3.420 \(\int \frac{(a+b x)^{5/2} (A+B x)}{x^6} \, dx\)

Optimal. Leaf size=177 \[ \frac{b^3 \sqrt{a+b x} (3 A b-10 a B)}{128 a^2 x}-\frac{b^4 (3 A b-10 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{128 a^{5/2}}+\frac{b^2 \sqrt{a+b x} (3 A b-10 a B)}{64 a x^2}+\frac{b (a+b x)^{3/2} (3 A b-10 a B)}{48 a x^3}+\frac{(a+b x)^{5/2} (3 A b-10 a B)}{40 a x^4}-\frac{A (a+b x)^{7/2}}{5 a x^5} \]

[Out]

(b^2*(3*A*b - 10*a*B)*Sqrt[a + b*x])/(64*a*x^2) + (b^3*(3*A*b - 10*a*B)*Sqrt[a + b*x])/(128*a^2*x) + (b*(3*A*b
- 10*a*B)*(a + b*x)^(3/2))/(48*a*x^3) + ((3*A*b - 10*a*B)*(a + b*x)^(5/2))/(40*a*x^4) - (A*(a + b*x)^(7/2))/(
5*a*x^5) - (b^4*(3*A*b - 10*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(128*a^(5/2))

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Rubi [A]  time = 0.0809073, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {78, 47, 51, 63, 208} \[ \frac{b^3 \sqrt{a+b x} (3 A b-10 a B)}{128 a^2 x}-\frac{b^4 (3 A b-10 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{128 a^{5/2}}+\frac{b^2 \sqrt{a+b x} (3 A b-10 a B)}{64 a x^2}+\frac{b (a+b x)^{3/2} (3 A b-10 a B)}{48 a x^3}+\frac{(a+b x)^{5/2} (3 A b-10 a B)}{40 a x^4}-\frac{A (a+b x)^{7/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^6,x]

[Out]

(b^2*(3*A*b - 10*a*B)*Sqrt[a + b*x])/(64*a*x^2) + (b^3*(3*A*b - 10*a*B)*Sqrt[a + b*x])/(128*a^2*x) + (b*(3*A*b
- 10*a*B)*(a + b*x)^(3/2))/(48*a*x^3) + ((3*A*b - 10*a*B)*(a + b*x)^(5/2))/(40*a*x^4) - (A*(a + b*x)^(7/2))/(
5*a*x^5) - (b^4*(3*A*b - 10*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(128*a^(5/2))

Rule 78

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] || LtQ
[p, n]))))

Rule 47

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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^6} \, dx &=-\frac{A (a+b x)^{7/2}}{5 a x^5}+\frac{\left (-\frac{3 A b}{2}+5 a B\right ) \int \frac{(a+b x)^{5/2}}{x^5} \, dx}{5 a}\\ &=\frac{(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac{A (a+b x)^{7/2}}{5 a x^5}-\frac{(b (3 A b-10 a B)) \int \frac{(a+b x)^{3/2}}{x^4} \, dx}{16 a}\\ &=\frac{b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac{(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac{A (a+b x)^{7/2}}{5 a x^5}-\frac{\left (b^2 (3 A b-10 a B)\right ) \int \frac{\sqrt{a+b x}}{x^3} \, dx}{32 a}\\ &=\frac{b^2 (3 A b-10 a B) \sqrt{a+b x}}{64 a x^2}+\frac{b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac{(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac{A (a+b x)^{7/2}}{5 a x^5}-\frac{\left (b^3 (3 A b-10 a B)\right ) \int \frac{1}{x^2 \sqrt{a+b x}} \, dx}{128 a}\\ &=\frac{b^2 (3 A b-10 a B) \sqrt{a+b x}}{64 a x^2}+\frac{b^3 (3 A b-10 a B) \sqrt{a+b x}}{128 a^2 x}+\frac{b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac{(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac{A (a+b x)^{7/2}}{5 a x^5}+\frac{\left (b^4 (3 A b-10 a B)\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx}{256 a^2}\\ &=\frac{b^2 (3 A b-10 a B) \sqrt{a+b x}}{64 a x^2}+\frac{b^3 (3 A b-10 a B) \sqrt{a+b x}}{128 a^2 x}+\frac{b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac{(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac{A (a+b x)^{7/2}}{5 a x^5}+\frac{\left (b^3 (3 A b-10 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{128 a^2}\\ &=\frac{b^2 (3 A b-10 a B) \sqrt{a+b x}}{64 a x^2}+\frac{b^3 (3 A b-10 a B) \sqrt{a+b x}}{128 a^2 x}+\frac{b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac{(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac{A (a+b x)^{7/2}}{5 a x^5}-\frac{b^4 (3 A b-10 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{128 a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0229107, size = 58, normalized size = 0.33 \[ -\frac{(a+b x)^{7/2} \left (7 a^5 A+b^4 x^5 (10 a B-3 A b) \, _2F_1\left (\frac{7}{2},5;\frac{9}{2};\frac{b x}{a}+1\right )\right )}{35 a^6 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^6,x]

[Out]

-((a + b*x)^(7/2)*(7*a^5*A + b^4*(-3*A*b + 10*a*B)*x^5*Hypergeometric2F1[7/2, 5, 9/2, 1 + (b*x)/a]))/(35*a^6*x
^5)

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Maple [A]  time = 0.011, size = 140, normalized size = 0.8 \begin{align*} 2\,{b}^{4} \left ({\frac{1}{{b}^{5}{x}^{5}} \left ({\frac{ \left ( 3\,Ab-10\,Ba \right ) \left ( bx+a \right ) ^{9/2}}{256\,{a}^{2}}}-{\frac{ \left ( 21\,Ab+58\,Ba \right ) \left ( bx+a \right ) ^{7/2}}{384\,a}}+ \left ( -1/10\,Ab+1/3\,Ba \right ) \left ( bx+a \right ) ^{5/2}+{\frac{7\,a \left ( 3\,Ab-10\,Ba \right ) \left ( bx+a \right ) ^{3/2}}{384}}-{\frac{{a}^{2} \left ( 3\,Ab-10\,Ba \right ) \sqrt{bx+a}}{256}} \right ) }-{\frac{3\,Ab-10\,Ba}{256\,{a}^{5/2}}{\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^6,x)

[Out]

2*b^4*((1/256*(3*A*b-10*B*a)/a^2*(b*x+a)^(9/2)-1/384*(21*A*b+58*B*a)/a*(b*x+a)^(7/2)+(-1/10*A*b+1/3*B*a)*(b*x+
a)^(5/2)+7/384*a*(3*A*b-10*B*a)*(b*x+a)^(3/2)-1/256*a^2*(3*A*b-10*B*a)*(b*x+a)^(1/2))/b^5/x^5-1/256*(3*A*b-10*
B*a)/a^(5/2)*arctanh((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^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.3359, size = 736, normalized size = 4.16 \begin{align*} \left [-\frac{15 \,{\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt{a} x^{5} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (384 \, A a^{5} + 15 \,{\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + 10 \,{\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{3} + 8 \,{\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{2} + 48 \,{\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x\right )} \sqrt{b x + a}}{3840 \, a^{3} x^{5}}, -\frac{15 \,{\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt{-a} x^{5} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (384 \, A a^{5} + 15 \,{\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + 10 \,{\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{3} + 8 \,{\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{2} + 48 \,{\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x\right )} \sqrt{b x + a}}{1920 \, a^{3} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^6,x, algorithm="fricas")

[Out]

[-1/3840*(15*(10*B*a*b^4 - 3*A*b^5)*sqrt(a)*x^5*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(384*A*a^5 +
15*(10*B*a^2*b^3 - 3*A*a*b^4)*x^4 + 10*(118*B*a^3*b^2 + 3*A*a^2*b^3)*x^3 + 8*(170*B*a^4*b + 93*A*a^3*b^2)*x^2
+ 48*(10*B*a^5 + 21*A*a^4*b)*x)*sqrt(b*x + a))/(a^3*x^5), -1/1920*(15*(10*B*a*b^4 - 3*A*b^5)*sqrt(-a)*x^5*arct
an(sqrt(b*x + a)*sqrt(-a)/a) + (384*A*a^5 + 15*(10*B*a^2*b^3 - 3*A*a*b^4)*x^4 + 10*(118*B*a^3*b^2 + 3*A*a^2*b^
3)*x^3 + 8*(170*B*a^4*b + 93*A*a^3*b^2)*x^2 + 48*(10*B*a^5 + 21*A*a^4*b)*x)*sqrt(b*x + a))/(a^3*x^5)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(B*x+A)/x**6,x)

[Out]

Timed out

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Giac [A]  time = 1.19247, size = 281, normalized size = 1.59 \begin{align*} -\frac{\frac{15 \,{\left (10 \, B a b^{5} - 3 \, A b^{6}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{150 \,{\left (b x + a\right )}^{\frac{9}{2}} B a b^{5} + 580 \,{\left (b x + a\right )}^{\frac{7}{2}} B a^{2} b^{5} - 1280 \,{\left (b x + a\right )}^{\frac{5}{2}} B a^{3} b^{5} + 700 \,{\left (b x + a\right )}^{\frac{3}{2}} B a^{4} b^{5} - 150 \, \sqrt{b x + a} B a^{5} b^{5} - 45 \,{\left (b x + a\right )}^{\frac{9}{2}} A b^{6} + 210 \,{\left (b x + a\right )}^{\frac{7}{2}} A a b^{6} + 384 \,{\left (b x + a\right )}^{\frac{5}{2}} A a^{2} b^{6} - 210 \,{\left (b x + a\right )}^{\frac{3}{2}} A a^{3} b^{6} + 45 \, \sqrt{b x + a} A a^{4} b^{6}}{a^{2} b^{5} x^{5}}}{1920 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^6,x, algorithm="giac")

[Out]

-1/1920*(15*(10*B*a*b^5 - 3*A*b^6)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (150*(b*x + a)^(9/2)*B*a*b^
5 + 580*(b*x + a)^(7/2)*B*a^2*b^5 - 1280*(b*x + a)^(5/2)*B*a^3*b^5 + 700*(b*x + a)^(3/2)*B*a^4*b^5 - 150*sqrt(
b*x + a)*B*a^5*b^5 - 45*(b*x + a)^(9/2)*A*b^6 + 210*(b*x + a)^(7/2)*A*a*b^6 + 384*(b*x + a)^(5/2)*A*a^2*b^6 -
210*(b*x + a)^(3/2)*A*a^3*b^6 + 45*sqrt(b*x + a)*A*a^4*b^6)/(a^2*b^5*x^5))/b