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

Optimal. Leaf size=98 \[ -\frac{5 A b-2 a B}{a^3 \sqrt{a+b x}}-\frac{5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}+\frac{(5 A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{7/2}}-\frac{A}{a x (a+b x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0443182, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 51, 63, 208} \[ -\frac{5 A b-2 a B}{a^3 \sqrt{a+b x}}-\frac{5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}+\frac{(5 A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{7/2}}-\frac{A}{a x (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(5*A*b - 2*a*B)/(3*a^2*(a + b*x)^(3/2)) - A/(a*x*(a + b*x)^(3/2)) - (5*A*b - 2*a*B)/(a^3*Sqrt[a + b*x]) + ((5
*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/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 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^2 (a+b x)^{5/2}} \, dx &=-\frac{A}{a x (a+b x)^{3/2}}+\frac{\left (-\frac{5 A b}{2}+a B\right ) \int \frac{1}{x (a+b x)^{5/2}} \, dx}{a}\\ &=-\frac{5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}-\frac{A}{a x (a+b x)^{3/2}}-\frac{(5 A b-2 a B) \int \frac{1}{x (a+b x)^{3/2}} \, dx}{2 a^2}\\ &=-\frac{5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}-\frac{A}{a x (a+b x)^{3/2}}-\frac{5 A b-2 a B}{a^3 \sqrt{a+b x}}-\frac{(5 A b-2 a B) \int \frac{1}{x \sqrt{a+b x}} \, dx}{2 a^3}\\ &=-\frac{5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}-\frac{A}{a x (a+b x)^{3/2}}-\frac{5 A b-2 a B}{a^3 \sqrt{a+b x}}-\frac{(5 A b-2 a B) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{a^3 b}\\ &=-\frac{5 A b-2 a B}{3 a^2 (a+b x)^{3/2}}-\frac{A}{a x (a+b x)^{3/2}}-\frac{5 A b-2 a B}{a^3 \sqrt{a+b x}}+\frac{(5 A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0158223, size = 52, normalized size = 0.53 \[ \frac{\, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b x}{a}+1\right ) (2 a B x-5 A b x)-3 a A}{3 a^2 x (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 88, normalized size = 0.9 \begin{align*} -2\,{\frac{1}{{a}^{3}} \left ( 1/2\,{\frac{A\sqrt{bx+a}}{x}}-1/2\,{\frac{5\,Ab-2\,Ba}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }-{\frac{2\,Ab-2\,Ba}{3\,{a}^{2}} \left ( bx+a \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{2\,Ab-Ba}{{a}^{3}\sqrt{bx+a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/a^3*(1/2*A*(b*x+a)^(1/2)/x-1/2*(5*A*b-2*B*a)/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))-2/3*(A*b-B*a)/a^2/(b*x
+a)^(3/2)-2*(2*A*b-B*a)/a^3/(b*x+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)/x^2/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.39521, size = 724, normalized size = 7.39 \begin{align*} \left [-\frac{3 \,{\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \,{\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} +{\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{a} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (3 \, A a^{3} - 3 \,{\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} - 4 \,{\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{6 \,{\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}, \frac{3 \,{\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \,{\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} +{\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (3 \, A a^{3} - 3 \,{\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} - 4 \,{\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{3 \,{\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*((2*B*a*b^2 - 5*A*b^3)*x^3 + 2*(2*B*a^2*b - 5*A*a*b^2)*x^2 + (2*B*a^3 - 5*A*a^2*b)*x)*sqrt(a)*log((b*
x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(3*A*a^3 - 3*(2*B*a^2*b - 5*A*a*b^2)*x^2 - 4*(2*B*a^3 - 5*A*a^2*b)*x
)*sqrt(b*x + a))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x), 1/3*(3*((2*B*a*b^2 - 5*A*b^3)*x^3 + 2*(2*B*a^2*b - 5*A*a
*b^2)*x^2 + (2*B*a^3 - 5*A*a^2*b)*x)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (3*A*a^3 - 3*(2*B*a^2*b - 5*A
*a*b^2)*x^2 - 4*(2*B*a^3 - 5*A*a^2*b)*x)*sqrt(b*x + a))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)]

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Sympy [B]  time = 45.5914, size = 1520, normalized size = 15.51 \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)/x**2/(b*x+a)**(5/2),x)

[Out]

A*(-6*a**17*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) - 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**(3
3/2)*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 + 1
8*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**(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) - 30*a**14*b**3*x**3*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**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**(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**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*log(sqrt(1 + b*x/a) + 1)/(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)) + B*(8*a**7*sqrt(1 + b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9
*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 3*a**7*log(b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)
*b**2*x**2 + 3*a**(13/2)*b**3*x**3) - 6*a**7*log(sqrt(1 + b*x/a) + 1)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(1
5/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 14*a**6*b*x*sqrt(1 + b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(1
5/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 9*a**6*b*x*log(b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b
**2*x**2 + 3*a**(13/2)*b**3*x**3) - 18*a**6*b*x*log(sqrt(1 + b*x/a) + 1)/(3*a**(19/2) + 9*a**(17/2)*b*x + 9*a*
*(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 6*a**5*b**2*x**2*sqrt(1 + b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x +
9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 9*a**5*b**2*x**2*log(b*x/a)/(3*a**(19/2) + 9*a**(17/2)*b*x +
9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) - 18*a**5*b**2*x**2*log(sqrt(1 + b*x/a) + 1)/(3*a**(19/2) + 9*
a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) + 3*a**4*b**3*x**3*log(b*x/a)/(3*a**(19/2) + 9*
a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3) - 6*a**4*b**3*x**3*log(sqrt(1 + b*x/a) + 1)/(3*
a**(19/2) + 9*a**(17/2)*b*x + 9*a**(15/2)*b**2*x**2 + 3*a**(13/2)*b**3*x**3))

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Giac [A]  time = 1.18447, size = 122, normalized size = 1.24 \begin{align*} \frac{{\left (2 \, B a - 5 \, A b\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} - \frac{\sqrt{b x + a} A}{a^{3} x} + \frac{2 \,{\left (3 \,{\left (b x + a\right )} B a + B a^{2} - 6 \,{\left (b x + a\right )} A b - A a b\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*a - 5*A*b)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) - sqrt(b*x + a)*A/(a^3*x) + 2/3*(3*(b*x + a)*B*a
+ B*a^2 - 6*(b*x + a)*A*b - A*a*b)/((b*x + a)^(3/2)*a^3)