3.442 \(\int \frac{A+B x}{x^4 (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=143 \[ -\frac{5 b^2 (7 A b-6 a B)}{8 a^4 \sqrt{a+b x}}+\frac{5 b^2 (7 A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{9/2}}+\frac{7 A b-6 a B}{12 a^2 x^2 \sqrt{a+b x}}-\frac{5 b (7 A b-6 a B)}{24 a^3 x \sqrt{a+b x}}-\frac{A}{3 a x^3 \sqrt{a+b x}} \]

[Out]

(-5*b^2*(7*A*b - 6*a*B))/(8*a^4*Sqrt[a + b*x]) - A/(3*a*x^3*Sqrt[a + b*x]) + (7*A*b - 6*a*B)/(12*a^2*x^2*Sqrt[
a + b*x]) - (5*b*(7*A*b - 6*a*B))/(24*a^3*x*Sqrt[a + b*x]) + (5*b^2*(7*A*b - 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt
[a]])/(8*a^(9/2))

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Rubi [A]  time = 0.0605418, antiderivative size = 143, normalized size of antiderivative = 1., 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} \[ \frac{5 b^2 (7 A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{9/2}}+\frac{5 \sqrt{a+b x} (7 A b-6 a B)}{12 a^3 x^2}-\frac{7 A b-6 a B}{3 a^2 x^2 \sqrt{a+b x}}-\frac{5 b \sqrt{a+b x} (7 A b-6 a B)}{8 a^4 x}-\frac{A}{3 a x^3 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(3*a*x^3*Sqrt[a + b*x]) - (7*A*b - 6*a*B)/(3*a^2*x^2*Sqrt[a + b*x]) + (5*(7*A*b - 6*a*B)*Sqrt[a + b*x])/(12
*a^3*x^2) - (5*b*(7*A*b - 6*a*B)*Sqrt[a + b*x])/(8*a^4*x) + (5*b^2*(7*A*b - 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[
a]])/(8*a^(9/2))

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

Mathematica [C]  time = 0.0170816, size = 58, normalized size = 0.41 \[ \frac{b^2 x^3 (6 a B-7 A b) \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{b x}{a}+1\right )-a^3 A}{3 a^4 x^3 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.015, size = 126, normalized size = 0.9 \begin{align*} 2\,{b}^{2} \left ( -{\frac{1}{{a}^{4}} \left ({\frac{1}{{b}^{3}{x}^{3}} \left ( \left ({\frac{19\,Ab}{16}}-{\frac{7\,Ba}{8}} \right ) \left ( bx+a \right ) ^{5/2}+ \left ( -{\frac{17\,Aba}{6}}+2\,B{a}^{2} \right ) \left ( bx+a \right ) ^{3/2}+ \left ({\frac{29\,Ab{a}^{2}}{16}}-{\frac{9\,B{a}^{3}}{8}} \right ) \sqrt{bx+a} \right ) }-{\frac{35\,Ab-30\,Ba}{16\,\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }-{\frac{Ab-Ba}{{a}^{4}\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)^(3/2),x)

[Out]

2*b^2*(-1/a^4*(((19/16*A*b-7/8*B*a)*(b*x+a)^(5/2)+(-17/6*A*b*a+2*B*a^2)*(b*x+a)^(3/2)+(29/16*A*b*a^2-9/8*B*a^3
)*(b*x+a)^(1/2))/b^3/x^3-5/16*(7*A*b-6*B*a)/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))-(A*b-B*a)/a^4/(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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.61293, size = 721, normalized size = 5.04 \begin{align*} \left [-\frac{15 \,{\left ({\left (6 \, B a b^{3} - 7 \, A b^{4}\right )} x^{4} +{\left (6 \, B a^{2} b^{2} - 7 \, A a 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^{4} - 15 \,{\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} - 5 \,{\left (6 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 2 \,{\left (6 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt{b x + a}}{48 \,{\left (a^{5} b x^{4} + a^{6} x^{3}\right )}}, \frac{15 \,{\left ({\left (6 \, B a b^{3} - 7 \, A b^{4}\right )} x^{4} +{\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (8 \, A a^{4} - 15 \,{\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} - 5 \,{\left (6 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 2 \,{\left (6 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt{b x + a}}{24 \,{\left (a^{5} b x^{4} + a^{6} 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)^(3/2),x, algorithm="fricas")

[Out]

[-1/48*(15*((6*B*a*b^3 - 7*A*b^4)*x^4 + (6*B*a^2*b^2 - 7*A*a*b^3)*x^3)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt
(a) + 2*a)/x) + 2*(8*A*a^4 - 15*(6*B*a^2*b^2 - 7*A*a*b^3)*x^3 - 5*(6*B*a^3*b - 7*A*a^2*b^2)*x^2 + 2*(6*B*a^4 -
7*A*a^3*b)*x)*sqrt(b*x + a))/(a^5*b*x^4 + a^6*x^3), 1/24*(15*((6*B*a*b^3 - 7*A*b^4)*x^4 + (6*B*a^2*b^2 - 7*A*
a*b^3)*x^3)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (8*A*a^4 - 15*(6*B*a^2*b^2 - 7*A*a*b^3)*x^3 - 5*(6*B*a
^3*b - 7*A*a^2*b^2)*x^2 + 2*(6*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x + a))/(a^5*b*x^4 + a^6*x^3)]

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Sympy [A]  time = 130.558, size = 246, normalized size = 1.72 \begin{align*} A \left (- \frac{1}{3 a \sqrt{b} x^{\frac{7}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{7 \sqrt{b}}{12 a^{2} x^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{35 b^{\frac{3}{2}}}{24 a^{3} x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} - \frac{35 b^{\frac{5}{2}}}{8 a^{4} \sqrt{x} \sqrt{\frac{a}{b x} + 1}} + \frac{35 b^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{8 a^{\frac{9}{2}}}\right ) + B \left (- \frac{1}{2 a \sqrt{b} x^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{5 \sqrt{b}}{4 a^{2} x^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}} + \frac{15 b^{\frac{3}{2}}}{4 a^{3} \sqrt{x} \sqrt{\frac{a}{b x} + 1}} - \frac{15 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{4 a^{\frac{7}{2}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**4/(b*x+a)**(3/2),x)

[Out]

A*(-1/(3*a*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) + 7*sqrt(b)/(12*a**2*x**(5/2)*sqrt(a/(b*x) + 1)) - 35*b**(3/2)/
(24*a**3*x**(3/2)*sqrt(a/(b*x) + 1)) - 35*b**(5/2)/(8*a**4*sqrt(x)*sqrt(a/(b*x) + 1)) + 35*b**3*asinh(sqrt(a)/
(sqrt(b)*sqrt(x)))/(8*a**(9/2))) + B*(-1/(2*a*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) + 5*sqrt(b)/(4*a**2*x**(3/2)
*sqrt(a/(b*x) + 1)) + 15*b**(3/2)/(4*a**3*sqrt(x)*sqrt(a/(b*x) + 1)) - 15*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x))
)/(4*a**(7/2)))

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Giac [A]  time = 1.13821, size = 223, normalized size = 1.56 \begin{align*} \frac{5 \,{\left (6 \, B a b^{2} - 7 \, A b^{3}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a^{4}} + \frac{2 \,{\left (B a b^{2} - A b^{3}\right )}}{\sqrt{b x + a} a^{4}} + \frac{42 \,{\left (b x + a\right )}^{\frac{5}{2}} B a b^{2} - 96 \,{\left (b x + a\right )}^{\frac{3}{2}} B a^{2} b^{2} + 54 \, \sqrt{b x + a} B a^{3} b^{2} - 57 \,{\left (b x + a\right )}^{\frac{5}{2}} A b^{3} + 136 \,{\left (b x + a\right )}^{\frac{3}{2}} A a b^{3} - 87 \, \sqrt{b x + a} A a^{2} b^{3}}{24 \, a^{4} b^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

5/8*(6*B*a*b^2 - 7*A*b^3)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4) + 2*(B*a*b^2 - A*b^3)/(sqrt(b*x + a)*a
^4) + 1/24*(42*(b*x + a)^(5/2)*B*a*b^2 - 96*(b*x + a)^(3/2)*B*a^2*b^2 + 54*sqrt(b*x + a)*B*a^3*b^2 - 57*(b*x +
a)^(5/2)*A*b^3 + 136*(b*x + a)^(3/2)*A*a*b^3 - 87*sqrt(b*x + a)*A*a^2*b^3)/(a^4*b^3*x^3)