\(\int (d x)^m (a^2+2 a b x^3+b^2 x^6)^{3/2} \, dx\) [119]

Optimal result

Integrand size = 28, antiderivative size = 205 \[ \int (d x)^m \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {a^3 (d x)^{1+m} \sqrt {a^2+2 a b x^3+b^2 x^6}}{d (1+m) \left (a+b x^3\right )}+\frac {3 a^2 b (d x)^{4+m} \sqrt {a^2+2 a b x^3+b^2 x^6}}{d^4 (4+m) \left (a+b x^3\right )}+\frac {3 a b^2 (d x)^{7+m} \sqrt {a^2+2 a b x^3+b^2 x^6}}{d^7 (7+m) \left (a+b x^3\right )}+\frac {b^3 (d x)^{10+m} \sqrt {a^2+2 a b x^3+b^2 x^6}}{d^{10} (10+m) \left (a+b x^3\right )} \]

[Out]

a^3*(d*x)^(1+m)*((b*x^3+a)^2)^(1/2)/d/(1+m)/(b*x^3+a)+3*a^2*b*(d*x)^(4+m)*((b*x^3+a)^2)^(1/2)/d^4/(4+m)/(b*x^3
+a)+3*a*b^2*(d*x)^(7+m)*((b*x^3+a)^2)^(1/2)/d^7/(7+m)/(b*x^3+a)+b^3*(d*x)^(10+m)*((b*x^3+a)^2)^(1/2)/d^10/(10+
m)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1369, 276} \[ \int (d x)^m \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {3 a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6} (d x)^{m+7}}{d^7 (m+7) \left (a+b x^3\right )}+\frac {3 a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6} (d x)^{m+4}}{d^4 (m+4) \left (a+b x^3\right )}+\frac {b^3 \sqrt {a^2+2 a b x^3+b^2 x^6} (d x)^{m+10}}{d^{10} (m+10) \left (a+b x^3\right )}+\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6} (d x)^{m+1}}{d (m+1) \left (a+b x^3\right )} \]

[In]

Int[(d*x)^m*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(a^3*(d*x)^(1 + m)*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(d*(1 + m)*(a + b*x^3)) + (3*a^2*b*(d*x)^(4 + m)*Sqrt[a^2
+ 2*a*b*x^3 + b^2*x^6])/(d^4*(4 + m)*(a + b*x^3)) + (3*a*b^2*(d*x)^(7 + m)*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(d
^7*(7 + m)*(a + b*x^3)) + (b^3*(d*x)^(10 + m)*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(d^10*(10 + m)*(a + b*x^3))

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int (d x)^m \left (a b+b^2 x^3\right )^3 \, dx}{b^2 \left (a b+b^2 x^3\right )} \\ & = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (a^3 b^3 (d x)^m+\frac {3 a^2 b^4 (d x)^{3+m}}{d^3}+\frac {3 a b^5 (d x)^{6+m}}{d^6}+\frac {b^6 (d x)^{9+m}}{d^9}\right ) \, dx}{b^2 \left (a b+b^2 x^3\right )} \\ & = \frac {a^3 (d x)^{1+m} \sqrt {a^2+2 a b x^3+b^2 x^6}}{d (1+m) \left (a+b x^3\right )}+\frac {3 a^2 b (d x)^{4+m} \sqrt {a^2+2 a b x^3+b^2 x^6}}{d^4 (4+m) \left (a+b x^3\right )}+\frac {3 a b^2 (d x)^{7+m} \sqrt {a^2+2 a b x^3+b^2 x^6}}{d^7 (7+m) \left (a+b x^3\right )}+\frac {b^3 (d x)^{10+m} \sqrt {a^2+2 a b x^3+b^2 x^6}}{d^{10} (10+m) \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.64 \[ \int (d x)^m \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {x (d x)^m \sqrt {\left (a+b x^3\right )^2} \left (a^3 \left (280+138 m+21 m^2+m^3\right )+3 a^2 b \left (70+87 m+18 m^2+m^3\right ) x^3+3 a b^2 \left (40+54 m+15 m^2+m^3\right ) x^6+b^3 \left (28+39 m+12 m^2+m^3\right ) x^9\right )}{(1+m) (4+m) (7+m) (10+m) \left (a+b x^3\right )} \]

[In]

Integrate[(d*x)^m*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(x*(d*x)^m*Sqrt[(a + b*x^3)^2]*(a^3*(280 + 138*m + 21*m^2 + m^3) + 3*a^2*b*(70 + 87*m + 18*m^2 + m^3)*x^3 + 3*
a*b^2*(40 + 54*m + 15*m^2 + m^3)*x^6 + b^3*(28 + 39*m + 12*m^2 + m^3)*x^9))/((1 + m)*(4 + m)*(7 + m)*(10 + m)*
(a + b*x^3))

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.97

method	result	size
gosper	\(\frac {x \left (b^{3} m^{3} x^{9}+12 b^{3} m^{2} x^{9}+39 m \,x^{9} b^{3}+3 a \,b^{2} m^{3} x^{6}+28 b^{3} x^{9}+45 a \,b^{2} m^{2} x^{6}+162 m \,x^{6} b^{2} a +3 a^{2} b \,m^{3} x^{3}+120 b^{2} x^{6} a +54 a^{2} b \,m^{2} x^{3}+261 m \,x^{3} a^{2} b +a^{3} m^{3}+210 a^{2} b \,x^{3}+21 a^{3} m^{2}+138 m \,a^{3}+280 a^{3}\right ) \left (d x \right )^{m} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{\left (10+m \right ) \left (7+m \right ) \left (4+m \right ) \left (1+m \right ) \left (b \,x^{3}+a \right )^{3}}\)	\(199\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (b^{3} m^{3} x^{9}+12 b^{3} m^{2} x^{9}+39 m \,x^{9} b^{3}+3 a \,b^{2} m^{3} x^{6}+28 b^{3} x^{9}+45 a \,b^{2} m^{2} x^{6}+162 m \,x^{6} b^{2} a +3 a^{2} b \,m^{3} x^{3}+120 b^{2} x^{6} a +54 a^{2} b \,m^{2} x^{3}+261 m \,x^{3} a^{2} b +a^{3} m^{3}+210 a^{2} b \,x^{3}+21 a^{3} m^{2}+138 m \,a^{3}+280 a^{3}\right ) x \left (d x \right )^{m}}{\left (b \,x^{3}+a \right ) \left (10+m \right ) \left (7+m \right ) \left (4+m \right ) \left (1+m \right )}\)	\(199\)

[In]

int((d*x)^m*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

x*(b^3*m^3*x^9+12*b^3*m^2*x^9+39*b^3*m*x^9+3*a*b^2*m^3*x^6+28*b^3*x^9+45*a*b^2*m^2*x^6+162*a*b^2*m*x^6+3*a^2*b
*m^3*x^3+120*a*b^2*x^6+54*a^2*b*m^2*x^3+261*a^2*b*m*x^3+a^3*m^3+210*a^2*b*x^3+21*a^3*m^2+138*a^3*m+280*a^3)*(d
*x)^m*((b*x^3+a)^2)^(3/2)/(10+m)/(7+m)/(4+m)/(1+m)/(b*x^3+a)^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.78 \[ \int (d x)^m \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {{\left ({\left (b^{3} m^{3} + 12 \, b^{3} m^{2} + 39 \, b^{3} m + 28 \, b^{3}\right )} x^{10} + 3 \, {\left (a b^{2} m^{3} + 15 \, a b^{2} m^{2} + 54 \, a b^{2} m + 40 \, a b^{2}\right )} x^{7} + 3 \, {\left (a^{2} b m^{3} + 18 \, a^{2} b m^{2} + 87 \, a^{2} b m + 70 \, a^{2} b\right )} x^{4} + {\left (a^{3} m^{3} + 21 \, a^{3} m^{2} + 138 \, a^{3} m + 280 \, a^{3}\right )} x\right )} \left (d x\right )^{m}}{m^{4} + 22 \, m^{3} + 159 \, m^{2} + 418 \, m + 280} \]

[In]

integrate((d*x)^m*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")

[Out]

((b^3*m^3 + 12*b^3*m^2 + 39*b^3*m + 28*b^3)*x^10 + 3*(a*b^2*m^3 + 15*a*b^2*m^2 + 54*a*b^2*m + 40*a*b^2)*x^7 +
3*(a^2*b*m^3 + 18*a^2*b*m^2 + 87*a^2*b*m + 70*a^2*b)*x^4 + (a^3*m^3 + 21*a^3*m^2 + 138*a^3*m + 280*a^3)*x)*(d*
x)^m/(m^4 + 22*m^3 + 159*m^2 + 418*m + 280)

Sympy [F]

\[ \int (d x)^m \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int \left (d x\right )^{m} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}\, dx \]

[In]

integrate((d*x)**m*(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)

[Out]

Integral((d*x)**m*((a + b*x**3)**2)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.58 \[ \int (d x)^m \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {{\left ({\left (m^{3} + 12 \, m^{2} + 39 \, m + 28\right )} b^{3} d^{m} x^{10} + 3 \, {\left (m^{3} + 15 \, m^{2} + 54 \, m + 40\right )} a b^{2} d^{m} x^{7} + 3 \, {\left (m^{3} + 18 \, m^{2} + 87 \, m + 70\right )} a^{2} b d^{m} x^{4} + {\left (m^{3} + 21 \, m^{2} + 138 \, m + 280\right )} a^{3} d^{m} x\right )} x^{m}}{m^{4} + 22 \, m^{3} + 159 \, m^{2} + 418 \, m + 280} \]

[In]

integrate((d*x)^m*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")

[Out]

((m^3 + 12*m^2 + 39*m + 28)*b^3*d^m*x^10 + 3*(m^3 + 15*m^2 + 54*m + 40)*a*b^2*d^m*x^7 + 3*(m^3 + 18*m^2 + 87*m
+ 70)*a^2*b*d^m*x^4 + (m^3 + 21*m^2 + 138*m + 280)*a^3*d^m*x)*x^m/(m^4 + 22*m^3 + 159*m^2 + 418*m + 280)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (161) = 322\).

Time = 0.31 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.87 \[ \int (d x)^m \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\frac {\left (d x\right )^{m} b^{3} m^{3} x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) + 12 \, \left (d x\right )^{m} b^{3} m^{2} x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) + 39 \, \left (d x\right )^{m} b^{3} m x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) + 3 \, \left (d x\right )^{m} a b^{2} m^{3} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + 28 \, \left (d x\right )^{m} b^{3} x^{10} \mathrm {sgn}\left (b x^{3} + a\right ) + 45 \, \left (d x\right )^{m} a b^{2} m^{2} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + 162 \, \left (d x\right )^{m} a b^{2} m x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + 3 \, \left (d x\right )^{m} a^{2} b m^{3} x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + 120 \, \left (d x\right )^{m} a b^{2} x^{7} \mathrm {sgn}\left (b x^{3} + a\right ) + 54 \, \left (d x\right )^{m} a^{2} b m^{2} x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + 261 \, \left (d x\right )^{m} a^{2} b m x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + \left (d x\right )^{m} a^{3} m^{3} x \mathrm {sgn}\left (b x^{3} + a\right ) + 210 \, \left (d x\right )^{m} a^{2} b x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + 21 \, \left (d x\right )^{m} a^{3} m^{2} x \mathrm {sgn}\left (b x^{3} + a\right ) + 138 \, \left (d x\right )^{m} a^{3} m x \mathrm {sgn}\left (b x^{3} + a\right ) + 280 \, \left (d x\right )^{m} a^{3} x \mathrm {sgn}\left (b x^{3} + a\right )}{m^{4} + 22 \, m^{3} + 159 \, m^{2} + 418 \, m + 280} \]

[In]

integrate((d*x)^m*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")

[Out]

((d*x)^m*b^3*m^3*x^10*sgn(b*x^3 + a) + 12*(d*x)^m*b^3*m^2*x^10*sgn(b*x^3 + a) + 39*(d*x)^m*b^3*m*x^10*sgn(b*x^
3 + a) + 3*(d*x)^m*a*b^2*m^3*x^7*sgn(b*x^3 + a) + 28*(d*x)^m*b^3*x^10*sgn(b*x^3 + a) + 45*(d*x)^m*a*b^2*m^2*x^
7*sgn(b*x^3 + a) + 162*(d*x)^m*a*b^2*m*x^7*sgn(b*x^3 + a) + 3*(d*x)^m*a^2*b*m^3*x^4*sgn(b*x^3 + a) + 120*(d*x)
^m*a*b^2*x^7*sgn(b*x^3 + a) + 54*(d*x)^m*a^2*b*m^2*x^4*sgn(b*x^3 + a) + 261*(d*x)^m*a^2*b*m*x^4*sgn(b*x^3 + a)
+ (d*x)^m*a^3*m^3*x*sgn(b*x^3 + a) + 210*(d*x)^m*a^2*b*x^4*sgn(b*x^3 + a) + 21*(d*x)^m*a^3*m^2*x*sgn(b*x^3 +
a) + 138*(d*x)^m*a^3*m*x*sgn(b*x^3 + a) + 280*(d*x)^m*a^3*x*sgn(b*x^3 + a))/(m^4 + 22*m^3 + 159*m^2 + 418*m +
280)

Mupad [F(-1)]

Timed out. \[ \int (d x)^m \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx=\int {\left (d\,x\right )}^m\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2} \,d x \]

[In]

int((d*x)^m*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2),x)

[Out]

int((d*x)^m*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2), x)