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

Optimal. Leaf size=192 \[ \frac{a^3 \sqrt{x} \sqrt{a+b x} (10 A b-3 a B)}{128 b^2}-\frac{a^4 (10 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{128 b^{5/2}}+\frac{a^2 x^{3/2} \sqrt{a+b x} (10 A b-3 a B)}{64 b}+\frac{a x^{3/2} (a+b x)^{3/2} (10 A b-3 a B)}{48 b}+\frac{x^{3/2} (a+b x)^{5/2} (10 A b-3 a B)}{40 b}+\frac{B x^{3/2} (a+b x)^{7/2}}{5 b} \]

[Out]

(a^3*(10*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(128*b^2) + (a^2*(10*A*b - 3*a*B)*x^(3/2)*Sqrt[a + b*x])/(64*b) +
(a*(10*A*b - 3*a*B)*x^(3/2)*(a + b*x)^(3/2))/(48*b) + ((10*A*b - 3*a*B)*x^(3/2)*(a + b*x)^(5/2))/(40*b) + (B*
x^(3/2)*(a + b*x)^(7/2))/(5*b) - (a^4*(10*A*b - 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(128*b^(5/2))

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Rubi [A]  time = 0.0823472, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{a^3 \sqrt{x} \sqrt{a+b x} (10 A b-3 a B)}{128 b^2}-\frac{a^4 (10 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{128 b^{5/2}}+\frac{a^2 x^{3/2} \sqrt{a+b x} (10 A b-3 a B)}{64 b}+\frac{a x^{3/2} (a+b x)^{3/2} (10 A b-3 a B)}{48 b}+\frac{x^{3/2} (a+b x)^{5/2} (10 A b-3 a B)}{40 b}+\frac{B x^{3/2} (a+b x)^{7/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(10*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(128*b^2) + (a^2*(10*A*b - 3*a*B)*x^(3/2)*Sqrt[a + b*x])/(64*b) +
(a*(10*A*b - 3*a*B)*x^(3/2)*(a + b*x)^(3/2))/(48*b) + ((10*A*b - 3*a*B)*x^(3/2)*(a + b*x)^(5/2))/(40*b) + (B*
x^(3/2)*(a + b*x)^(7/2))/(5*b) - (a^4*(10*A*b - 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(128*b^(5/2))

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 \sqrt{x} (a+b x)^{5/2} (A+B x) \, dx &=\frac{B x^{3/2} (a+b x)^{7/2}}{5 b}+\frac{\left (5 A b-\frac{3 a B}{2}\right ) \int \sqrt{x} (a+b x)^{5/2} \, dx}{5 b}\\ &=\frac{(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac{B x^{3/2} (a+b x)^{7/2}}{5 b}+\frac{(a (10 A b-3 a B)) \int \sqrt{x} (a+b x)^{3/2} \, dx}{16 b}\\ &=\frac{a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac{(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac{B x^{3/2} (a+b x)^{7/2}}{5 b}+\frac{\left (a^2 (10 A b-3 a B)\right ) \int \sqrt{x} \sqrt{a+b x} \, dx}{32 b}\\ &=\frac{a^2 (10 A b-3 a B) x^{3/2} \sqrt{a+b x}}{64 b}+\frac{a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac{(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac{B x^{3/2} (a+b x)^{7/2}}{5 b}+\frac{\left (a^3 (10 A b-3 a B)\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x}} \, dx}{128 b}\\ &=\frac{a^3 (10 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{128 b^2}+\frac{a^2 (10 A b-3 a B) x^{3/2} \sqrt{a+b x}}{64 b}+\frac{a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac{(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac{B x^{3/2} (a+b x)^{7/2}}{5 b}-\frac{\left (a^4 (10 A b-3 a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{256 b^2}\\ &=\frac{a^3 (10 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{128 b^2}+\frac{a^2 (10 A b-3 a B) x^{3/2} \sqrt{a+b x}}{64 b}+\frac{a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac{(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac{B x^{3/2} (a+b x)^{7/2}}{5 b}-\frac{\left (a^4 (10 A b-3 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{128 b^2}\\ &=\frac{a^3 (10 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{128 b^2}+\frac{a^2 (10 A b-3 a B) x^{3/2} \sqrt{a+b x}}{64 b}+\frac{a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac{(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac{B x^{3/2} (a+b x)^{7/2}}{5 b}-\frac{\left (a^4 (10 A b-3 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )}{128 b^2}\\ &=\frac{a^3 (10 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{128 b^2}+\frac{a^2 (10 A b-3 a B) x^{3/2} \sqrt{a+b x}}{64 b}+\frac{a (10 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{48 b}+\frac{(10 A b-3 a B) x^{3/2} (a+b x)^{5/2}}{40 b}+\frac{B x^{3/2} (a+b x)^{7/2}}{5 b}-\frac{a^4 (10 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{128 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.255093, size = 145, normalized size = 0.76 \[ \frac{\sqrt{a+b x} \left (\sqrt{b} \sqrt{x} \left (4 a^2 b^2 x (295 A+186 B x)+30 a^3 b (5 A+B x)-45 a^4 B+16 a b^3 x^2 (85 A+63 B x)+96 b^4 x^3 (5 A+4 B x)\right )+\frac{15 a^{7/2} (3 a B-10 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{1920 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(-45*a^4*B + 30*a^3*b*(5*A + B*x) + 96*b^4*x^3*(5*A + 4*B*x) + 16*a*b^3*x^2*(8
5*A + 63*B*x) + 4*a^2*b^2*x*(295*A + 186*B*x)) + (15*a^(7/2)*(-10*A*b + 3*a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[
a]])/Sqrt[1 + (b*x)/a]))/(1920*b^(5/2))

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Maple [A]  time = 0.008, size = 260, normalized size = 1.4 \begin{align*} -{\frac{1}{3840}\sqrt{bx+a}\sqrt{x} \left ( -768\,B{x}^{4}{b}^{9/2}\sqrt{x \left ( bx+a \right ) }-960\,A{x}^{3}{b}^{9/2}\sqrt{x \left ( bx+a \right ) }-2016\,B{x}^{3}a{b}^{7/2}\sqrt{x \left ( bx+a \right ) }-2720\,A{x}^{2}a{b}^{7/2}\sqrt{x \left ( bx+a \right ) }-1488\,B{x}^{2}{a}^{2}{b}^{5/2}\sqrt{x \left ( bx+a \right ) }-2360\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}x{a}^{2}-60\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}x{a}^{3}+150\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{4}b-300\,A\sqrt{x \left ( bx+a \right ) }{b}^{3/2}{a}^{3}-45\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{5}+90\,B\sqrt{x \left ( bx+a \right ) }\sqrt{b}{a}^{4} \right ){b}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3840*(b*x+a)^(1/2)*x^(1/2)/b^(5/2)*(-768*B*x^4*b^(9/2)*(x*(b*x+a))^(1/2)-960*A*x^3*b^(9/2)*(x*(b*x+a))^(1/2
)-2016*B*x^3*a*b^(7/2)*(x*(b*x+a))^(1/2)-2720*A*x^2*a*b^(7/2)*(x*(b*x+a))^(1/2)-1488*B*x^2*a^2*b^(5/2)*(x*(b*x
+a))^(1/2)-2360*A*(x*(b*x+a))^(1/2)*b^(5/2)*x*a^2-60*B*(x*(b*x+a))^(1/2)*b^(3/2)*x*a^3+150*A*ln(1/2*(2*(x*(b*x
+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^4*b-300*A*(x*(b*x+a))^(1/2)*b^(3/2)*a^3-45*B*ln(1/2*(2*(x*(b*x+a))^(1/2
)*b^(1/2)+2*b*x+a)/b^(1/2))*a^5+90*B*(x*(b*x+a))^(1/2)*b^(1/2)*a^4)/(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((b*x+a)^(5/2)*(B*x+A)*x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.64408, size = 741, normalized size = 3.86 \begin{align*} \left [-\frac{15 \,{\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt{b} \log \left (2 \, b x - 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (384 \, B b^{5} x^{4} - 45 \, B a^{4} b + 150 \, A a^{3} b^{2} + 48 \,{\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \,{\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{2} + 10 \,{\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{3840 \, b^{3}}, -\frac{15 \,{\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (384 \, B b^{5} x^{4} - 45 \, B a^{4} b + 150 \, A a^{3} b^{2} + 48 \,{\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \,{\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{2} + 10 \,{\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{1920 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3840*(15*(3*B*a^5 - 10*A*a^4*b)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(384*B*b^5*x^
4 - 45*B*a^4*b + 150*A*a^3*b^2 + 48*(21*B*a*b^4 + 10*A*b^5)*x^3 + 8*(93*B*a^2*b^3 + 170*A*a*b^4)*x^2 + 10*(3*B
*a^3*b^2 + 118*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^3, -1/1920*(15*(3*B*a^5 - 10*A*a^4*b)*sqrt(-b)*arctan(sq
rt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (384*B*b^5*x^4 - 45*B*a^4*b + 150*A*a^3*b^2 + 48*(21*B*a*b^4 + 10*A*b^5)*x
^3 + 8*(93*B*a^2*b^3 + 170*A*a*b^4)*x^2 + 10*(3*B*a^3*b^2 + 118*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^3]

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Sympy [B]  time = 84.3795, size = 359, normalized size = 1.87 \begin{align*} \frac{5 A a^{\frac{7}{2}} \sqrt{x}}{64 b \sqrt{1 + \frac{b x}{a}}} + \frac{133 A a^{\frac{5}{2}} x^{\frac{3}{2}}}{192 \sqrt{1 + \frac{b x}{a}}} + \frac{127 A a^{\frac{3}{2}} b x^{\frac{5}{2}}}{96 \sqrt{1 + \frac{b x}{a}}} + \frac{23 A \sqrt{a} b^{2} x^{\frac{7}{2}}}{24 \sqrt{1 + \frac{b x}{a}}} - \frac{5 A a^{4} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{64 b^{\frac{3}{2}}} + \frac{A b^{3} x^{\frac{9}{2}}}{4 \sqrt{a} \sqrt{1 + \frac{b x}{a}}} - \frac{3 B a^{\frac{9}{2}} \sqrt{x}}{128 b^{2} \sqrt{1 + \frac{b x}{a}}} - \frac{B a^{\frac{7}{2}} x^{\frac{3}{2}}}{128 b \sqrt{1 + \frac{b x}{a}}} + \frac{129 B a^{\frac{5}{2}} x^{\frac{5}{2}}}{320 \sqrt{1 + \frac{b x}{a}}} + \frac{73 B a^{\frac{3}{2}} b x^{\frac{7}{2}}}{80 \sqrt{1 + \frac{b x}{a}}} + \frac{29 B \sqrt{a} b^{2} x^{\frac{9}{2}}}{40 \sqrt{1 + \frac{b x}{a}}} + \frac{3 B a^{5} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{128 b^{\frac{5}{2}}} + \frac{B b^{3} x^{\frac{11}{2}}}{5 \sqrt{a} \sqrt{1 + \frac{b x}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*A*a**(7/2)*sqrt(x)/(64*b*sqrt(1 + b*x/a)) + 133*A*a**(5/2)*x**(3/2)/(192*sqrt(1 + b*x/a)) + 127*A*a**(3/2)*b
*x**(5/2)/(96*sqrt(1 + b*x/a)) + 23*A*sqrt(a)*b**2*x**(7/2)/(24*sqrt(1 + b*x/a)) - 5*A*a**4*asinh(sqrt(b)*sqrt
(x)/sqrt(a))/(64*b**(3/2)) + A*b**3*x**(9/2)/(4*sqrt(a)*sqrt(1 + b*x/a)) - 3*B*a**(9/2)*sqrt(x)/(128*b**2*sqrt
(1 + b*x/a)) - B*a**(7/2)*x**(3/2)/(128*b*sqrt(1 + b*x/a)) + 129*B*a**(5/2)*x**(5/2)/(320*sqrt(1 + b*x/a)) + 7
3*B*a**(3/2)*b*x**(7/2)/(80*sqrt(1 + b*x/a)) + 29*B*sqrt(a)*b**2*x**(9/2)/(40*sqrt(1 + b*x/a)) + 3*B*a**5*asin
h(sqrt(b)*sqrt(x)/sqrt(a))/(128*b**(5/2)) + B*b**3*x**(11/2)/(5*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^(1/2),x, algorithm="giac")

[Out]

Timed out