3.793 \(\int (d x)^m \sqrt{a^2+2 a b x^2+b^2 x^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0337345, 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 = {1112, 14} \[ \frac{b \sqrt{a^2+2 a b x^2+b^2 x^4} (d x)^{m+3}}{d^3 (m+3) \left (a+b x^2\right )}+\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4} (d x)^{m+1}}{d (m+1) \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && 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^2+b^2 x^4} \, dx &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int (d x)^m \left (a b+b^2 x^2\right ) \, dx}{a b+b^2 x^2}\\ &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \left (a b (d x)^m+\frac{b^2 (d x)^{2+m}}{d^2}\right ) \, dx}{a b+b^2 x^2}\\ &=\frac{a (d x)^{1+m} \sqrt{a^2+2 a b x^2+b^2 x^4}}{d (1+m) \left (a+b x^2\right )}+\frac{b (d x)^{3+m} \sqrt{a^2+2 a b x^2+b^2 x^4}}{d^3 (3+m) \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0247809, size = 53, normalized size = 0.55 \[ \frac{x \sqrt{\left (a+b x^2\right )^2} (d x)^m \left (a (m+3)+b (m+1) x^2\right )}{(m+1) (m+3) \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

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Maple [A]  time = 0.166, size = 56, normalized size = 0.6 \begin{align*}{\frac{ \left ( bm{x}^{2}+b{x}^{2}+am+3\,a \right ) x \left ( dx \right ) ^{m}}{ \left ( 3+m \right ) \left ( 1+m \right ) \left ( b{x}^{2}+a \right ) }\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.971085, size = 47, normalized size = 0.48 \begin{align*} \frac{{\left (b d^{m}{\left (m + 1\right )} x^{3} + a d^{m}{\left (m + 3\right )} x\right )} x^{m}}{m^{2} + 4 \, m + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.54057, size = 77, normalized size = 0.79 \begin{align*} \frac{{\left ({\left (b m + b\right )} x^{3} +{\left (a m + 3 \, a\right )} x\right )} \left (d x\right )^{m}}{m^{2} + 4 \, m + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(1/2),x, algorithm="fricas")

[Out]

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

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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^{2}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*(b**2*x**4+2*a*b*x**2+a**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.28775, size = 112, normalized size = 1.15 \begin{align*} \frac{\left (d x\right )^{m} b m x^{3} \mathrm{sgn}\left (b x^{2} + a\right ) + \left (d x\right )^{m} b x^{3} \mathrm{sgn}\left (b x^{2} + a\right ) + \left (d x\right )^{m} a m x \mathrm{sgn}\left (b x^{2} + a\right ) + 3 \, \left (d x\right )^{m} a x \mathrm{sgn}\left (b x^{2} + a\right )}{m^{2} + 4 \, m + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(1/2),x, algorithm="giac")

[Out]

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