3.451 \(\int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=136 \[ -\frac {5 b (7 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{9/2}}+\frac {5 b (7 A b-4 a B)}{4 a^4 \sqrt {a+b x}}+\frac {5 b (7 A b-4 a B)}{12 a^3 (a+b x)^{3/2}}+\frac {7 A b-4 a B}{4 a^2 x (a+b x)^{3/2}}-\frac {A}{2 a x^2 (a+b x)^{3/2}} \]

[Out]

5/12*b*(7*A*b-4*B*a)/a^3/(b*x+a)^(3/2)-1/2*A/a/x^2/(b*x+a)^(3/2)+1/4*(7*A*b-4*B*a)/a^2/x/(b*x+a)^(3/2)-5/4*b*(
7*A*b-4*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(9/2)+5/4*b*(7*A*b-4*B*a)/a^4/(b*x+a)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 140, normalized size of antiderivative = 1.03, 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 = {78, 51, 63, 208} \[ \frac {5 \sqrt {a+b x} (7 A b-4 a B)}{4 a^4 x}-\frac {5 (7 A b-4 a B)}{6 a^3 x \sqrt {a+b x}}-\frac {7 A b-4 a B}{6 a^2 x (a+b x)^{3/2}}-\frac {5 b (7 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{9/2}}-\frac {A}{2 a x^2 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^3*(a + b*x)^(5/2)),x]

[Out]

-A/(2*a*x^2*(a + b*x)^(3/2)) - (7*A*b - 4*a*B)/(6*a^2*x*(a + b*x)^(3/2)) - (5*(7*A*b - 4*a*B))/(6*a^3*x*Sqrt[a
+ b*x]) + (5*(7*A*b - 4*a*B)*Sqrt[a + b*x])/(4*a^4*x) - (5*b*(7*A*b - 4*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/
(4*a^(9/2))

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^3 (a+b x)^{5/2}} \, dx &=-\frac {A}{2 a x^2 (a+b x)^{3/2}}+\frac {\left (-\frac {7 A b}{2}+2 a B\right ) \int \frac {1}{x^2 (a+b x)^{5/2}} \, dx}{2 a}\\ &=-\frac {A}{2 a x^2 (a+b x)^{3/2}}-\frac {7 A b-4 a B}{6 a^2 x (a+b x)^{3/2}}-\frac {(5 (7 A b-4 a B)) \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx}{12 a^2}\\ &=-\frac {A}{2 a x^2 (a+b x)^{3/2}}-\frac {7 A b-4 a B}{6 a^2 x (a+b x)^{3/2}}-\frac {5 (7 A b-4 a B)}{6 a^3 x \sqrt {a+b x}}-\frac {(5 (7 A b-4 a B)) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{4 a^3}\\ &=-\frac {A}{2 a x^2 (a+b x)^{3/2}}-\frac {7 A b-4 a B}{6 a^2 x (a+b x)^{3/2}}-\frac {5 (7 A b-4 a B)}{6 a^3 x \sqrt {a+b x}}+\frac {5 (7 A b-4 a B) \sqrt {a+b x}}{4 a^4 x}+\frac {(5 b (7 A b-4 a B)) \int \frac {1}{x \sqrt {a+b x}} \, dx}{8 a^4}\\ &=-\frac {A}{2 a x^2 (a+b x)^{3/2}}-\frac {7 A b-4 a B}{6 a^2 x (a+b x)^{3/2}}-\frac {5 (7 A b-4 a B)}{6 a^3 x \sqrt {a+b x}}+\frac {5 (7 A b-4 a B) \sqrt {a+b x}}{4 a^4 x}+\frac {(5 (7 A b-4 a B)) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{4 a^4}\\ &=-\frac {A}{2 a x^2 (a+b x)^{3/2}}-\frac {7 A b-4 a B}{6 a^2 x (a+b x)^{3/2}}-\frac {5 (7 A b-4 a B)}{6 a^3 x \sqrt {a+b x}}+\frac {5 (7 A b-4 a B) \sqrt {a+b x}}{4 a^4 x}-\frac {5 b (7 A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 56, normalized size = 0.41 \[ \frac {b x^2 (7 A b-4 a B) \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\frac {b x}{a}+1\right )-3 a^2 A}{6 a^3 x^2 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^3*(a + b*x)^(5/2)),x]

[Out]

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

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fricas [A]  time = 0.67, size = 395, normalized size = 2.90 \[ \left [-\frac {15 \, {\left ({\left (4 \, B a b^{3} - 7 \, A b^{4}\right )} x^{4} + 2 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (6 \, A a^{4} + 15 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + 20 \, {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left (4 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{24 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}, -\frac {15 \, {\left ({\left (4 \, B a b^{3} - 7 \, A b^{4}\right )} x^{4} + 2 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (6 \, A a^{4} + 15 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + 20 \, {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left (4 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{12 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/24*(15*((4*B*a*b^3 - 7*A*b^4)*x^4 + 2*(4*B*a^2*b^2 - 7*A*a*b^3)*x^3 + (4*B*a^3*b - 7*A*a^2*b^2)*x^2)*sqrt(
a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(6*A*a^4 + 15*(4*B*a^2*b^2 - 7*A*a*b^3)*x^3 + 20*(4*B*a^3*
b - 7*A*a^2*b^2)*x^2 + 3*(4*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x + a))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2), -1/12*
(15*((4*B*a*b^3 - 7*A*b^4)*x^4 + 2*(4*B*a^2*b^2 - 7*A*a*b^3)*x^3 + (4*B*a^3*b - 7*A*a^2*b^2)*x^2)*sqrt(-a)*arc
tan(sqrt(b*x + a)*sqrt(-a)/a) + (6*A*a^4 + 15*(4*B*a^2*b^2 - 7*A*a*b^3)*x^3 + 20*(4*B*a^3*b - 7*A*a^2*b^2)*x^2
+ 3*(4*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x + a))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)]

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giac [A]  time = 1.33, size = 149, normalized size = 1.10 \[ -\frac {5 \, {\left (4 \, B a b - 7 \, A b^{2}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{4}} - \frac {2 \, {\left (6 \, {\left (b x + a\right )} B a b + B a^{2} b - 9 \, {\left (b x + a\right )} A b^{2} - A a b^{2}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4}} - \frac {4 \, {\left (b x + a\right )}^{\frac {3}{2}} B a b - 4 \, \sqrt {b x + a} B a^{2} b - 11 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{2} + 13 \, \sqrt {b x + a} A a b^{2}}{4 \, a^{4} b^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

-5/4*(4*B*a*b - 7*A*b^2)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4) - 2/3*(6*(b*x + a)*B*a*b + B*a^2*b - 9*
(b*x + a)*A*b^2 - A*a*b^2)/((b*x + a)^(3/2)*a^4) - 1/4*(4*(b*x + a)^(3/2)*B*a*b - 4*sqrt(b*x + a)*B*a^2*b - 11
*(b*x + a)^(3/2)*A*b^2 + 13*sqrt(b*x + a)*A*a*b^2)/(a^4*b^2*x^2)

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maple [A]  time = 0.02, size = 122, normalized size = 0.90 \[ 2 \left (-\frac {-A b +B a}{3 \left (b x +a \right )^{\frac {3}{2}} a^{3}}-\frac {-3 A b +2 B a}{\sqrt {b x +a}\, a^{4}}+\frac {-\frac {5 \left (7 A b -4 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}+\frac {\left (\frac {11 A b}{8}-\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (-\frac {13}{8} A a b +\frac {1}{2} B \,a^{2}\right ) \sqrt {b x +a}}{b^{2} x^{2}}}{a^{4}}\right ) b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^3/(b*x+a)^(5/2),x)

[Out]

2*b*(-(-3*A*b+2*B*a)/a^4/(b*x+a)^(1/2)-1/3*(-A*b+B*a)/a^3/(b*x+a)^(3/2)+1/a^4*(((11/8*A*b-1/2*B*a)*(b*x+a)^(3/
2)+(-13/8*A*a*b+1/2*B*a^2)*(b*x+a)^(1/2))/x^2/b^2-5/8*(7*A*b-4*B*a)/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))))

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maxima [A]  time = 1.97, size = 167, normalized size = 1.23 \[ -\frac {1}{24} \, b^{2} {\left (\frac {2 \, {\left (8 \, B a^{4} - 8 \, A a^{3} b + 15 \, {\left (4 \, B a - 7 \, A b\right )} {\left (b x + a\right )}^{3} - 25 \, {\left (4 \, B a^{2} - 7 \, A a b\right )} {\left (b x + a\right )}^{2} + 8 \, {\left (4 \, B a^{3} - 7 \, A a^{2} b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {7}{2}} a^{4} b - 2 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} b + {\left (b x + a\right )}^{\frac {3}{2}} a^{6} b} + \frac {15 \, {\left (4 \, B a - 7 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {9}{2}} b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

-1/24*b^2*(2*(8*B*a^4 - 8*A*a^3*b + 15*(4*B*a - 7*A*b)*(b*x + a)^3 - 25*(4*B*a^2 - 7*A*a*b)*(b*x + a)^2 + 8*(4
*B*a^3 - 7*A*a^2*b)*(b*x + a))/((b*x + a)^(7/2)*a^4*b - 2*(b*x + a)^(5/2)*a^5*b + (b*x + a)^(3/2)*a^6*b) + 15*
(4*B*a - 7*A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(9/2)*b))

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mupad [B]  time = 0.45, size = 147, normalized size = 1.08 \[ \frac {\frac {2\,\left (A\,b^2-B\,a\,b\right )}{3\,a}+\frac {2\,\left (7\,A\,b^2-4\,B\,a\,b\right )\,\left (a+b\,x\right )}{3\,a^2}-\frac {25\,\left (7\,A\,b^2-4\,B\,a\,b\right )\,{\left (a+b\,x\right )}^2}{12\,a^3}+\frac {5\,\left (7\,A\,b^2-4\,B\,a\,b\right )\,{\left (a+b\,x\right )}^3}{4\,a^4}}{{\left (a+b\,x\right )}^{7/2}-2\,a\,{\left (a+b\,x\right )}^{5/2}+a^2\,{\left (a+b\,x\right )}^{3/2}}-\frac {5\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (7\,A\,b-4\,B\,a\right )}{4\,a^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^3*(a + b*x)^(5/2)),x)

[Out]

((2*(A*b^2 - B*a*b))/(3*a) + (2*(7*A*b^2 - 4*B*a*b)*(a + b*x))/(3*a^2) - (25*(7*A*b^2 - 4*B*a*b)*(a + b*x)^2)/
(12*a^3) + (5*(7*A*b^2 - 4*B*a*b)*(a + b*x)^3)/(4*a^4))/((a + b*x)^(7/2) - 2*a*(a + b*x)^(5/2) + a^2*(a + b*x)
^(3/2)) - (5*b*atanh((a + b*x)^(1/2)/a^(1/2))*(7*A*b - 4*B*a))/(4*a^(9/2))

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sympy [B]  time = 122.38, size = 1287, normalized size = 9.46 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**3/(b*x+a)**(5/2),x)

[Out]

A*(-6*a**(89/2)*b**75*x**75/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x*
*(157/2)*sqrt(a/(b*x) + 1)) + 21*a**(87/2)*b**76*x**76/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) + 1) +
12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1)) + 140*a**(85/2)*b**77*x**77/(12*a**(93/2)*b**(151/2)*x*
*(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1)) + 105*a**(83/2)*b**78*x**78
/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1))
- 105*a**42*b**(155/2)*x**(155/2)*sqrt(a/(b*x) + 1)*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(12*a**(93/2)*b**(151/2)
*x**(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1)) - 105*a**41*b**(157/2)*x
**(157/2)*sqrt(a/(b*x) + 1)*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x)
+ 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1))) + B*(-6*a**17*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 1
8*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 46*a**16*b*x*sqrt(1 + b*x/a)/(6*a**(39/
2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 15*a**16*b*x*log(b*x/a)/(6*a**(
39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 30*a**16*b*x*log(sqrt(1 + b*
x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 70*a**15*b*
*2*x**2*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4)
- 45*a**15*b**2*x**2*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b
**3*x**4) + 90*a**15*b**2*x**2*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b*
*2*x**3 + 6*a**(33/2)*b**3*x**4) - 30*a**14*b**3*x**3*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 1
8*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 45*a**14*b**3*x**3*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b
*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 90*a**14*b**3*x**3*log(sqrt(1 + b*x/a) + 1)/(6*a**(3
9/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 15*a**13*b**4*x**4*log(b*x/a)
/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 30*a**13*b**4*x**4*l
og(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4)
)

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