a^{2} b + 3 \, {\left (15 \, B a b^{2} + 7 \, A b^{3}\right )} x^{2} + {\left (45 \, B a^{2} b + 77 \, A a b^{2}\right )} x\right )} \sqrt {b x + a}}{105 \, b}, \frac {2 \, {\left (105 \, A \sqrt {-a} a^{2} b \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, B b^{3} x^{3} + 15 \, B a^{3} + 161 \, A a^{2} b + 3 \, {\left (15 \, B a b^{2} + 7 \, A b^{3}\right )} x^{2} + {\left (45 \, B a^{2} b + 77 \, A a b^{2}\right )} x\right )} \sqrt {b x + a}\right )}}{105 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/105*(105*A*a^(5/2)*b*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(15*B*b^3*x^3 + 15*B*a^3 + 161*A*a^2*
b + 3*(15*B*a*b^2 + 7*A*b^3)*x^2 + (45*B*a^2*b + 77*A*a*b^2)*x)*sqrt(b*x + a))/b, 2/105*(105*A*sqrt(-a)*a^2*b*
arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*B*b^3*x^3 + 15*B*a^3 + 161*A*a^2*b + 3*(15*B*a*b^2 + 7*A*b^3)*x^2 + (45
*B*a^2*b + 77*A*a*b^2)*x)*sqrt(b*x + a))/b]

________________________________________________________________________________________

giac [A]  time = 1.33, size = 88, normalized size = 1.02 \begin {gather*} \frac {2 \, A a^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (15 \, {\left (b x + a\right )}^{\frac {7}{2}} B b^{6} + 21 \, {\left (b x + a\right )}^{\frac {5}{2}} A b^{7} + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b^{7} + 105 \, \sqrt {b x + a} A a^{2} b^{7}\right )}}{105 \, b^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 2/105*(15*(b*x + a)^(7/2)*B*b^6 + 21*(b*x + a)^(5/2)*A*b^7 +
35*(b*x + a)^(3/2)*A*a*b^7 + 105*sqrt(b*x + a)*A*a^2*b^7)/b^7

________________________________________________________________________________________

maple [A]  time = 0.01, size = 72, normalized size = 0.84 \begin {gather*} \frac {-2 A \,a^{\frac {5}{2}} b \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 \sqrt {b x +a}\, A \,a^{2} b +\frac {2 \left (b x +a \right )^{\frac {3}{2}} A a b}{3}+\frac {2 \left (b x +a \right )^{\frac {5}{2}} A b}{5}+\frac {2 \left (b x +a \right )^{\frac {7}{2}} B}{7}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.00, size = 87, normalized size = 1.01 \begin {gather*} A a^{\frac {5}{2}} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2 \, {\left (15 \, {\left (b x + a\right )}^{\frac {7}{2}} B + 21 \, {\left (b x + a\right )}^{\frac {5}{2}} A b + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b + 105 \, \sqrt {b x + a} A a^{2} b\right )}}{105 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a^(5/2)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a))) + 2/105*(15*(b*x + a)^(7/2)*B + 21*(b*x + a
)^(5/2)*A*b + 35*(b*x + a)^(3/2)*A*a*b + 105*sqrt(b*x + a)*A*a^2*b)/b

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 128, normalized size = 1.49 \begin {gather*} \left (\frac {2\,A\,b-2\,B\,a}{5\,b}+\frac {2\,B\,a}{5\,b}\right )\,{\left (a+b\,x\right )}^{5/2}+a^2\,\left (\frac {2\,A\,b-2\,B\,a}{b}+\frac {2\,B\,a}{b}\right )\,\sqrt {a+b\,x}+\frac {2\,B\,{\left (a+b\,x\right )}^{7/2}}{7\,b}+\frac {a\,\left (\frac {2\,A\,b-2\,B\,a}{b}+\frac {2\,B\,a}{b}\right )\,{\left (a+b\,x\right )}^{3/2}}{3}+A\,a^{5/2}\,\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)^(5/2))/x,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 42.97, size = 88, normalized size = 1.02 \begin {gather*} \frac {2 A a^{3} \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A a^{2} \sqrt {a + b x} + \frac {2 A a \left (a + b x\right )^{\frac {3}{2}}}{3} + \frac {2 A \left (a + b x\right )^{\frac {5}{2}}}{5} + \frac {2 B \left (a + b x\right )^{\frac {7}{2}}}{7 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________