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

Optimal. Leaf size=97 \[ \frac{b \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+4}}{d^4 (m+4) \left (a+b x^3\right )}+\frac{a \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+1}}{d (m+1) \left (a+b x^3\right )} \]

[Out]

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

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Rubi [A]  time = 0.0378413, antiderivative size = 97, 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, 14} \[ \frac{b \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+4}}{d^4 (m+4) \left (a+b x^3\right )}+\frac{a \sqrt{a^2+2 a b x^3+b^2 x^6} (d x)^{m+1}}{d (m+1) \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int (d x)^m \sqrt{a^2+2 a b x^3+b^2 x^6} \, 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 ) \, dx}{a b+b^2 x^3}\\ &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \int \left (a b (d x)^m+\frac{b^2 (d x)^{3+m}}{d^3}\right ) \, dx}{a b+b^2 x^3}\\ &=\frac{a (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{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 )}\\ \end{align*}

Mathematica [A]  time = 0.0232985, size = 53, normalized size = 0.55 \[ \frac{x \sqrt{\left (a+b x^3\right )^2} (d x)^m \left (a (m+4)+b (m+1) x^3\right )}{(m+1) (m+4) \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 56, normalized size = 0.6 \begin{align*}{\frac{ \left ( bm{x}^{3}+b{x}^{3}+am+4\,a \right ) x \left ( dx \right ) ^{m}}{ \left ( 4+m \right ) \left ( 1+m \right ) \left ( b{x}^{3}+a \right ) }\sqrt{ \left ( b{x}^{3}+a \right ) ^{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)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.01596, size = 47, normalized size = 0.48 \begin{align*} \frac{{\left (b d^{m}{\left (m + 1\right )} x^{4} + a d^{m}{\left (m + 4\right )} x\right )} x^{m}}{m^{2} + 5 \, m + 4} \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)^(1/2),x, algorithm="maxima")

[Out]

(b*d^m*(m + 1)*x^4 + a*d^m*(m + 4)*x)*x^m/(m^2 + 5*m + 4)

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Fricas [A]  time = 1.51837, size = 77, normalized size = 0.79 \begin{align*} \frac{{\left ({\left (b m + b\right )} x^{4} +{\left (a m + 4 \, a\right )} x\right )} \left (d x\right )^{m}}{m^{2} + 5 \, m + 4} \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)^(1/2),x, algorithm="fricas")

[Out]

((b*m + b)*x^4 + (a*m + 4*a)*x)*(d*x)^m/(m^2 + 5*m + 4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \sqrt{\left (a + b x^{3}\right )^{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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.12445, size = 112, normalized size = 1.15 \begin{align*} \frac{\left (d x\right )^{m} b m x^{4} \mathrm{sgn}\left (b x^{3} + a\right ) + \left (d x\right )^{m} b x^{4} \mathrm{sgn}\left (b x^{3} + a\right ) + \left (d x\right )^{m} a m x \mathrm{sgn}\left (b x^{3} + a\right ) + 4 \, \left (d x\right )^{m} a x \mathrm{sgn}\left (b x^{3} + a\right )}{m^{2} + 5 \, m + 4} \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)^(1/2),x, algorithm="giac")

[Out]

((d*x)^m*b*m*x^4*sgn(b*x^3 + a) + (d*x)^m*b*x^4*sgn(b*x^3 + a) + (d*x)^m*a*m*x*sgn(b*x^3 + a) + 4*(d*x)^m*a*x*
sgn(b*x^3 + a))/(m^2 + 5*m + 4)