3.816 \(\int \frac{A+B x}{x^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=190 \[ -\frac{2 b (a+b x) (A b-a B)}{a^3 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (A b-a B)}{3 a^2 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 b^{3/2} (a+b x) (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 A (a+b x)}{5 a x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0938857, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {770, 78, 51, 63, 205} \[ -\frac{2 b (a+b x) (A b-a B)}{a^3 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (A b-a B)}{3 a^2 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 b^{3/2} (a+b x) (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 A (a+b x)}{5 a x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [C]  time = 0.0228595, size = 60, normalized size = 0.32 \[ -\frac{2 (a+b x) \left (\, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{b x}{a}\right ) (5 a B x-5 A b x)+3 a A\right )}{15 a^2 x^{5/2} \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

(-2*(a + b*x)*(3*a*A + (-5*A*b*x + 5*a*B*x)*Hypergeometric2F1[-3/2, 1, -1/2, -((b*x)/a)]))/(15*a^2*x^(5/2)*Sqr
t[(a + b*x)^2])

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Maple [A]  time = 0.011, size = 131, normalized size = 0.7 \begin{align*} -{\frac{2\,bx+2\,a}{15\,{a}^{3}} \left ( 15\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{5/2}{b}^{3}-15\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{5/2}a{b}^{2}+15\,A\sqrt{ab}{x}^{2}{b}^{2}-15\,B\sqrt{ab}{x}^{2}ab-5\,A\sqrt{ab}xab+5\,B\sqrt{ab}x{a}^{2}+3\,A{a}^{2}\sqrt{ab} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}{x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x)

[Out]

-2/15*(b*x+a)*(15*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(5/2)*b^3-15*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(5/2)*a*b^2
+15*A*(a*b)^(1/2)*x^2*b^2-15*B*(a*b)^(1/2)*x^2*a*b-5*A*(a*b)^(1/2)*x*a*b+5*B*(a*b)^(1/2)*x*a^2+3*A*a^2*(a*b)^(
1/2))/((b*x+a)^2)^(1/2)/a^3/x^(5/2)/(a*b)^(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^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38453, size = 441, normalized size = 2.32 \begin{align*} \left [-\frac{15 \,{\left (B a b - A b^{2}\right )} x^{3} \sqrt{-\frac{b}{a}} \log \left (\frac{b x - 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) + 2 \,{\left (3 \, A a^{2} - 15 \,{\left (B a b - A b^{2}\right )} x^{2} + 5 \,{\left (B a^{2} - A a b\right )} x\right )} \sqrt{x}}{15 \, a^{3} x^{3}}, -\frac{2 \,{\left (15 \,{\left (B a b - A b^{2}\right )} x^{3} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) +{\left (3 \, A a^{2} - 15 \,{\left (B a b - A b^{2}\right )} x^{2} + 5 \,{\left (B a^{2} - A a b\right )} x\right )} \sqrt{x}\right )}}{15 \, a^{3} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/15*(15*(B*a*b - A*b^2)*x^3*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(3*A*a^2 - 15*
(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b)*x)*sqrt(x))/(a^3*x^3), -2/15*(15*(B*a*b - A*b^2)*x^3*sqrt(b/a)*arctan(
a*sqrt(b/a)/(b*sqrt(x))) + (3*A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b)*x)*sqrt(x))/(a^3*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**(7/2)/((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.18502, size = 165, normalized size = 0.87 \begin{align*} \frac{2 \,{\left (B a b^{2} \mathrm{sgn}\left (b x + a\right ) - A b^{3} \mathrm{sgn}\left (b x + a\right )\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a^{3}} + \frac{2 \,{\left (15 \, B a b x^{2} \mathrm{sgn}\left (b x + a\right ) - 15 \, A b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) - 5 \, B a^{2} x \mathrm{sgn}\left (b x + a\right ) + 5 \, A a b x \mathrm{sgn}\left (b x + a\right ) - 3 \, A a^{2} \mathrm{sgn}\left (b x + a\right )\right )}}{15 \, a^{3} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2*(B*a*b^2*sgn(b*x + a) - A*b^3*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + 2/15*(15*B*a*b*x^2
*sgn(b*x + a) - 15*A*b^2*x^2*sgn(b*x + a) - 5*B*a^2*x*sgn(b*x + a) + 5*A*a*b*x*sgn(b*x + a) - 3*A*a^2*sgn(b*x
+ a))/(a^3*x^(5/2))