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

Optimal. Leaf size=202 \[ \frac{105 b^3 (11 A b-8 a B)}{64 a^6 \sqrt{a+b x}}+\frac{35 b^3 (11 A b-8 a B)}{64 a^5 (a+b x)^{3/2}}+\frac{21 b^2 (11 A b-8 a B)}{64 a^4 x (a+b x)^{3/2}}-\frac{105 b^3 (11 A b-8 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{64 a^{13/2}}-\frac{3 b (11 A b-8 a B)}{32 a^3 x^2 (a+b x)^{3/2}}+\frac{11 A b-8 a B}{24 a^2 x^3 (a+b x)^{3/2}}-\frac{A}{4 a x^4 (a+b x)^{3/2}} \]

[Out]

(35*b^3*(11*A*b - 8*a*B))/(64*a^5*(a + b*x)^(3/2)) - A/(4*a*x^4*(a + b*x)^(3/2)) + (11*A*b - 8*a*B)/(24*a^2*x^
3*(a + b*x)^(3/2)) - (3*b*(11*A*b - 8*a*B))/(32*a^3*x^2*(a + b*x)^(3/2)) + (21*b^2*(11*A*b - 8*a*B))/(64*a^4*x
*(a + b*x)^(3/2)) + (105*b^3*(11*A*b - 8*a*B))/(64*a^6*Sqrt[a + b*x]) - (105*b^3*(11*A*b - 8*a*B)*ArcTanh[Sqrt
[a + b*x]/Sqrt[a]])/(64*a^(13/2))

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Rubi [A]  time = 0.0971598, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 51, 63, 208} \[ \frac{105 b^2 \sqrt{a+b x} (11 A b-8 a B)}{64 a^6 x}-\frac{105 b^3 (11 A b-8 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{64 a^{13/2}}-\frac{35 b \sqrt{a+b x} (11 A b-8 a B)}{32 a^5 x^2}+\frac{7 \sqrt{a+b x} (11 A b-8 a B)}{8 a^4 x^3}-\frac{3 (11 A b-8 a B)}{4 a^3 x^3 \sqrt{a+b x}}-\frac{11 A b-8 a B}{12 a^2 x^3 (a+b x)^{3/2}}-\frac{A}{4 a x^4 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(4*a*x^4*(a + b*x)^(3/2)) - (11*A*b - 8*a*B)/(12*a^2*x^3*(a + b*x)^(3/2)) - (3*(11*A*b - 8*a*B))/(4*a^3*x^3
*Sqrt[a + b*x]) + (7*(11*A*b - 8*a*B)*Sqrt[a + b*x])/(8*a^4*x^3) - (35*b*(11*A*b - 8*a*B)*Sqrt[a + b*x])/(32*a
^5*x^2) + (105*b^2*(11*A*b - 8*a*B)*Sqrt[a + b*x])/(64*a^6*x) - (105*b^3*(11*A*b - 8*a*B)*ArcTanh[Sqrt[a + b*x
]/Sqrt[a]])/(64*a^(13/2))

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 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}{x^5 (a+b x)^{5/2}} \, dx &=-\frac{A}{4 a x^4 (a+b x)^{3/2}}+\frac{\left (-\frac{11 A b}{2}+4 a B\right ) \int \frac{1}{x^4 (a+b x)^{5/2}} \, dx}{4 a}\\ &=-\frac{A}{4 a x^4 (a+b x)^{3/2}}-\frac{11 A b-8 a B}{12 a^2 x^3 (a+b x)^{3/2}}-\frac{(3 (11 A b-8 a B)) \int \frac{1}{x^4 (a+b x)^{3/2}} \, dx}{8 a^2}\\ &=-\frac{A}{4 a x^4 (a+b x)^{3/2}}-\frac{11 A b-8 a B}{12 a^2 x^3 (a+b x)^{3/2}}-\frac{3 (11 A b-8 a B)}{4 a^3 x^3 \sqrt{a+b x}}-\frac{(21 (11 A b-8 a B)) \int \frac{1}{x^4 \sqrt{a+b x}} \, dx}{8 a^3}\\ &=-\frac{A}{4 a x^4 (a+b x)^{3/2}}-\frac{11 A b-8 a B}{12 a^2 x^3 (a+b x)^{3/2}}-\frac{3 (11 A b-8 a B)}{4 a^3 x^3 \sqrt{a+b x}}+\frac{7 (11 A b-8 a B) \sqrt{a+b x}}{8 a^4 x^3}+\frac{(35 b (11 A b-8 a B)) \int \frac{1}{x^3 \sqrt{a+b x}} \, dx}{16 a^4}\\ &=-\frac{A}{4 a x^4 (a+b x)^{3/2}}-\frac{11 A b-8 a B}{12 a^2 x^3 (a+b x)^{3/2}}-\frac{3 (11 A b-8 a B)}{4 a^3 x^3 \sqrt{a+b x}}+\frac{7 (11 A b-8 a B) \sqrt{a+b x}}{8 a^4 x^3}-\frac{35 b (11 A b-8 a B) \sqrt{a+b x}}{32 a^5 x^2}-\frac{\left (105 b^2 (11 A b-8 a B)\right ) \int \frac{1}{x^2 \sqrt{a+b x}} \, dx}{64 a^5}\\ &=-\frac{A}{4 a x^4 (a+b x)^{3/2}}-\frac{11 A b-8 a B}{12 a^2 x^3 (a+b x)^{3/2}}-\frac{3 (11 A b-8 a B)}{4 a^3 x^3 \sqrt{a+b x}}+\frac{7 (11 A b-8 a B) \sqrt{a+b x}}{8 a^4 x^3}-\frac{35 b (11 A b-8 a B) \sqrt{a+b x}}{32 a^5 x^2}+\frac{105 b^2 (11 A b-8 a B) \sqrt{a+b x}}{64 a^6 x}+\frac{\left (105 b^3 (11 A b-8 a B)\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx}{128 a^6}\\ &=-\frac{A}{4 a x^4 (a+b x)^{3/2}}-\frac{11 A b-8 a B}{12 a^2 x^3 (a+b x)^{3/2}}-\frac{3 (11 A b-8 a B)}{4 a^3 x^3 \sqrt{a+b x}}+\frac{7 (11 A b-8 a B) \sqrt{a+b x}}{8 a^4 x^3}-\frac{35 b (11 A b-8 a B) \sqrt{a+b x}}{32 a^5 x^2}+\frac{105 b^2 (11 A b-8 a B) \sqrt{a+b x}}{64 a^6 x}+\frac{\left (105 b^2 (11 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^6}\\ &=-\frac{A}{4 a x^4 (a+b x)^{3/2}}-\frac{11 A b-8 a B}{12 a^2 x^3 (a+b x)^{3/2}}-\frac{3 (11 A b-8 a B)}{4 a^3 x^3 \sqrt{a+b x}}+\frac{7 (11 A b-8 a B) \sqrt{a+b x}}{8 a^4 x^3}-\frac{35 b (11 A b-8 a B) \sqrt{a+b x}}{32 a^5 x^2}+\frac{105 b^2 (11 A b-8 a B) \sqrt{a+b x}}{64 a^6 x}-\frac{105 b^3 (11 A b-8 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{64 a^{13/2}}\\ \end{align*}

Mathematica [C]  time = 0.0192855, size = 58, normalized size = 0.29 \[ \frac{b^3 x^4 (11 A b-8 a B) \, _2F_1\left (-\frac{3}{2},4;-\frac{1}{2};\frac{b x}{a}+1\right )-3 a^4 A}{12 a^5 x^4 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^4*A + b^3*(11*A*b - 8*a*B)*x^4*Hypergeometric2F1[-3/2, 4, -1/2, 1 + (b*x)/a])/(12*a^5*x^4*(a + b*x)^(3/2
))

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Maple [A]  time = 0.018, size = 168, normalized size = 0.8 \begin{align*} 2\,{b}^{3} \left ({\frac{1}{{a}^{6}} \left ({\frac{1}{{b}^{4}{x}^{4}} \left ( \left ({\frac{515\,Ab}{128}}-{\frac{41\,Ba}{16}} \right ) \left ( bx+a \right ) ^{7/2}+ \left ( -{\frac{5153\,Aba}{384}}+{\frac{403\,B{a}^{2}}{48}} \right ) \left ( bx+a \right ) ^{5/2}+ \left ({\frac{5855\,Ab{a}^{2}}{384}}-{\frac{445\,B{a}^{3}}{48}} \right ) \left ( bx+a \right ) ^{3/2}+ \left ( -{\frac{765\,A{a}^{3}b}{128}}+{\frac{55\,B{a}^{4}}{16}} \right ) \sqrt{bx+a} \right ) }-{\frac{1155\,Ab-840\,Ba}{128\,\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }-{\frac{-5\,Ab+4\,Ba}{{a}^{6}\sqrt{bx+a}}}-1/3\,{\frac{-Ab+Ba}{{a}^{5} \left ( bx+a \right ) ^{3/2}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^3*(1/a^6*(((515/128*A*b-41/16*B*a)*(b*x+a)^(7/2)+(-5153/384*A*b*a+403/48*B*a^2)*(b*x+a)^(5/2)+(5855/384*A*
b*a^2-445/48*B*a^3)*(b*x+a)^(3/2)+(-765/128*A*a^3*b+55/16*B*a^4)*(b*x+a)^(1/2))/b^4/x^4-105/128*(11*A*b-8*B*a)
/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))-(-5*A*b+4*B*a)/a^6/(b*x+a)^(1/2)-1/3*(-A*b+B*a)/a^5/(b*x+a)^(3/2))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.55939, size = 1100, normalized size = 5.45 \begin{align*} \left [-\frac{315 \,{\left ({\left (8 \, B a b^{5} - 11 \, A b^{6}\right )} x^{6} + 2 \,{\left (8 \, B a^{2} b^{4} - 11 \, A a b^{5}\right )} x^{5} +{\left (8 \, B a^{3} b^{3} - 11 \, A a^{2} b^{4}\right )} x^{4}\right )} \sqrt{a} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (48 \, A a^{6} + 315 \,{\left (8 \, B a^{2} b^{4} - 11 \, A a b^{5}\right )} x^{5} + 420 \,{\left (8 \, B a^{3} b^{3} - 11 \, A a^{2} b^{4}\right )} x^{4} + 63 \,{\left (8 \, B a^{4} b^{2} - 11 \, A a^{3} b^{3}\right )} x^{3} - 18 \,{\left (8 \, B a^{5} b - 11 \, A a^{4} b^{2}\right )} x^{2} + 8 \,{\left (8 \, B a^{6} - 11 \, A a^{5} b\right )} x\right )} \sqrt{b x + a}}{384 \,{\left (a^{7} b^{2} x^{6} + 2 \, a^{8} b x^{5} + a^{9} x^{4}\right )}}, -\frac{315 \,{\left ({\left (8 \, B a b^{5} - 11 \, A b^{6}\right )} x^{6} + 2 \,{\left (8 \, B a^{2} b^{4} - 11 \, A a b^{5}\right )} x^{5} +{\left (8 \, B a^{3} b^{3} - 11 \, A a^{2} b^{4}\right )} x^{4}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (48 \, A a^{6} + 315 \,{\left (8 \, B a^{2} b^{4} - 11 \, A a b^{5}\right )} x^{5} + 420 \,{\left (8 \, B a^{3} b^{3} - 11 \, A a^{2} b^{4}\right )} x^{4} + 63 \,{\left (8 \, B a^{4} b^{2} - 11 \, A a^{3} b^{3}\right )} x^{3} - 18 \,{\left (8 \, B a^{5} b - 11 \, A a^{4} b^{2}\right )} x^{2} + 8 \,{\left (8 \, B a^{6} - 11 \, A a^{5} b\right )} x\right )} \sqrt{b x + a}}{192 \,{\left (a^{7} b^{2} x^{6} + 2 \, a^{8} b x^{5} + a^{9} x^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(315*((8*B*a*b^5 - 11*A*b^6)*x^6 + 2*(8*B*a^2*b^4 - 11*A*a*b^5)*x^5 + (8*B*a^3*b^3 - 11*A*a^2*b^4)*x^4
)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(48*A*a^6 + 315*(8*B*a^2*b^4 - 11*A*a*b^5)*x^5 + 42
0*(8*B*a^3*b^3 - 11*A*a^2*b^4)*x^4 + 63*(8*B*a^4*b^2 - 11*A*a^3*b^3)*x^3 - 18*(8*B*a^5*b - 11*A*a^4*b^2)*x^2 +
8*(8*B*a^6 - 11*A*a^5*b)*x)*sqrt(b*x + a))/(a^7*b^2*x^6 + 2*a^8*b*x^5 + a^9*x^4), -1/192*(315*((8*B*a*b^5 - 1
1*A*b^6)*x^6 + 2*(8*B*a^2*b^4 - 11*A*a*b^5)*x^5 + (8*B*a^3*b^3 - 11*A*a^2*b^4)*x^4)*sqrt(-a)*arctan(sqrt(b*x +
a)*sqrt(-a)/a) + (48*A*a^6 + 315*(8*B*a^2*b^4 - 11*A*a*b^5)*x^5 + 420*(8*B*a^3*b^3 - 11*A*a^2*b^4)*x^4 + 63*(
8*B*a^4*b^2 - 11*A*a^3*b^3)*x^3 - 18*(8*B*a^5*b - 11*A*a^4*b^2)*x^2 + 8*(8*B*a^6 - 11*A*a^5*b)*x)*sqrt(b*x + a
))/(a^7*b^2*x^6 + 2*a^8*b*x^5 + a^9*x^4)]

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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)/x**5/(b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.21031, size = 301, normalized size = 1.49 \begin{align*} -\frac{105 \,{\left (8 \, B a b^{3} - 11 \, A b^{4}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{64 \, \sqrt{-a} a^{6}} - \frac{2 \,{\left (12 \,{\left (b x + a\right )} B a b^{3} + B a^{2} b^{3} - 15 \,{\left (b x + a\right )} A b^{4} - A a b^{4}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{6}} - \frac{984 \,{\left (b x + a\right )}^{\frac{7}{2}} B a b^{3} - 3224 \,{\left (b x + a\right )}^{\frac{5}{2}} B a^{2} b^{3} + 3560 \,{\left (b x + a\right )}^{\frac{3}{2}} B a^{3} b^{3} - 1320 \, \sqrt{b x + a} B a^{4} b^{3} - 1545 \,{\left (b x + a\right )}^{\frac{7}{2}} A b^{4} + 5153 \,{\left (b x + a\right )}^{\frac{5}{2}} A a b^{4} - 5855 \,{\left (b x + a\right )}^{\frac{3}{2}} A a^{2} b^{4} + 2295 \, \sqrt{b x + a} A a^{3} b^{4}}{192 \, a^{6} b^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-105/64*(8*B*a*b^3 - 11*A*b^4)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^6) - 2/3*(12*(b*x + a)*B*a*b^3 + B*a
^2*b^3 - 15*(b*x + a)*A*b^4 - A*a*b^4)/((b*x + a)^(3/2)*a^6) - 1/192*(984*(b*x + a)^(7/2)*B*a*b^3 - 3224*(b*x
+ a)^(5/2)*B*a^2*b^3 + 3560*(b*x + a)^(3/2)*B*a^3*b^3 - 1320*sqrt(b*x + a)*B*a^4*b^3 - 1545*(b*x + a)^(7/2)*A*
b^4 + 5153*(b*x + a)^(5/2)*A*a*b^4 - 5855*(b*x + a)^(3/2)*A*a^2*b^4 + 2295*sqrt(b*x + a)*A*a^3*b^4)/(a^6*b^4*x
^4)