\(\int (a+b \sqrt {x})^{10} x^2 \, dx\) [2155]

Optimal result

Integrand size = 15, antiderivative size = 122 \[ \int \left (a+b \sqrt {x}\right )^{10} x^2 \, dx=-\frac {2 a^5 \left (a+b \sqrt {x}\right )^{11}}{11 b^6}+\frac {5 a^4 \left (a+b \sqrt {x}\right )^{12}}{6 b^6}-\frac {20 a^3 \left (a+b \sqrt {x}\right )^{13}}{13 b^6}+\frac {10 a^2 \left (a+b \sqrt {x}\right )^{14}}{7 b^6}-\frac {2 a \left (a+b \sqrt {x}\right )^{15}}{3 b^6}+\frac {\left (a+b \sqrt {x}\right )^{16}}{8 b^6} \]

[Out]

-2/11*a^5*(a+b*x^(1/2))^11/b^6+5/6*a^4*(a+b*x^(1/2))^12/b^6-20/13*a^3*(a+b*x^(1/2))^13/b^6+10/7*a^2*(a+b*x^(1/
2))^14/b^6-2/3*a*(a+b*x^(1/2))^15/b^6+1/8*(a+b*x^(1/2))^16/b^6

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \left (a+b \sqrt {x}\right )^{10} x^2 \, dx=-\frac {2 a^5 \left (a+b \sqrt {x}\right )^{11}}{11 b^6}+\frac {5 a^4 \left (a+b \sqrt {x}\right )^{12}}{6 b^6}-\frac {20 a^3 \left (a+b \sqrt {x}\right )^{13}}{13 b^6}+\frac {10 a^2 \left (a+b \sqrt {x}\right )^{14}}{7 b^6}+\frac {\left (a+b \sqrt {x}\right )^{16}}{8 b^6}-\frac {2 a \left (a+b \sqrt {x}\right )^{15}}{3 b^6} \]

[In]

Int[(a + b*Sqrt[x])^10*x^2,x]

[Out]

(-2*a^5*(a + b*Sqrt[x])^11)/(11*b^6) + (5*a^4*(a + b*Sqrt[x])^12)/(6*b^6) - (20*a^3*(a + b*Sqrt[x])^13)/(13*b^
6) + (10*a^2*(a + b*Sqrt[x])^14)/(7*b^6) - (2*a*(a + b*Sqrt[x])^15)/(3*b^6) + (a + b*Sqrt[x])^16/(8*b^6)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x^5 (a+b x)^{10} \, dx,x,\sqrt {x}\right ) \\ & = 2 \text {Subst}\left (\int \left (-\frac {a^5 (a+b x)^{10}}{b^5}+\frac {5 a^4 (a+b x)^{11}}{b^5}-\frac {10 a^3 (a+b x)^{12}}{b^5}+\frac {10 a^2 (a+b x)^{13}}{b^5}-\frac {5 a (a+b x)^{14}}{b^5}+\frac {(a+b x)^{15}}{b^5}\right ) \, dx,x,\sqrt {x}\right ) \\ & = -\frac {2 a^5 \left (a+b \sqrt {x}\right )^{11}}{11 b^6}+\frac {5 a^4 \left (a+b \sqrt {x}\right )^{12}}{6 b^6}-\frac {20 a^3 \left (a+b \sqrt {x}\right )^{13}}{13 b^6}+\frac {10 a^2 \left (a+b \sqrt {x}\right )^{14}}{7 b^6}-\frac {2 a \left (a+b \sqrt {x}\right )^{15}}{3 b^6}+\frac {\left (a+b \sqrt {x}\right )^{16}}{8 b^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.03 \[ \int \left (a+b \sqrt {x}\right )^{10} x^2 \, dx=\frac {8008 a^{10} x^3+68640 a^9 b x^{7/2}+270270 a^8 b^2 x^4+640640 a^7 b^3 x^{9/2}+1009008 a^6 b^4 x^5+1100736 a^5 b^5 x^{11/2}+840840 a^4 b^6 x^6+443520 a^3 b^7 x^{13/2}+154440 a^2 b^8 x^7+32032 a b^9 x^{15/2}+3003 b^{10} x^8}{24024} \]

[In]

Integrate[(a + b*Sqrt[x])^10*x^2,x]

[Out]

(8008*a^10*x^3 + 68640*a^9*b*x^(7/2) + 270270*a^8*b^2*x^4 + 640640*a^7*b^3*x^(9/2) + 1009008*a^6*b^4*x^5 + 110
0736*a^5*b^5*x^(11/2) + 840840*a^4*b^6*x^6 + 443520*a^3*b^7*x^(13/2) + 154440*a^2*b^8*x^7 + 32032*a*b^9*x^(15/
2) + 3003*b^10*x^8)/24024

Maple [A] (verified)

Time = 3.47 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {b^{10} x^{8}}{8}+\frac {4 a \,b^{9} x^{\frac {15}{2}}}{3}+\frac {45 a^{2} b^{8} x^{7}}{7}+\frac {240 x^{\frac {13}{2}} a^{3} b^{7}}{13}+35 a^{4} b^{6} x^{6}+\frac {504 x^{\frac {11}{2}} a^{5} b^{5}}{11}+42 a^{6} b^{4} x^{5}+\frac {80 a^{7} b^{3} x^{\frac {9}{2}}}{3}+\frac {45 a^{8} b^{2} x^{4}}{4}+\frac {20 a^{9} b \,x^{\frac {7}{2}}}{7}+\frac {a^{10} x^{3}}{3}\)	\(113\)
default	\(\frac {b^{10} x^{8}}{8}+\frac {4 a \,b^{9} x^{\frac {15}{2}}}{3}+\frac {45 a^{2} b^{8} x^{7}}{7}+\frac {240 x^{\frac {13}{2}} a^{3} b^{7}}{13}+35 a^{4} b^{6} x^{6}+\frac {504 x^{\frac {11}{2}} a^{5} b^{5}}{11}+42 a^{6} b^{4} x^{5}+\frac {80 a^{7} b^{3} x^{\frac {9}{2}}}{3}+\frac {45 a^{8} b^{2} x^{4}}{4}+\frac {20 a^{9} b \,x^{\frac {7}{2}}}{7}+\frac {a^{10} x^{3}}{3}\)	\(113\)
trager	\(\frac {\left (21 b^{10} x^{7}+1080 a^{2} b^{8} x^{6}+21 b^{10} x^{6}+5880 a^{4} b^{6} x^{5}+1080 a^{2} b^{8} x^{5}+21 b^{10} x^{5}+7056 a^{6} b^{4} x^{4}+5880 x^{4} a^{4} b^{6}+1080 a^{2} b^{8} x^{4}+21 b^{10} x^{4}+1890 a^{8} b^{2} x^{3}+7056 a^{6} b^{4} x^{3}+5880 a^{4} b^{6} x^{3}+1080 a^{2} b^{8} x^{3}+21 b^{10} x^{3}+56 a^{10} x^{2}+1890 a^{8} b^{2} x^{2}+7056 x^{2} a^{6} b^{4}+5880 a^{4} b^{6} x^{2}+1080 b^{8} x^{2} a^{2}+21 x^{2} b^{10}+56 a^{10} x +1890 a^{8} b^{2} x +7056 a^{6} b^{4} x +5880 a^{4} b^{6} x +1080 b^{8} x \,a^{2}+21 b^{10} x +56 a^{10}+1890 a^{8} b^{2}+7056 a^{6} b^{4}+5880 a^{4} b^{6}+1080 a^{2} b^{8}+21 b^{10}\right ) \left (-1+x \right )}{168}+\frac {4 a b \,x^{\frac {7}{2}} \left (1001 b^{8} x^{4}+13860 a^{2} b^{6} x^{3}+34398 a^{4} b^{4} x^{2}+20020 a^{6} b^{2} x +2145 a^{8}\right )}{3003}\)	\(360\)

[In]

int(x^2*(a+b*x^(1/2))^10,x,method=_RETURNVERBOSE)

[Out]

1/8*b^10*x^8+4/3*a*b^9*x^(15/2)+45/7*a^2*b^8*x^7+240/13*x^(13/2)*a^3*b^7+35*a^4*b^6*x^6+504/11*x^(11/2)*a^5*b^
5+42*a^6*b^4*x^5+80/3*a^7*b^3*x^(9/2)+45/4*a^8*b^2*x^4+20/7*a^9*b*x^(7/2)+1/3*a^10*x^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.97 \[ \int \left (a+b \sqrt {x}\right )^{10} x^2 \, dx=\frac {1}{8} \, b^{10} x^{8} + \frac {45}{7} \, a^{2} b^{8} x^{7} + 35 \, a^{4} b^{6} x^{6} + 42 \, a^{6} b^{4} x^{5} + \frac {45}{4} \, a^{8} b^{2} x^{4} + \frac {1}{3} \, a^{10} x^{3} + \frac {4}{3003} \, {\left (1001 \, a b^{9} x^{7} + 13860 \, a^{3} b^{7} x^{6} + 34398 \, a^{5} b^{5} x^{5} + 20020 \, a^{7} b^{3} x^{4} + 2145 \, a^{9} b x^{3}\right )} \sqrt {x} \]

[In]

integrate(x^2*(a+b*x^(1/2))^10,x, algorithm="fricas")

[Out]

1/8*b^10*x^8 + 45/7*a^2*b^8*x^7 + 35*a^4*b^6*x^6 + 42*a^6*b^4*x^5 + 45/4*a^8*b^2*x^4 + 1/3*a^10*x^3 + 4/3003*(
1001*a*b^9*x^7 + 13860*a^3*b^7*x^6 + 34398*a^5*b^5*x^5 + 20020*a^7*b^3*x^4 + 2145*a^9*b*x^3)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.14 \[ \int \left (a+b \sqrt {x}\right )^{10} x^2 \, dx=\frac {a^{10} x^{3}}{3} + \frac {20 a^{9} b x^{\frac {7}{2}}}{7} + \frac {45 a^{8} b^{2} x^{4}}{4} + \frac {80 a^{7} b^{3} x^{\frac {9}{2}}}{3} + 42 a^{6} b^{4} x^{5} + \frac {504 a^{5} b^{5} x^{\frac {11}{2}}}{11} + 35 a^{4} b^{6} x^{6} + \frac {240 a^{3} b^{7} x^{\frac {13}{2}}}{13} + \frac {45 a^{2} b^{8} x^{7}}{7} + \frac {4 a b^{9} x^{\frac {15}{2}}}{3} + \frac {b^{10} x^{8}}{8} \]

[In]

integrate(x**2*(a+b*x**(1/2))**10,x)

[Out]

a**10*x**3/3 + 20*a**9*b*x**(7/2)/7 + 45*a**8*b**2*x**4/4 + 80*a**7*b**3*x**(9/2)/3 + 42*a**6*b**4*x**5 + 504*
a**5*b**5*x**(11/2)/11 + 35*a**4*b**6*x**6 + 240*a**3*b**7*x**(13/2)/13 + 45*a**2*b**8*x**7/7 + 4*a*b**9*x**(1
5/2)/3 + b**10*x**8/8

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int \left (a+b \sqrt {x}\right )^{10} x^2 \, dx=\frac {{\left (b \sqrt {x} + a\right )}^{16}}{8 \, b^{6}} - \frac {2 \, {\left (b \sqrt {x} + a\right )}^{15} a}{3 \, b^{6}} + \frac {10 \, {\left (b \sqrt {x} + a\right )}^{14} a^{2}}{7 \, b^{6}} - \frac {20 \, {\left (b \sqrt {x} + a\right )}^{13} a^{3}}{13 \, b^{6}} + \frac {5 \, {\left (b \sqrt {x} + a\right )}^{12} a^{4}}{6 \, b^{6}} - \frac {2 \, {\left (b \sqrt {x} + a\right )}^{11} a^{5}}{11 \, b^{6}} \]

[In]

integrate(x^2*(a+b*x^(1/2))^10,x, algorithm="maxima")

[Out]

1/8*(b*sqrt(x) + a)^16/b^6 - 2/3*(b*sqrt(x) + a)^15*a/b^6 + 10/7*(b*sqrt(x) + a)^14*a^2/b^6 - 20/13*(b*sqrt(x)
+ a)^13*a^3/b^6 + 5/6*(b*sqrt(x) + a)^12*a^4/b^6 - 2/11*(b*sqrt(x) + a)^11*a^5/b^6

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.92 \[ \int \left (a+b \sqrt {x}\right )^{10} x^2 \, dx=\frac {1}{8} \, b^{10} x^{8} + \frac {4}{3} \, a b^{9} x^{\frac {15}{2}} + \frac {45}{7} \, a^{2} b^{8} x^{7} + \frac {240}{13} \, a^{3} b^{7} x^{\frac {13}{2}} + 35 \, a^{4} b^{6} x^{6} + \frac {504}{11} \, a^{5} b^{5} x^{\frac {11}{2}} + 42 \, a^{6} b^{4} x^{5} + \frac {80}{3} \, a^{7} b^{3} x^{\frac {9}{2}} + \frac {45}{4} \, a^{8} b^{2} x^{4} + \frac {20}{7} \, a^{9} b x^{\frac {7}{2}} + \frac {1}{3} \, a^{10} x^{3} \]

[In]

integrate(x^2*(a+b*x^(1/2))^10,x, algorithm="giac")

[Out]

1/8*b^10*x^8 + 4/3*a*b^9*x^(15/2) + 45/7*a^2*b^8*x^7 + 240/13*a^3*b^7*x^(13/2) + 35*a^4*b^6*x^6 + 504/11*a^5*b
^5*x^(11/2) + 42*a^6*b^4*x^5 + 80/3*a^7*b^3*x^(9/2) + 45/4*a^8*b^2*x^4 + 20/7*a^9*b*x^(7/2) + 1/3*a^10*x^3

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.92 \[ \int \left (a+b \sqrt {x}\right )^{10} x^2 \, dx=\frac {a^{10}\,x^3}{3}+\frac {b^{10}\,x^8}{8}+\frac {20\,a^9\,b\,x^{7/2}}{7}+\frac {4\,a\,b^9\,x^{15/2}}{3}+\frac {45\,a^8\,b^2\,x^4}{4}+42\,a^6\,b^4\,x^5+35\,a^4\,b^6\,x^6+\frac {45\,a^2\,b^8\,x^7}{7}+\frac {80\,a^7\,b^3\,x^{9/2}}{3}+\frac {504\,a^5\,b^5\,x^{11/2}}{11}+\frac {240\,a^3\,b^7\,x^{13/2}}{13} \]

[In]

int(x^2*(a + b*x^(1/2))^10,x)

[Out]

(a^10*x^3)/3 + (b^10*x^8)/8 + (20*a^9*b*x^(7/2))/7 + (4*a*b^9*x^(15/2))/3 + (45*a^8*b^2*x^4)/4 + 42*a^6*b^4*x^
5 + 35*a^4*b^6*x^6 + (45*a^2*b^8*x^7)/7 + (80*a^7*b^3*x^(9/2))/3 + (504*a^5*b^5*x^(11/2))/11 + (240*a^3*b^7*x^
(13/2))/13