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

Optimal. Leaf size=205 \[ \frac{3 a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+4}}{d^4 (m+4) \left (a+b x^3\right )}+\frac{3 a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+7}}{d^7 (m+7) \left (a+b x^3\right )}+\frac{b^3 \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+10}}{d^{10} (m+10) \left (a+b x^3\right )}+\frac{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+1}}{d (m+1) \left (a+b x^3\right )} \]

[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))

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Rubi [A]  time = 0.0851968, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {1355, 270} \[ \frac{3 a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+4}}{d^4 (m+4) \left (a+b x^3\right )}+\frac{3 a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+7}}{d^7 (m+7) \left (a+b x^3\right )}+\frac{b^3 \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+10}}{d^{10} (m+10) \left (a+b x^3\right )}+\frac{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+1}}{d (m+1) \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[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 1355

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]

Rule 270

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]

Rubi steps

\begin{align*} \int (d x)^m \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx &=\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]  time = 0.0709297, size = 131, normalized size = 0.64 \[ \frac{x \sqrt{\left (a+b x^3\right )^2} (d x)^m \left (3 a^2 b \left (m^3+18 m^2+87 m+70\right ) x^3+a^3 \left (m^3+21 m^2+138 m+280\right )+3 a b^2 \left (m^3+15 m^2+54 m+40\right ) x^6+b^3 \left (m^3+12 m^2+39 m+28\right ) x^9\right )}{(m+1) (m+4) (m+7) (m+10) \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.006, size = 199, normalized size = 1. \begin{align*}{\frac{ \left ({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}bm{x}^{3}+{a}^{3}{m}^{3}+210\,{a}^{2}b{x}^{3}+21\,{a}^{3}{m}^{2}+138\,{a}^{3}m+280\,{a}^{3} \right ) x \left ( dx \right ) ^{m}}{ \left ( 10+m \right ) \left ( 7+m \right ) \left ( 4+m \right ) \left ( 1+m \right ) \left ( b{x}^{3}+a \right ) ^{3}} \left ( \left ( b{x}^{3}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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Maxima [A]  time = 1.09974, size = 161, normalized size = 0.79 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.58441, size = 358, normalized size = 1.75 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [B]  time = 1.12653, size = 518, normalized size = 2.53 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)