3.527 \(\int \frac{x^{5/2} (A+B x)}{(a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=167 \[ \frac{5 a^2 (6 A b-7 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{9/2}}-\frac{x^{5/2} \sqrt{a+b x} (6 A b-7 a B)}{3 a b^2}+\frac{5 x^{3/2} \sqrt{a+b x} (6 A b-7 a B)}{12 b^3}-\frac{5 a \sqrt{x} \sqrt{a+b x} (6 A b-7 a B)}{8 b^4}+\frac{2 x^{7/2} (A b-a B)}{a b \sqrt{a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.0690089, antiderivative size = 167, 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, 50, 63, 217, 206} \[ \frac{5 a^2 (6 A b-7 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{9/2}}-\frac{x^{5/2} \sqrt{a+b x} (6 A b-7 a B)}{3 a b^2}+\frac{5 x^{3/2} \sqrt{a+b x} (6 A b-7 a B)}{12 b^3}-\frac{5 a \sqrt{x} \sqrt{a+b x} (6 A b-7 a B)}{8 b^4}+\frac{2 x^{7/2} (A b-a B)}{a b \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x))/(a + b*x)^(3/2),x]

[Out]

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

Mathematica [A]  time = 0.167263, size = 140, normalized size = 0.84 \[ \frac{\frac{(a+b x) (7 a B-6 A b) \left (b x \sqrt{\frac{b x}{a}+1} \left (15 a^2-10 a b x+8 b^2 x^2\right )-15 a^{5/2} \sqrt{b} \sqrt{x} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )\right )}{3 \sqrt{\frac{b x}{a}+1}}+16 b^4 x^4 (A b-a B)}{8 a b^5 \sqrt{x} \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x))/(a + b*x)^(3/2),x]

[Out]

(16*b^4*(A*b - a*B)*x^4 + ((-6*A*b + 7*a*B)*(a + b*x)*(b*x*Sqrt[1 + (b*x)/a]*(15*a^2 - 10*a*b*x + 8*b^2*x^2) -
15*a^(5/2)*Sqrt[b]*Sqrt[x]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(3*Sqrt[1 + (b*x)/a]))/(8*a*b^5*Sqrt[x]*Sqrt[
a + b*x])

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Maple [B]  time = 0.016, size = 288, normalized size = 1.7 \begin{align*}{\frac{1}{48} \left ( 16\,B{x}^{3}{b}^{7/2}\sqrt{x \left ( bx+a \right ) }+24\,A{x}^{2}{b}^{7/2}\sqrt{x \left ( bx+a \right ) }-28\,B{x}^{2}a{b}^{5/2}\sqrt{x \left ( bx+a \right ) }+90\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) x{a}^{2}{b}^{2}-60\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}xa-105\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) x{a}^{3}b+70\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}x{a}^{2}+90\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{3}b-180\,A\sqrt{x \left ( bx+a \right ) }{b}^{3/2}{a}^{2}-105\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{4}+210\,B\sqrt{x \left ( bx+a \right ) }\sqrt{b}{a}^{3} \right ) \sqrt{x}{b}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)/(b*x+a)^(3/2),x)

[Out]

1/48*(16*B*x^3*b^(7/2)*(x*(b*x+a))^(1/2)+24*A*x^2*b^(7/2)*(x*(b*x+a))^(1/2)-28*B*x^2*a*b^(5/2)*(x*(b*x+a))^(1/
2)+90*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x*a^2*b^2-60*A*(x*(b*x+a))^(1/2)*b^(5/2)*x*a-105
*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x*a^3*b+70*B*(x*(b*x+a))^(1/2)*b^(3/2)*x*a^2+90*A*ln(
1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^3*b-180*A*(x*(b*x+a))^(1/2)*b^(3/2)*a^2-105*B*ln(1/2*(2*(
x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^4+210*B*(x*(b*x+a))^(1/2)*b^(1/2)*a^3)/b^(9/2)*x^(1/2)/(x*(b*x+a)
)^(1/2)/(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(x^(5/2)*(B*x+A)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.82028, size = 718, normalized size = 4.3 \begin{align*} \left [-\frac{15 \,{\left (7 \, B a^{4} - 6 \, A a^{3} b +{\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (8 \, B b^{4} x^{3} + 105 \, B a^{3} b - 90 \, A a^{2} b^{2} - 2 \,{\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{2} + 5 \,{\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{48 \,{\left (b^{6} x + a b^{5}\right )}}, \frac{15 \,{\left (7 \, B a^{4} - 6 \, A a^{3} b +{\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (8 \, B b^{4} x^{3} + 105 \, B a^{3} b - 90 \, A a^{2} b^{2} - 2 \,{\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{2} + 5 \,{\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{24 \,{\left (b^{6} x + a b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/48*(15*(7*B*a^4 - 6*A*a^3*b + (7*B*a^3*b - 6*A*a^2*b^2)*x)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqr
t(x) + a) - 2*(8*B*b^4*x^3 + 105*B*a^3*b - 90*A*a^2*b^2 - 2*(7*B*a*b^3 - 6*A*b^4)*x^2 + 5*(7*B*a^2*b^2 - 6*A*a
*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^6*x + a*b^5), 1/24*(15*(7*B*a^4 - 6*A*a^3*b + (7*B*a^3*b - 6*A*a^2*b^2)*x)*
sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*B*b^4*x^3 + 105*B*a^3*b - 90*A*a^2*b^2 - 2*(7*B*a*b^3
- 6*A*b^4)*x^2 + 5*(7*B*a^2*b^2 - 6*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^6*x + a*b^5)]

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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(x**(5/2)*(B*x+A)/(b*x+a)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 82.7259, size = 293, normalized size = 1.75 \begin{align*} \frac{1}{24} \, \sqrt{{\left (b x + a\right )} b - a b} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )} B{\left | b \right |}}{b^{6}} - \frac{19 \, B a b^{17}{\left | b \right |} - 6 \, A b^{18}{\left | b \right |}}{b^{23}}\right )} + \frac{3 \,{\left (29 \, B a^{2} b^{17}{\left | b \right |} - 18 \, A a b^{18}{\left | b \right |}\right )}}{b^{23}}\right )} + \frac{5 \,{\left (7 \, B a^{3} \sqrt{b}{\left | b \right |} - 6 \, A a^{2} b^{\frac{3}{2}}{\left | b \right |}\right )} \log \left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{16 \, b^{6}} + \frac{4 \,{\left (B a^{4} \sqrt{b}{\left | b \right |} - A a^{3} b^{\frac{3}{2}}{\left | b \right |}\right )}}{{\left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

1/24*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)*B*abs(b)/b^6 - (19*B*a*b^17*abs(b) - 6*A*
b^18*abs(b))/b^23) + 3*(29*B*a^2*b^17*abs(b) - 18*A*a*b^18*abs(b))/b^23) + 5/16*(7*B*a^3*sqrt(b)*abs(b) - 6*A*
a^2*b^(3/2)*abs(b))*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^6 + 4*(B*a^4*sqrt(b)*abs(b) - A
*a^3*b^(3/2)*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)*b^5)