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

Optimal. Leaf size=159 \[ \frac{5 a^2 \sqrt{x} \sqrt{a+b x} (8 A b-7 a B)}{64 b^4}-\frac{5 a^3 (8 A b-7 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^{9/2}}+\frac{x^{5/2} \sqrt{a+b x} (8 A b-7 a B)}{24 b^2}-\frac{5 a x^{3/2} \sqrt{a+b x} (8 A b-7 a B)}{96 b^3}+\frac{B x^{7/2} \sqrt{a+b x}}{4 b} \]

[Out]

(5*a^2*(8*A*b - 7*a*B)*Sqrt[x]*Sqrt[a + b*x])/(64*b^4) - (5*a*(8*A*b - 7*a*B)*x^(3/2)*Sqrt[a + b*x])/(96*b^3)
+ ((8*A*b - 7*a*B)*x^(5/2)*Sqrt[a + b*x])/(24*b^2) + (B*x^(7/2)*Sqrt[a + b*x])/(4*b) - (5*a^3*(8*A*b - 7*a*B)*
ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*b^(9/2))

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Rubi [A]  time = 0.0677438, antiderivative size = 159, 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 = {80, 50, 63, 217, 206} \[ \frac{5 a^2 \sqrt{x} \sqrt{a+b x} (8 A b-7 a B)}{64 b^4}-\frac{5 a^3 (8 A b-7 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^{9/2}}+\frac{x^{5/2} \sqrt{a+b x} (8 A b-7 a B)}{24 b^2}-\frac{5 a x^{3/2} \sqrt{a+b x} (8 A b-7 a B)}{96 b^3}+\frac{B x^{7/2} \sqrt{a+b x}}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^2*(8*A*b - 7*a*B)*Sqrt[x]*Sqrt[a + b*x])/(64*b^4) - (5*a*(8*A*b - 7*a*B)*x^(3/2)*Sqrt[a + b*x])/(96*b^3)
+ ((8*A*b - 7*a*B)*x^(5/2)*Sqrt[a + b*x])/(24*b^2) + (B*x^(7/2)*Sqrt[a + b*x])/(4*b) - (5*a^3*(8*A*b - 7*a*B)*
ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*b^(9/2))

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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)}{\sqrt{a+b x}} \, dx &=\frac{B x^{7/2} \sqrt{a+b x}}{4 b}+\frac{\left (4 A b-\frac{7 a B}{2}\right ) \int \frac{x^{5/2}}{\sqrt{a+b x}} \, dx}{4 b}\\ &=\frac{(8 A b-7 a B) x^{5/2} \sqrt{a+b x}}{24 b^2}+\frac{B x^{7/2} \sqrt{a+b x}}{4 b}-\frac{(5 a (8 A b-7 a B)) \int \frac{x^{3/2}}{\sqrt{a+b x}} \, dx}{48 b^2}\\ &=-\frac{5 a (8 A b-7 a B) x^{3/2} \sqrt{a+b x}}{96 b^3}+\frac{(8 A b-7 a B) x^{5/2} \sqrt{a+b x}}{24 b^2}+\frac{B x^{7/2} \sqrt{a+b x}}{4 b}+\frac{\left (5 a^2 (8 A b-7 a B)\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x}} \, dx}{64 b^3}\\ &=\frac{5 a^2 (8 A b-7 a B) \sqrt{x} \sqrt{a+b x}}{64 b^4}-\frac{5 a (8 A b-7 a B) x^{3/2} \sqrt{a+b x}}{96 b^3}+\frac{(8 A b-7 a B) x^{5/2} \sqrt{a+b x}}{24 b^2}+\frac{B x^{7/2} \sqrt{a+b x}}{4 b}-\frac{\left (5 a^3 (8 A b-7 a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{128 b^4}\\ &=\frac{5 a^2 (8 A b-7 a B) \sqrt{x} \sqrt{a+b x}}{64 b^4}-\frac{5 a (8 A b-7 a B) x^{3/2} \sqrt{a+b x}}{96 b^3}+\frac{(8 A b-7 a B) x^{5/2} \sqrt{a+b x}}{24 b^2}+\frac{B x^{7/2} \sqrt{a+b x}}{4 b}-\frac{\left (5 a^3 (8 A b-7 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{64 b^4}\\ &=\frac{5 a^2 (8 A b-7 a B) \sqrt{x} \sqrt{a+b x}}{64 b^4}-\frac{5 a (8 A b-7 a B) x^{3/2} \sqrt{a+b x}}{96 b^3}+\frac{(8 A b-7 a B) x^{5/2} \sqrt{a+b x}}{24 b^2}+\frac{B x^{7/2} \sqrt{a+b x}}{4 b}-\frac{\left (5 a^3 (8 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 )}{64 b^4}\\ &=\frac{5 a^2 (8 A b-7 a B) \sqrt{x} \sqrt{a+b x}}{64 b^4}-\frac{5 a (8 A b-7 a B) x^{3/2} \sqrt{a+b x}}{96 b^3}+\frac{(8 A b-7 a B) x^{5/2} \sqrt{a+b x}}{24 b^2}+\frac{B x^{7/2} \sqrt{a+b x}}{4 b}-\frac{5 a^3 (8 A b-7 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.26734, size = 122, normalized size = 0.77 \[ \frac{\sqrt{a+b x} \left (\frac{(8 A b-7 a B) \left (b x \sqrt{\frac{b x}{a}+1} \left (15 a^2-10 a b x+8 b^2 x^2\right )-15 a^{5/2} \sqrt{b} \sqrt{x} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )\right )}{\sqrt{\frac{b x}{a}+1}}+48 b^4 B x^4\right )}{192 b^5 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(48*b^4*B*x^4 + ((8*A*b - 7*a*B)*(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]]))/Sqrt[1 + (b*x)/a]))/(192*b^5*Sqrt[x])

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Maple [A]  time = 0.011, size = 218, normalized size = 1.4 \begin{align*} -{\frac{1}{384}\sqrt{x}\sqrt{bx+a} \left ( -96\,B{x}^{3}{b}^{7/2}\sqrt{x \left ( bx+a \right ) }-128\,A{x}^{2}{b}^{7/2}\sqrt{x \left ( bx+a \right ) }+112\,B{x}^{2}a{b}^{5/2}\sqrt{x \left ( bx+a \right ) }+160\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}xa-140\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}x{a}^{2}+120\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{3}b-240\,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 ){b}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)/(b*x+a)^(1/2),x)

[Out]

-1/384*x^(1/2)*(b*x+a)^(1/2)/b^(9/2)*(-96*B*x^3*b^(7/2)*(x*(b*x+a))^(1/2)-128*A*x^2*b^(7/2)*(x*(b*x+a))^(1/2)+
112*B*x^2*a*b^(5/2)*(x*(b*x+a))^(1/2)+160*A*(x*(b*x+a))^(1/2)*b^(5/2)*x*a-140*B*(x*(b*x+a))^(1/2)*b^(3/2)*x*a^
2+120*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^3*b-240*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)/(x*(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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.7944, size = 617, normalized size = 3.88 \begin{align*} \left [-\frac{15 \,{\left (7 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt{b} \log \left (2 \, b x - 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (48 \, B b^{4} x^{3} - 105 \, B a^{3} b + 120 \, A a^{2} b^{2} - 8 \,{\left (7 \, B a b^{3} - 8 \, A b^{4}\right )} x^{2} + 10 \,{\left (7 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{384 \, b^{5}}, -\frac{15 \,{\left (7 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (48 \, B b^{4} x^{3} - 105 \, B a^{3} b + 120 \, A a^{2} b^{2} - 8 \,{\left (7 \, B a b^{3} - 8 \, A b^{4}\right )} x^{2} + 10 \,{\left (7 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{192 \, b^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(7*B*a^4 - 8*A*a^3*b)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(48*B*b^4*x^3 -
105*B*a^3*b + 120*A*a^2*b^2 - 8*(7*B*a*b^3 - 8*A*b^4)*x^2 + 10*(7*B*a^2*b^2 - 8*A*a*b^3)*x)*sqrt(b*x + a)*sqr
t(x))/b^5, -1/192*(15*(7*B*a^4 - 8*A*a^3*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (48*B*b^4*x^
3 - 105*B*a^3*b + 120*A*a^2*b^2 - 8*(7*B*a*b^3 - 8*A*b^4)*x^2 + 10*(7*B*a^2*b^2 - 8*A*a*b^3)*x)*sqrt(b*x + a)*
sqrt(x))/b^5]

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Sympy [A]  time = 117.504, size = 303, normalized size = 1.91 \begin{align*} \frac{5 A a^{\frac{5}{2}} \sqrt{x}}{8 b^{3} \sqrt{1 + \frac{b x}{a}}} + \frac{5 A a^{\frac{3}{2}} x^{\frac{3}{2}}}{24 b^{2} \sqrt{1 + \frac{b x}{a}}} - \frac{A \sqrt{a} x^{\frac{5}{2}}}{12 b \sqrt{1 + \frac{b x}{a}}} - \frac{5 A a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{8 b^{\frac{7}{2}}} + \frac{A x^{\frac{7}{2}}}{3 \sqrt{a} \sqrt{1 + \frac{b x}{a}}} - \frac{35 B a^{\frac{7}{2}} \sqrt{x}}{64 b^{4} \sqrt{1 + \frac{b x}{a}}} - \frac{35 B a^{\frac{5}{2}} x^{\frac{3}{2}}}{192 b^{3} \sqrt{1 + \frac{b x}{a}}} + \frac{7 B a^{\frac{3}{2}} x^{\frac{5}{2}}}{96 b^{2} \sqrt{1 + \frac{b x}{a}}} - \frac{B \sqrt{a} x^{\frac{7}{2}}}{24 b \sqrt{1 + \frac{b x}{a}}} + \frac{35 B a^{4} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{64 b^{\frac{9}{2}}} + \frac{B x^{\frac{9}{2}}}{4 \sqrt{a} \sqrt{1 + \frac{b x}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x+A)/(b*x+a)**(1/2),x)

[Out]

5*A*a**(5/2)*sqrt(x)/(8*b**3*sqrt(1 + b*x/a)) + 5*A*a**(3/2)*x**(3/2)/(24*b**2*sqrt(1 + b*x/a)) - A*sqrt(a)*x*
*(5/2)/(12*b*sqrt(1 + b*x/a)) - 5*A*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(7/2)) + A*x**(7/2)/(3*sqrt(a)*s
qrt(1 + b*x/a)) - 35*B*a**(7/2)*sqrt(x)/(64*b**4*sqrt(1 + b*x/a)) - 35*B*a**(5/2)*x**(3/2)/(192*b**3*sqrt(1 +
b*x/a)) + 7*B*a**(3/2)*x**(5/2)/(96*b**2*sqrt(1 + b*x/a)) - B*sqrt(a)*x**(7/2)/(24*b*sqrt(1 + b*x/a)) + 35*B*a
**4*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(64*b**(9/2)) + B*x**(9/2)/(4*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(x^(5/2)*(B*x+A)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

Timed out