3.6.15 \(\int (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx\) [515]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 108, normalized size of antiderivative = 1.00, 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 = {1369, 14, 20, 30} \begin {gather*} \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 )} \end {gather*}

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 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]

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*} \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*}

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Mathematica [A]
time = 0.04, size = 55, normalized size = 0.51 \begin {gather*} \frac {x (d x)^m \sqrt {\left (a+b x^n\right )^2} \left (a (1+m+n)+b (1+m) x^n\right )}{(1+m) (1+m+n) \left (a+b x^n\right )} \end {gather*}

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))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.05, size = 132, normalized size = 1.22

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, x \left (m b \,x^{n}+a m +a n +b \,x^{n}+a \right ) {\mathrm e}^{\frac {m \left (i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i d x \right )^{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i d x \right ) \mathrm {csgn}\left (i d \right )-i \pi \mathrm {csgn}\left (i d x \right )^{3}+i \pi \mathrm {csgn}\left (i d x \right )^{2} \mathrm {csgn}\left (i d \right )+2 \ln \left (x \right )+2 \ln \left (d \right )\right )}{2}}}{\left (a +b \,x^{n}\right ) \left (1+m \right ) \left (1+m +n \right )}\)	\(132\)

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,method=_RETURNVERBOSE)

[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*Pi*csgn(I*x)*csgn(I*d*x)^
2-I*Pi*csgn(I*x)*csgn(I*d*x)*csgn(I*d)-I*Pi*csgn(I*d*x)^3+I*Pi*csgn(I*d*x)^2*csgn(I*d)+2*ln(x)+2*ln(d)))

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Maxima [A]
time = 0.31, size = 47, normalized size = 0.44 \begin {gather*} \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 {gather*}

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)

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Fricas [A]
time = 0.36, size = 57, normalized size = 0.53 \begin {gather*} \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 {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d x\right )^{m} \sqrt {\left (a + b x^{n}\right )^{2}}\, dx \end {gather*}

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)

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Giac [A]
time = 3.83, size = 173, normalized size = 1.60 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,x\right )}^m\,\sqrt {a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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