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

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

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Rubi [A]  time = 0.10, antiderivative size = 202, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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 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 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*}

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Mathematica [C]  time = 0.03, size = 58, normalized size = 0.29 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.31, size = 200, normalized size = 0.99 \begin {gather*} \frac {105 \left (8 a b^3 B-11 A b^4\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{13/2}}-\frac {128 a^6 B-128 a^5 A b+1024 a^5 B (a+b x)-1408 a^4 A b (a+b x)-6696 a^4 B (a+b x)^2+9207 a^3 A b (a+b x)^2+12264 a^3 B (a+b x)^3-16863 a^2 A b (a+b x)^3-9240 a^2 B (a+b x)^4+12705 a A b (a+b x)^4-3465 A b (a+b x)^5+2520 a B (a+b x)^5}{192 a^6 b x^4 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*(-128*a^5*A*b + 128*a^6*B - 1408*a^4*A*b*(a + b*x) + 1024*a^5*B*(a + b*x) + 9207*a^3*A*b*(a + b*x)^2 -
6696*a^4*B*(a + b*x)^2 - 16863*a^2*A*b*(a + b*x)^3 + 12264*a^3*B*(a + b*x)^3 + 12705*a*A*b*(a + b*x)^4 - 9240*
a^2*B*(a + b*x)^4 - 3465*A*b*(a + b*x)^5 + 2520*a*B*(a + b*x)^5)/(a^6*b*x^4*(a + b*x)^(3/2)) + (105*(-11*A*b^4
+ 8*a*b^3*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(13/2))

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fricas [A]  time = 1.29, size = 495, normalized size = 2.45 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.38, size = 223, normalized size = 1.10 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.02, size = 168, normalized size = 0.83 \begin {gather*} 2 \left (-\frac {-A b +B a}{3 \left (b x +a \right )^{\frac {3}{2}} a^{5}}-\frac {-5 A b +4 B a}{\sqrt {b x +a}\, a^{6}}+\frac {-\frac {105 \left (11 A b -8 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 \sqrt {a}}+\frac {\left (\frac {515 A b}{128}-\frac {41 B a}{16}\right ) \left (b x +a \right )^{\frac {7}{2}}+\left (-\frac {5153}{384} A a b +\frac {403}{48} B \,a^{2}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (\frac {5855}{384} A \,a^{2} b -\frac {445}{48} B \,a^{3}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (-\frac {765}{128} A \,a^{3} b +\frac {55}{16} B \,a^{4}\right ) \sqrt {b x +a}}{b^{4} x^{4}}}{a^{6}}\right ) b^{3} \end {gather*}

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

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maxima [A]  time = 2.03, size = 239, normalized size = 1.18 \begin {gather*} -\frac {1}{384} \, b^{4} {\left (\frac {2 \, {\left (128 \, B a^{6} - 128 \, A a^{5} b + 315 \, {\left (8 \, B a - 11 \, A b\right )} {\left (b x + a\right )}^{5} - 1155 \, {\left (8 \, B a^{2} - 11 \, A a b\right )} {\left (b x + a\right )}^{4} + 1533 \, {\left (8 \, B a^{3} - 11 \, A a^{2} b\right )} {\left (b x + a\right )}^{3} - 837 \, {\left (8 \, B a^{4} - 11 \, A a^{3} b\right )} {\left (b x + a\right )}^{2} + 128 \, {\left (8 \, B a^{5} - 11 \, A a^{4} b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {11}{2}} a^{6} b - 4 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{7} b + 6 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{8} b - 4 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{9} b + {\left (b x + a\right )}^{\frac {3}{2}} a^{10} b} + \frac {315 \, {\left (8 \, B a - 11 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {13}{2}} b}\right )} \end {gather*}

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]

-1/384*b^4*(2*(128*B*a^6 - 128*A*a^5*b + 315*(8*B*a - 11*A*b)*(b*x + a)^5 - 1155*(8*B*a^2 - 11*A*a*b)*(b*x + a
)^4 + 1533*(8*B*a^3 - 11*A*a^2*b)*(b*x + a)^3 - 837*(8*B*a^4 - 11*A*a^3*b)*(b*x + a)^2 + 128*(8*B*a^5 - 11*A*a
^4*b)*(b*x + a))/((b*x + a)^(11/2)*a^6*b - 4*(b*x + a)^(9/2)*a^7*b + 6*(b*x + a)^(7/2)*a^8*b - 4*(b*x + a)^(5/
2)*a^9*b + (b*x + a)^(3/2)*a^10*b) + 315*(8*B*a - 11*A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(
a)))/(a^(13/2)*b))

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mupad [B]  time = 0.47, size = 233, normalized size = 1.15 \begin {gather*} \frac {\frac {2\,\left (A\,b^4-B\,a\,b^3\right )}{3\,a}+\frac {2\,\left (11\,A\,b^4-8\,B\,a\,b^3\right )\,\left (a+b\,x\right )}{3\,a^2}-\frac {279\,\left (11\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^2}{64\,a^3}+\frac {511\,\left (11\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^3}{64\,a^4}-\frac {385\,\left (11\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^4}{64\,a^5}+\frac {105\,\left (11\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^5}{64\,a^6}}{{\left (a+b\,x\right )}^{11/2}-4\,a\,{\left (a+b\,x\right )}^{9/2}+a^4\,{\left (a+b\,x\right )}^{3/2}-4\,a^3\,{\left (a+b\,x\right )}^{5/2}+6\,a^2\,{\left (a+b\,x\right )}^{7/2}}-\frac {105\,b^3\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (11\,A\,b-8\,B\,a\right )}{64\,a^{13/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*(A*b^4 - B*a*b^3))/(3*a) + (2*(11*A*b^4 - 8*B*a*b^3)*(a + b*x))/(3*a^2) - (279*(11*A*b^4 - 8*B*a*b^3)*(a +
b*x)^2)/(64*a^3) + (511*(11*A*b^4 - 8*B*a*b^3)*(a + b*x)^3)/(64*a^4) - (385*(11*A*b^4 - 8*B*a*b^3)*(a + b*x)^
4)/(64*a^5) + (105*(11*A*b^4 - 8*B*a*b^3)*(a + b*x)^5)/(64*a^6))/((a + b*x)^(11/2) - 4*a*(a + b*x)^(9/2) + a^4
*(a + b*x)^(3/2) - 4*a^3*(a + b*x)^(5/2) + 6*a^2*(a + b*x)^(7/2)) - (105*b^3*atanh((a + b*x)^(1/2)/a^(1/2))*(1
1*A*b - 8*B*a))/(64*a^(13/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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