3.508 \(\int \frac{(a+b x)^{5/2} (A+B x)}{x^{9/2}} \, dx\)

Optimal. Leaf size=111 \[ -\frac{2 A (a+b x)^{7/2}}{7 a x^{7/2}}-\frac{2 b^2 B \sqrt{a+b x}}{\sqrt{x}}+2 b^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )-\frac{2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac{2 b B (a+b x)^{3/2}}{3 x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0365201, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {78, 47, 63, 217, 206} \[ -\frac{2 A (a+b x)^{7/2}}{7 a x^{7/2}}-\frac{2 b^2 B \sqrt{a+b x}}{\sqrt{x}}+2 b^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )-\frac{2 B (a+b x)^{5/2}}{5 x^{5/2}}-\frac{2 b B (a+b x)^{3/2}}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^(9/2),x]

[Out]

(-2*b^2*B*Sqrt[a + b*x])/Sqrt[x] - (2*b*B*(a + b*x)^(3/2))/(3*x^(3/2)) - (2*B*(a + b*x)^(5/2))/(5*x^(5/2)) - (
2*A*(a + b*x)^(7/2))/(7*a*x^(7/2)) + 2*b^(5/2)*B*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.0615031, size = 77, normalized size = 0.69 \[ \frac{2 \sqrt{a+b x} \left (-\frac{a^4 B \, _2F_1\left (-\frac{7}{2},-\frac{7}{2};-\frac{5}{2};-\frac{b x}{a}\right )}{\sqrt{\frac{b x}{a}+1}}-(a+b x)^3 (A b-a B)\right )}{7 a b x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^(9/2),x]

[Out]

(2*Sqrt[a + b*x]*(-((A*b - a*B)*(a + b*x)^3) - (a^4*B*Hypergeometric2F1[-7/2, -7/2, -5/2, -((b*x)/a)])/Sqrt[1
+ (b*x)/a]))/(7*a*b*x^(7/2))

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Maple [B]  time = 0.012, size = 198, normalized size = 1.8 \begin{align*} -{\frac{1}{105\,a}\sqrt{bx+a} \left ( -105\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){x}^{4}a{b}^{3}+30\,A\sqrt{x \left ( bx+a \right ) }{b}^{7/2}{x}^{3}+322\,B\sqrt{x \left ( bx+a \right ) }{b}^{5/2}{x}^{3}a+90\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}{x}^{2}a+154\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}{x}^{2}{a}^{2}+90\,Ax{a}^{2}{b}^{3/2}\sqrt{x \left ( bx+a \right ) }+42\,Bx{a}^{3}\sqrt{x \left ( bx+a \right ) }\sqrt{b}+30\,A{a}^{3}\sqrt{x \left ( bx+a \right ) }\sqrt{b} \right ){x}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(9/2),x)

[Out]

-1/105*(b*x+a)^(1/2)/x^(7/2)*(-105*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x^4*a*b^3+30*A*(x*(
b*x+a))^(1/2)*b^(7/2)*x^3+322*B*(x*(b*x+a))^(1/2)*b^(5/2)*x^3*a+90*A*(x*(b*x+a))^(1/2)*b^(5/2)*x^2*a+154*B*(x*
(b*x+a))^(1/2)*b^(3/2)*x^2*a^2+90*A*x*a^2*b^(3/2)*(x*(b*x+a))^(1/2)+42*B*x*a^3*(x*(b*x+a))^(1/2)*b^(1/2)+30*A*
a^3*(x*(b*x+a))^(1/2)*b^(1/2))/a/(x*(b*x+a))^(1/2)/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)^(5/2)*(B*x+A)/x^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.62486, size = 583, normalized size = 5.25 \begin{align*} \left [\frac{105 \, B a b^{\frac{5}{2}} x^{4} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (15 \, A a^{3} +{\left (161 \, B a b^{2} + 15 \, A b^{3}\right )} x^{3} +{\left (77 \, B a^{2} b + 45 \, A a b^{2}\right )} x^{2} + 3 \,{\left (7 \, B a^{3} + 15 \, A a^{2} b\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{105 \, a x^{4}}, -\frac{2 \,{\left (105 \, B a \sqrt{-b} b^{2} x^{4} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (15 \, A a^{3} +{\left (161 \, B a b^{2} + 15 \, A b^{3}\right )} x^{3} +{\left (77 \, B a^{2} b + 45 \, A a b^{2}\right )} x^{2} + 3 \,{\left (7 \, B a^{3} + 15 \, A a^{2} b\right )} x\right )} \sqrt{b x + a} \sqrt{x}\right )}}{105 \, a x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(9/2),x, algorithm="fricas")

[Out]

[1/105*(105*B*a*b^(5/2)*x^4*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(15*A*a^3 + (161*B*a*b^2 + 15
*A*b^3)*x^3 + (77*B*a^2*b + 45*A*a*b^2)*x^2 + 3*(7*B*a^3 + 15*A*a^2*b)*x)*sqrt(b*x + a)*sqrt(x))/(a*x^4), -2/1
05*(105*B*a*sqrt(-b)*b^2*x^4*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (15*A*a^3 + (161*B*a*b^2 + 15*A*b^3)
*x^3 + (77*B*a^2*b + 45*A*a*b^2)*x^2 + 3*(7*B*a^3 + 15*A*a^2*b)*x)*sqrt(b*x + a)*sqrt(x))/(a*x^4)]

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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)**(5/2)*(B*x+A)/x**(9/2),x)

[Out]

Timed out

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Giac [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)^(5/2)*(B*x+A)/x^(9/2),x, algorithm="giac")

[Out]

Timed out