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

Optimal. Leaf size=152 \[ -\frac{2 (a+b x)^{5/2} (3 a B+4 A b)}{3 a \sqrt{x}}+\frac{5 b \sqrt{x} (a+b x)^{3/2} (3 a B+4 A b)}{6 a}+\frac{5}{4} b \sqrt{x} \sqrt{a+b x} (3 a B+4 A b)+\frac{5}{4} a \sqrt{b} (3 a B+4 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )-\frac{2 A (a+b x)^{7/2}}{3 a x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0633792, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {78, 47, 50, 63, 217, 206} \[ -\frac{2 (a+b x)^{5/2} (3 a B+4 A b)}{3 a \sqrt{x}}+\frac{5 b \sqrt{x} (a+b x)^{3/2} (3 a B+4 A b)}{6 a}+\frac{5}{4} b \sqrt{x} \sqrt{a+b x} (3 a B+4 A b)+\frac{5}{4} a \sqrt{b} (3 a B+4 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )-\frac{2 A (a+b x)^{7/2}}{3 a x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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^{5/2}} \, dx &=-\frac{2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac{\left (2 \left (2 A b+\frac{3 a B}{2}\right )\right ) \int \frac{(a+b x)^{5/2}}{x^{3/2}} \, dx}{3 a}\\ &=-\frac{2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt{x}}-\frac{2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac{(5 b (4 A b+3 a B)) \int \frac{(a+b x)^{3/2}}{\sqrt{x}} \, dx}{3 a}\\ &=\frac{5 b (4 A b+3 a B) \sqrt{x} (a+b x)^{3/2}}{6 a}-\frac{2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt{x}}-\frac{2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac{1}{4} (5 b (4 A b+3 a B)) \int \frac{\sqrt{a+b x}}{\sqrt{x}} \, dx\\ &=\frac{5}{4} b (4 A b+3 a B) \sqrt{x} \sqrt{a+b x}+\frac{5 b (4 A b+3 a B) \sqrt{x} (a+b x)^{3/2}}{6 a}-\frac{2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt{x}}-\frac{2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac{1}{8} (5 a b (4 A b+3 a B)) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx\\ &=\frac{5}{4} b (4 A b+3 a B) \sqrt{x} \sqrt{a+b x}+\frac{5 b (4 A b+3 a B) \sqrt{x} (a+b x)^{3/2}}{6 a}-\frac{2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt{x}}-\frac{2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac{1}{4} (5 a b (4 A b+3 a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{5}{4} b (4 A b+3 a B) \sqrt{x} \sqrt{a+b x}+\frac{5 b (4 A b+3 a B) \sqrt{x} (a+b x)^{3/2}}{6 a}-\frac{2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt{x}}-\frac{2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac{1}{4} (5 a b (4 A b+3 a B)) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )\\ &=\frac{5}{4} b (4 A b+3 a B) \sqrt{x} \sqrt{a+b x}+\frac{5 b (4 A b+3 a B) \sqrt{x} (a+b x)^{3/2}}{6 a}-\frac{2 (4 A b+3 a B) (a+b x)^{5/2}}{3 a \sqrt{x}}-\frac{2 A (a+b x)^{7/2}}{3 a x^{3/2}}+\frac{5}{4} a \sqrt{b} (4 A b+3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0369134, size = 76, normalized size = 0.5 \[ \frac{2 \sqrt{a+b x} \left (-\frac{a^2 x (3 a B+4 A b) \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x}{a}\right )}{\sqrt{\frac{b x}{a}+1}}-A (a+b x)^3\right )}{3 a x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 207, normalized size = 1.4 \begin{align*}{\frac{1}{24}\sqrt{bx+a} \left ( 12\,B\sqrt{x \left ( bx+a \right ) }{b}^{5/2}{x}^{3}+60\,A{b}^{2}\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){x}^{2}a+24\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}{x}^{2}+45\,Bb\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){x}^{2}{a}^{2}+54\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}{x}^{2}a-112\,Axa{b}^{3/2}\sqrt{x \left ( bx+a \right ) }-48\,Bx{a}^{2}\sqrt{x \left ( bx+a \right ) }\sqrt{b}-16\,A{a}^{2}\sqrt{x \left ( bx+a \right ) }\sqrt{b} \right ){x}^{-{\frac{3}{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^(5/2),x)

[Out]

1/24*(b*x+a)^(1/2)/x^(3/2)*(12*B*(x*(b*x+a))^(1/2)*b^(5/2)*x^3+60*A*b^2*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*
b*x+a)/b^(1/2))*x^2*a+24*A*(x*(b*x+a))^(1/2)*b^(5/2)*x^2+45*B*b*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b
^(1/2))*x^2*a^2+54*B*(x*(b*x+a))^(1/2)*b^(3/2)*x^2*a-112*A*x*a*b^(3/2)*(x*(b*x+a))^(1/2)-48*B*x*a^2*(x*(b*x+a)
)^(1/2)*b^(1/2)-16*A*a^2*(x*(b*x+a))^(1/2)*b^(1/2))/(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^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.69887, size = 537, normalized size = 3.53 \begin{align*} \left [\frac{15 \,{\left (3 \, B a^{2} + 4 \, A a b\right )} \sqrt{b} x^{2} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (6 \, B b^{2} x^{3} - 8 \, A a^{2} + 3 \,{\left (9 \, B a b + 4 \, A b^{2}\right )} x^{2} - 8 \,{\left (3 \, B a^{2} + 7 \, A a b\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{24 \, x^{2}}, -\frac{15 \,{\left (3 \, B a^{2} + 4 \, A a b\right )} \sqrt{-b} x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (6 \, B b^{2} x^{3} - 8 \, A a^{2} + 3 \,{\left (9 \, B a b + 4 \, A b^{2}\right )} x^{2} - 8 \,{\left (3 \, B a^{2} + 7 \, A a b\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{12 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(15*(3*B*a^2 + 4*A*a*b)*sqrt(b)*x^2*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(6*B*b^2*x^3 -
8*A*a^2 + 3*(9*B*a*b + 4*A*b^2)*x^2 - 8*(3*B*a^2 + 7*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/x^2, -1/12*(15*(3*B*a^2
+ 4*A*a*b)*sqrt(-b)*x^2*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (6*B*b^2*x^3 - 8*A*a^2 + 3*(9*B*a*b + 4*A
*b^2)*x^2 - 8*(3*B*a^2 + 7*A*a*b)*x)*sqrt(b*x + a)*sqrt(x))/x^2]

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Sympy [A]  time = 133.843, size = 230, normalized size = 1.51 \begin{align*} A \left (- \frac{2 a^{2} \sqrt{b} \sqrt{\frac{a}{b x} + 1}}{3 x} - \frac{14 a b^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}}{3} - \frac{5 a b^{\frac{3}{2}} \log{\left (\frac{a}{b x} \right )}}{2} + 5 a b^{\frac{3}{2}} \log{\left (\sqrt{\frac{a}{b x} + 1} + 1 \right )} + b^{\frac{5}{2}} x \sqrt{\frac{a}{b x} + 1}\right ) + B \left (- \frac{2 a^{\frac{5}{2}}}{\sqrt{x} \sqrt{1 + \frac{b x}{a}}} + \frac{a^{\frac{3}{2}} b \sqrt{x}}{4 \sqrt{1 + \frac{b x}{a}}} + \frac{11 \sqrt{a} b^{2} x^{\frac{3}{2}}}{4 \sqrt{1 + \frac{b x}{a}}} + \frac{15 a^{2} \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4} + \frac{b^{3} x^{\frac{5}{2}}}{2 \sqrt{a} \sqrt{1 + \frac{b x}{a}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-2*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x) - 14*a*b**(3/2)*sqrt(a/(b*x) + 1)/3 - 5*a*b**(3/2)*log(a/(b*x))/2 +
5*a*b**(3/2)*log(sqrt(a/(b*x) + 1) + 1) + b**(5/2)*x*sqrt(a/(b*x) + 1)) + B*(-2*a**(5/2)/(sqrt(x)*sqrt(1 + b*
x/a)) + a**(3/2)*b*sqrt(x)/(4*sqrt(1 + b*x/a)) + 11*sqrt(a)*b**2*x**(3/2)/(4*sqrt(1 + b*x/a)) + 15*a**2*sqrt(b
)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/4 + b**3*x**(5/2)/(2*sqrt(a)*sqrt(1 + b*x/a)))

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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^(5/2),x, algorithm="giac")

[Out]

Timed out