3.452 \(\int \frac{A+B x}{x^4 (a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=171 \[ -\frac{35 b^2 (3 A b-2 a B)}{8 a^5 \sqrt{a+b x}}-\frac{35 b^2 (3 A b-2 a B)}{24 a^4 (a+b x)^{3/2}}+\frac{35 b^2 (3 A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{11/2}}+\frac{3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}-\frac{7 b (3 A b-2 a B)}{8 a^3 x (a+b x)^{3/2}}-\frac{A}{3 a x^3 (a+b x)^{3/2}} \]

[Out]

(-35*b^2*(3*A*b - 2*a*B))/(24*a^4*(a + b*x)^(3/2)) - A/(3*a*x^3*(a + b*x)^(3/2)) + (3*A*b - 2*a*B)/(4*a^2*x^2*
(a + b*x)^(3/2)) - (7*b*(3*A*b - 2*a*B))/(8*a^3*x*(a + b*x)^(3/2)) - (35*b^2*(3*A*b - 2*a*B))/(8*a^5*Sqrt[a +
b*x]) + (35*b^2*(3*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(11/2))

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Rubi [A]  time = 0.0754013, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 51, 63, 208} \[ \frac{35 b^2 (3 A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{11/2}}+\frac{35 \sqrt{a+b x} (3 A b-2 a B)}{12 a^4 x^2}-\frac{7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt{a+b x}}-\frac{3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac{35 b \sqrt{a+b x} (3 A b-2 a B)}{8 a^5 x}-\frac{A}{3 a x^3 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^4*(a + b*x)^(5/2)),x]

[Out]

-A/(3*a*x^3*(a + b*x)^(3/2)) - (3*A*b - 2*a*B)/(3*a^2*x^2*(a + b*x)^(3/2)) - (7*(3*A*b - 2*a*B))/(3*a^3*x^2*Sq
rt[a + b*x]) + (35*(3*A*b - 2*a*B)*Sqrt[a + b*x])/(12*a^4*x^2) - (35*b*(3*A*b - 2*a*B)*Sqrt[a + b*x])/(8*a^5*x
) + (35*b^2*(3*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(11/2))

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^4 (a+b x)^{5/2}} \, dx &=-\frac{A}{3 a x^3 (a+b x)^{3/2}}+\frac{\left (-\frac{9 A b}{2}+3 a B\right ) \int \frac{1}{x^3 (a+b x)^{5/2}} \, dx}{3 a}\\ &=-\frac{A}{3 a x^3 (a+b x)^{3/2}}-\frac{3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac{(7 (3 A b-2 a B)) \int \frac{1}{x^3 (a+b x)^{3/2}} \, dx}{6 a^2}\\ &=-\frac{A}{3 a x^3 (a+b x)^{3/2}}-\frac{3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac{7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt{a+b x}}-\frac{(35 (3 A b-2 a B)) \int \frac{1}{x^3 \sqrt{a+b x}} \, dx}{6 a^3}\\ &=-\frac{A}{3 a x^3 (a+b x)^{3/2}}-\frac{3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac{7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt{a+b x}}+\frac{35 (3 A b-2 a B) \sqrt{a+b x}}{12 a^4 x^2}+\frac{(35 b (3 A b-2 a B)) \int \frac{1}{x^2 \sqrt{a+b x}} \, dx}{8 a^4}\\ &=-\frac{A}{3 a x^3 (a+b x)^{3/2}}-\frac{3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac{7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt{a+b x}}+\frac{35 (3 A b-2 a B) \sqrt{a+b x}}{12 a^4 x^2}-\frac{35 b (3 A b-2 a B) \sqrt{a+b x}}{8 a^5 x}-\frac{\left (35 b^2 (3 A b-2 a B)\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx}{16 a^5}\\ &=-\frac{A}{3 a x^3 (a+b x)^{3/2}}-\frac{3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac{7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt{a+b x}}+\frac{35 (3 A b-2 a B) \sqrt{a+b x}}{12 a^4 x^2}-\frac{35 b (3 A b-2 a B) \sqrt{a+b x}}{8 a^5 x}-\frac{(35 b (3 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 )}{8 a^5}\\ &=-\frac{A}{3 a x^3 (a+b x)^{3/2}}-\frac{3 A b-2 a B}{3 a^2 x^2 (a+b x)^{3/2}}-\frac{7 (3 A b-2 a B)}{3 a^3 x^2 \sqrt{a+b x}}+\frac{35 (3 A b-2 a B) \sqrt{a+b x}}{12 a^4 x^2}-\frac{35 b (3 A b-2 a B) \sqrt{a+b x}}{8 a^5 x}+\frac{35 b^2 (3 A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0185498, size = 58, normalized size = 0.34 \[ \frac{b^2 x^3 (2 a B-3 A b) \, _2F_1\left (-\frac{3}{2},3;-\frac{1}{2};\frac{b x}{a}+1\right )-a^3 A}{3 a^4 x^3 (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^4*(a + b*x)^(5/2)),x]

[Out]

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

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Maple [A]  time = 0.016, size = 147, normalized size = 0.9 \begin{align*} 2\,{b}^{2} \left ( -{\frac{1}{{a}^{5}} \left ({\frac{1}{{b}^{3}{x}^{3}} \left ( \left ({\frac{41\,Ab}{16}}-{\frac{11\,Ba}{8}} \right ) \left ( bx+a \right ) ^{5/2}+ \left ( -{\frac{35\,Aba}{6}}+3\,B{a}^{2} \right ) \left ( bx+a \right ) ^{3/2}+ \left ({\frac{55\,Ab{a}^{2}}{16}}-{\frac{13\,B{a}^{3}}{8}} \right ) \sqrt{bx+a} \right ) }-{\frac{105\,Ab-70\,Ba}{16\,\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }-1/3\,{\frac{Ab-Ba}{{a}^{4} \left ( bx+a \right ) ^{3/2}}}-{\frac{4\,Ab-3\,Ba}{{a}^{5}\sqrt{bx+a}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^4/(b*x+a)^(5/2),x)

[Out]

2*b^2*(-1/a^5*(((41/16*A*b-11/8*B*a)*(b*x+a)^(5/2)+(-35/6*A*b*a+3*B*a^2)*(b*x+a)^(3/2)+(55/16*A*b*a^2-13/8*B*a
^3)*(b*x+a)^(1/2))/b^3/x^3-35/16*(3*A*b-2*B*a)/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))-1/3*(A*b-B*a)/a^4/(b*x+
a)^(3/2)-(4*A*b-3*B*a)/a^5/(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^4/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.48025, size = 972, normalized size = 5.68 \begin{align*} \left [-\frac{105 \,{\left ({\left (2 \, B a b^{4} - 3 \, A b^{5}\right )} x^{5} + 2 \,{\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} +{\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt{a} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (8 \, A a^{5} - 105 \,{\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} - 140 \,{\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3} - 21 \,{\left (2 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x^{2} + 6 \,{\left (2 \, B a^{5} - 3 \, A a^{4} b\right )} x\right )} \sqrt{b x + a}}{48 \,{\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}, \frac{105 \,{\left ({\left (2 \, B a b^{4} - 3 \, A b^{5}\right )} x^{5} + 2 \,{\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} +{\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (8 \, A a^{5} - 105 \,{\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} - 140 \,{\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3} - 21 \,{\left (2 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x^{2} + 6 \,{\left (2 \, B a^{5} - 3 \, A a^{4} b\right )} x\right )} \sqrt{b x + a}}{24 \,{\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/48*(105*((2*B*a*b^4 - 3*A*b^5)*x^5 + 2*(2*B*a^2*b^3 - 3*A*a*b^4)*x^4 + (2*B*a^3*b^2 - 3*A*a^2*b^3)*x^3)*sq
rt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*A*a^5 - 105*(2*B*a^2*b^3 - 3*A*a*b^4)*x^4 - 140*(2*B
*a^3*b^2 - 3*A*a^2*b^3)*x^3 - 21*(2*B*a^4*b - 3*A*a^3*b^2)*x^2 + 6*(2*B*a^5 - 3*A*a^4*b)*x)*sqrt(b*x + a))/(a^
6*b^2*x^5 + 2*a^7*b*x^4 + a^8*x^3), 1/24*(105*((2*B*a*b^4 - 3*A*b^5)*x^5 + 2*(2*B*a^2*b^3 - 3*A*a*b^4)*x^4 + (
2*B*a^3*b^2 - 3*A*a^2*b^3)*x^3)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (8*A*a^5 - 105*(2*B*a^2*b^3 - 3*A*
a*b^4)*x^4 - 140*(2*B*a^3*b^2 - 3*A*a^2*b^3)*x^3 - 21*(2*B*a^4*b - 3*A*a^3*b^2)*x^2 + 6*(2*B*a^5 - 3*A*a^4*b)*
x)*sqrt(b*x + a))/(a^6*b^2*x^5 + 2*a^7*b*x^4 + a^8*x^3)]

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

[Out]

Timed out

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Giac [A]  time = 1.21333, size = 270, normalized size = 1.58 \begin{align*} \frac{35 \,{\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a^{5}} + \frac{210 \,{\left (b x + a\right )}^{4} B a b^{2} - 560 \,{\left (b x + a\right )}^{3} B a^{2} b^{2} + 462 \,{\left (b x + a\right )}^{2} B a^{3} b^{2} - 96 \,{\left (b x + a\right )} B a^{4} b^{2} - 16 \, B a^{5} b^{2} - 315 \,{\left (b x + a\right )}^{4} A b^{3} + 840 \,{\left (b x + a\right )}^{3} A a b^{3} - 693 \,{\left (b x + a\right )}^{2} A a^{2} b^{3} + 144 \,{\left (b x + a\right )} A a^{3} b^{3} + 16 \, A a^{4} b^{3}}{24 \,{\left ({\left (b x + a\right )}^{\frac{3}{2}} - \sqrt{b x + a} a\right )}^{3} a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

35/8*(2*B*a*b^2 - 3*A*b^3)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5) + 1/24*(210*(b*x + a)^4*B*a*b^2 - 560
*(b*x + a)^3*B*a^2*b^2 + 462*(b*x + a)^2*B*a^3*b^2 - 96*(b*x + a)*B*a^4*b^2 - 16*B*a^5*b^2 - 315*(b*x + a)^4*A
*b^3 + 840*(b*x + a)^3*A*a*b^3 - 693*(b*x + a)^2*A*a^2*b^3 + 144*(b*x + a)*A*a^3*b^3 + 16*A*a^4*b^3)/(((b*x +
a)^(3/2) - sqrt(b*x + a)*a)^3*a^5)