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

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

[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]]

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Rubi [A]  time = 0.029255, antiderivative size = 86, normalized size of antiderivative = 1., 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} \[ 2 a^2 A \sqrt{a+b x}-2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{2}{5} A (a+b x)^{5/2}+\frac{2}{3} a A (a+b x)^{3/2}+\frac{2 B (a+b x)^{7/2}}{7 b} \]

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 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 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 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 = 0.103489, size = 78, normalized size = 0.91 \[ -2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{2}{5} A (a+b x)^{5/2}+\frac{2}{3} a A (4 a+b x) \sqrt{a+b x}+\frac{2 B (a+b x)^{7/2}}{7 b} \]

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]]

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Maple [A]  time = 0.004, size = 72, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{b} \left ( 1/7\,B \left ( bx+a \right ) ^{7/2}+1/5\,Ab \left ( bx+a \right ) ^{5/2}+1/3\,Aba \left ( bx+a \right ) ^{3/2}+{a}^{2}bA\sqrt{bx+a}-A{a}^{5/2}b{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) \right ) } \end{align*}

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*b*a*(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)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.34322, size = 509, normalized size = 5.92 \begin{align*} \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{align*}

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]

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Sympy [A]  time = 42.1884, size = 88, normalized size = 1.02 \begin{align*} \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{align*}

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)

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Giac [A]  time = 1.20601, size = 119, normalized size = 1.38 \begin{align*} \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{align*}

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