3.396 \(\int \frac{\sqrt{a+b x} (A+B x)}{x^5} \, dx\)

Optimal. Leaf size=146 \[ -\frac{b^2 \sqrt{a+b x} (5 A b-8 a B)}{64 a^3 x}+\frac{b^3 (5 A b-8 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{64 a^{7/2}}+\frac{b \sqrt{a+b x} (5 A b-8 a B)}{96 a^2 x^2}+\frac{\sqrt{a+b x} (5 A b-8 a B)}{24 a x^3}-\frac{A (a+b x)^{3/2}}{4 a x^4} \]

[Out]

((5*A*b - 8*a*B)*Sqrt[a + b*x])/(24*a*x^3) + (b*(5*A*b - 8*a*B)*Sqrt[a + b*x])/(96*a^2*x^2) - (b^2*(5*A*b - 8*
a*B)*Sqrt[a + b*x])/(64*a^3*x) - (A*(a + b*x)^(3/2))/(4*a*x^4) + (b^3*(5*A*b - 8*a*B)*ArcTanh[Sqrt[a + b*x]/Sq
rt[a]])/(64*a^(7/2))

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Rubi [A]  time = 0.0654546, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{b^2 \sqrt{a+b x} (5 A b-8 a B)}{64 a^3 x}+\frac{b^3 (5 A b-8 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{64 a^{7/2}}+\frac{b \sqrt{a+b x} (5 A b-8 a B)}{96 a^2 x^2}+\frac{\sqrt{a+b x} (5 A b-8 a B)}{24 a x^3}-\frac{A (a+b x)^{3/2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x]*(A + B*x))/x^5,x]

[Out]

((5*A*b - 8*a*B)*Sqrt[a + b*x])/(24*a*x^3) + (b*(5*A*b - 8*a*B)*Sqrt[a + b*x])/(96*a^2*x^2) - (b^2*(5*A*b - 8*
a*B)*Sqrt[a + b*x])/(64*a^3*x) - (A*(a + b*x)^(3/2))/(4*a*x^4) + (b^3*(5*A*b - 8*a*B)*ArcTanh[Sqrt[a + b*x]/Sq
rt[a]])/(64*a^(7/2))

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{\sqrt{a+b x} (A+B x)}{x^5} \, dx &=-\frac{A (a+b x)^{3/2}}{4 a x^4}+\frac{\left (-\frac{5 A b}{2}+4 a B\right ) \int \frac{\sqrt{a+b x}}{x^4} \, dx}{4 a}\\ &=\frac{(5 A b-8 a B) \sqrt{a+b x}}{24 a x^3}-\frac{A (a+b x)^{3/2}}{4 a x^4}-\frac{(b (5 A b-8 a B)) \int \frac{1}{x^3 \sqrt{a+b x}} \, dx}{48 a}\\ &=\frac{(5 A b-8 a B) \sqrt{a+b x}}{24 a x^3}+\frac{b (5 A b-8 a B) \sqrt{a+b x}}{96 a^2 x^2}-\frac{A (a+b x)^{3/2}}{4 a x^4}+\frac{\left (b^2 (5 A b-8 a B)\right ) \int \frac{1}{x^2 \sqrt{a+b x}} \, dx}{64 a^2}\\ &=\frac{(5 A b-8 a B) \sqrt{a+b x}}{24 a x^3}+\frac{b (5 A b-8 a B) \sqrt{a+b x}}{96 a^2 x^2}-\frac{b^2 (5 A b-8 a B) \sqrt{a+b x}}{64 a^3 x}-\frac{A (a+b x)^{3/2}}{4 a x^4}-\frac{\left (b^3 (5 A b-8 a B)\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx}{128 a^3}\\ &=\frac{(5 A b-8 a B) \sqrt{a+b x}}{24 a x^3}+\frac{b (5 A b-8 a B) \sqrt{a+b x}}{96 a^2 x^2}-\frac{b^2 (5 A b-8 a B) \sqrt{a+b x}}{64 a^3 x}-\frac{A (a+b x)^{3/2}}{4 a x^4}-\frac{\left (b^2 (5 A b-8 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{64 a^3}\\ &=\frac{(5 A b-8 a B) \sqrt{a+b x}}{24 a x^3}+\frac{b (5 A b-8 a B) \sqrt{a+b x}}{96 a^2 x^2}-\frac{b^2 (5 A b-8 a B) \sqrt{a+b x}}{64 a^3 x}-\frac{A (a+b x)^{3/2}}{4 a x^4}+\frac{b^3 (5 A b-8 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{64 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.018406, size = 58, normalized size = 0.4 \[ -\frac{(a+b x)^{3/2} \left (3 a^4 A+b^3 x^4 (5 A b-8 a B) \, _2F_1\left (\frac{3}{2},4;\frac{5}{2};\frac{b x}{a}+1\right )\right )}{12 a^5 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*x)^(3/2)*(3*a^4*A + b^3*(5*A*b - 8*a*B)*x^4*Hypergeometric2F1[3/2, 4, 5/2, 1 + (b*x)/a]))/(12*a^5*x^4
)

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Maple [A]  time = 0.013, size = 121, normalized size = 0.8 \begin{align*} 2\,{b}^{3} \left ({\frac{1}{{b}^{4}{x}^{4}} \left ( -{\frac{ \left ( 5\,Ab-8\,Ba \right ) \left ( bx+a \right ) ^{7/2}}{128\,{a}^{3}}}+{\frac{ \left ( 55\,Ab-88\,Ba \right ) \left ( bx+a \right ) ^{5/2}}{384\,{a}^{2}}}-{\frac{ \left ( 73\,Ab-40\,Ba \right ) \left ( bx+a \right ) ^{3/2}}{384\,a}}+ \left ( -{\frac{5\,Ab}{128}}+1/16\,Ba \right ) \sqrt{bx+a} \right ) }+{\frac{5\,Ab-8\,Ba}{128\,{a}^{7/2}}{\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)*(b*x+a)^(1/2)/x^5,x)

[Out]

2*b^3*((-1/128*(5*A*b-8*B*a)/a^3*(b*x+a)^(7/2)+11/384/a^2*(5*A*b-8*B*a)*(b*x+a)^(5/2)-1/384*(73*A*b-40*B*a)/a*
(b*x+a)^(3/2)+(-5/128*A*b+1/16*B*a)*(b*x+a)^(1/2))/b^4/x^4+1/128*(5*A*b-8*B*a)/a^(7/2)*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)*(b*x+a)^(1/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.31583, size = 591, normalized size = 4.05 \begin{align*} \left [-\frac{3 \,{\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} \sqrt{a} x^{4} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (48 \, A a^{4} - 3 \,{\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{3} + 2 \,{\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + 8 \,{\left (8 \, B a^{4} + A a^{3} b\right )} x\right )} \sqrt{b x + a}}{384 \, a^{4} x^{4}}, \frac{3 \,{\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} \sqrt{-a} x^{4} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (48 \, A a^{4} - 3 \,{\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{3} + 2 \,{\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + 8 \,{\left (8 \, B a^{4} + A a^{3} b\right )} x\right )} \sqrt{b x + a}}{192 \, a^{4} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^5,x, algorithm="fricas")

[Out]

[-1/384*(3*(8*B*a*b^3 - 5*A*b^4)*sqrt(a)*x^4*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(48*A*a^4 - 3*(8
*B*a^2*b^2 - 5*A*a*b^3)*x^3 + 2*(8*B*a^3*b - 5*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + A*a^3*b)*x)*sqrt(b*x + a))/(a^4*x
^4), 1/192*(3*(8*B*a*b^3 - 5*A*b^4)*sqrt(-a)*x^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (48*A*a^4 - 3*(8*B*a^2*b^2
- 5*A*a*b^3)*x^3 + 2*(8*B*a^3*b - 5*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + A*a^3*b)*x)*sqrt(b*x + a))/(a^4*x^4)]

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Sympy [B]  time = 44.1345, size = 1001, normalized size = 6.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)**(1/2)/x**5,x)

[Out]

-558*A*a**4*b**4*sqrt(a + b*x)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 +
384*a**4*(a + b*x)**4) + 1022*A*a**3*b**4*(a + b*x)**(3/2)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)*
*2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) - 770*A*a**2*b**4*(a + b*x)**(5/2)/(-1152*a**8 - 1536*a**
7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) - 66*A*a**2*b**4*sqrt(a + b*x
)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) + 210*A*a*b**4*(a + b*x)**(7/2)/(-11
52*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) + 80*A*a*b*
*4*(a + b*x)**(3/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) + 35*A*a*b**4*sqrt
(a**(-9))*log(-a**5*sqrt(a**(-9)) + sqrt(a + b*x))/128 - 35*A*a*b**4*sqrt(a**(-9))*log(a**5*sqrt(a**(-9)) + sq
rt(a + b*x))/128 - 30*A*b**4*(a + b*x)**(5/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b
*x)**3) - 5*A*b**4*sqrt(a**(-7))*log(-a**4*sqrt(a**(-7)) + sqrt(a + b*x))/16 + 5*A*b**4*sqrt(a**(-7))*log(a**4
*sqrt(a**(-7)) + sqrt(a + b*x))/16 - 66*B*a**3*b**3*sqrt(a + b*x)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)
**2 + 48*a**3*(a + b*x)**3) + 80*B*a**2*b**3*(a + b*x)**(3/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2
+ 48*a**3*(a + b*x)**3) - 30*B*a*b**3*(a + b*x)**(5/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a*
*3*(a + b*x)**3) - 10*B*a*b**3*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) - 5*B*a*b**3*sqrt(a
**(-7))*log(-a**4*sqrt(a**(-7)) + sqrt(a + b*x))/16 + 5*B*a*b**3*sqrt(a**(-7))*log(a**4*sqrt(a**(-7)) + sqrt(a
+ b*x))/16 + 6*B*b**3*(a + b*x)**(3/2)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 3*B*b**3*sqrt(a**(-5))
*log(-a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*B*b**3*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8

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Giac [A]  time = 1.17264, size = 238, normalized size = 1.63 \begin{align*} \frac{\frac{3 \,{\left (8 \, B a b^{4} - 5 \, A b^{5}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{24 \,{\left (b x + a\right )}^{\frac{7}{2}} B a b^{4} - 88 \,{\left (b x + a\right )}^{\frac{5}{2}} B a^{2} b^{4} + 40 \,{\left (b x + a\right )}^{\frac{3}{2}} B a^{3} b^{4} + 24 \, \sqrt{b x + a} B a^{4} b^{4} - 15 \,{\left (b x + a\right )}^{\frac{7}{2}} A b^{5} + 55 \,{\left (b x + a\right )}^{\frac{5}{2}} A a b^{5} - 73 \,{\left (b x + a\right )}^{\frac{3}{2}} A a^{2} b^{5} - 15 \, \sqrt{b x + a} A a^{3} b^{5}}{a^{3} b^{4} x^{4}}}{192 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x+a)^(1/2)/x^5,x, algorithm="giac")

[Out]

1/192*(3*(8*B*a*b^4 - 5*A*b^5)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (24*(b*x + a)^(7/2)*B*a*b^4 - 8
8*(b*x + a)^(5/2)*B*a^2*b^4 + 40*(b*x + a)^(3/2)*B*a^3*b^4 + 24*sqrt(b*x + a)*B*a^4*b^4 - 15*(b*x + a)^(7/2)*A
*b^5 + 55*(b*x + a)^(5/2)*A*a*b^5 - 73*(b*x + a)^(3/2)*A*a^2*b^5 - 15*sqrt(b*x + a)*A*a^3*b^5)/(a^3*b^4*x^4))/
b