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

Optimal. Leaf size=108 \[ \frac{b^2 x^{n+1} (d x)^m \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{(m+n+1) \left (a b+b^2 x^n\right )}+\frac{a (d x)^{m+1} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{d (m+1) \left (a+b x^n\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0415609, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {1355, 14, 20, 30} \[ \frac{b^2 x^{n+1} (d x)^m \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{(m+n+1) \left (a b+b^2 x^n\right )}+\frac{a (d x)^{m+1} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{d (m+1) \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]]

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int (d x)^m \sqrt{a^2+2 a b x^n+b^2 x^{2 n}} \, dx &=\frac{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}} \int (d x)^m \left (a b+b^2 x^n\right ) \, dx}{a b+b^2 x^n}\\ &=\frac{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}} \int \left (a b (d x)^m+b^2 x^n (d x)^m\right ) \, dx}{a b+b^2 x^n}\\ &=\frac{a (d x)^{1+m} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{d (1+m) \left (a+b x^n\right )}+\frac{\left (b^2 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}\right ) \int x^n (d x)^m \, dx}{a b+b^2 x^n}\\ &=\frac{a (d x)^{1+m} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{d (1+m) \left (a+b x^n\right )}+\frac{\left (b^2 x^{-m} (d x)^m \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}\right ) \int x^{m+n} \, dx}{a b+b^2 x^n}\\ &=\frac{a (d x)^{1+m} \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{d (1+m) \left (a+b x^n\right )}+\frac{b^2 x^{1+n} (d x)^m \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}{(1+m+n) \left (a b+b^2 x^n\right )}\\ \end{align*}

Mathematica [A]  time = 0.0318535, size = 55, normalized size = 0.51 \[ \frac{x (d x)^m \sqrt{\left (a+b x^n\right )^2} \left (a (m+n+1)+b (m+1) x^n\right )}{(m+1) (m+n+1) \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.041, size = 132, normalized size = 1.2 \begin{align*}{\frac{x \left ( mb{x}^{n}+am+an+b{x}^{n}+a \right ) }{ \left ( a+b{x}^{n} \right ) \left ( 1+m \right ) \left ( 1+m+n \right ) }\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}{{\rm e}^{{\frac{m \left ( -i \left ({\it csgn} \left ( idx \right ) \right ) ^{3}\pi +i \left ({\it csgn} \left ( idx \right ) \right ) ^{2}{\it csgn} \left ( id \right ) \pi +i \left ({\it csgn} \left ( idx \right ) \right ) ^{2}{\it csgn} \left ( ix \right ) \pi -i{\it csgn} \left ( idx \right ){\it csgn} \left ( id \right ){\it csgn} \left ( ix \right ) \pi +2\,\ln \left ( x \right ) +2\,\ln \left ( d \right ) \right ) }{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a+b*x^n)^2)^(1/2)/(a+b*x^n)*x*(m*b*x^n+a*m+a*n+b*x^n+a)/(1+m)/(1+m+n)*exp(1/2*m*(-I*csgn(I*d*x)^3*Pi+I*csgn(
I*d*x)^2*csgn(I*d)*Pi+I*csgn(I*d*x)^2*csgn(I*x)*Pi-I*csgn(I*d*x)*csgn(I*d)*csgn(I*x)*Pi+2*ln(x)+2*ln(d)))

________________________________________________________________________________________

Maxima [A]  time = 1.05472, size = 63, normalized size = 0.58 \begin{align*} \frac{a d^{m}{\left (m + n + 1\right )} x x^{m} + b d^{m}{\left (m + 1\right )} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m^{2} + m{\left (n + 2\right )} + n + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*d^m*(m + n + 1)*x*x^m + b*d^m*(m + 1)*x*e^(m*log(x) + n*log(x)))/(m^2 + m*(n + 2) + n + 1)

________________________________________________________________________________________

Fricas [A]  time = 1.63823, size = 155, normalized size = 1.44 \begin{align*} \frac{{\left (b m + b\right )} x x^{n} e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} +{\left (a m + a n + a\right )} x e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )}}{m^{2} +{\left (m + 1\right )} n + 2 \, m + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*m + b)*x*x^n*e^(m*log(d) + m*log(x)) + (a*m + a*n + a)*x*e^(m*log(d) + m*log(x)))/(m^2 + (m + 1)*n + 2*m +
1)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.11743, size = 234, normalized size = 2.17 \begin{align*} \frac{b m x x^{n} e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} \mathrm{sgn}\left (b x^{n} + a\right ) + a m x e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} \mathrm{sgn}\left (b x^{n} + a\right ) + b m x e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} \mathrm{sgn}\left (b x^{n} + a\right ) + a n x e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} \mathrm{sgn}\left (b x^{n} + a\right ) + b x x^{n} e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} \mathrm{sgn}\left (b x^{n} + a\right ) + a x e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} \mathrm{sgn}\left (b x^{n} + a\right ) + b x e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} \mathrm{sgn}\left (b x^{n} + a\right )}{m^{2} + m n + 2 \, m + n + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*m*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + a*m*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + b*m*x*e^(m*
log(d) + m*log(x))*sgn(b*x^n + a) + a*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + b*x*x^n*e^(m*log(d) + m*log
(x))*sgn(b*x^n + a) + a*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + b*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a))
/(m^2 + m*n + 2*m + n + 1)