3.6.22 \(\int (d x)^m (a^2+2 a b x^n+b^2 x^{2 n})^{3/2} \, dx\) [522]

Optimal. Leaf size=238 \[ \frac {a^3 (d x)^{1+m} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{d (1+m) \left (a+b x^n\right )}+\frac {3 a^2 b^2 x^{1+n} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1+m+n) \left (a b+b^2 x^n\right )}+\frac {3 a b^3 x^{1+2 n} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1+m+2 n) \left (a b+b^2 x^n\right )}+\frac {b^4 x^{1+3 n} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1+m+3 n) \left (a b+b^2 x^n\right )} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1369, 276, 20, 30} \begin {gather*} \frac {3 a^2 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 {b^4 x^{3 n+1} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(m+3 n+1) \left (a b+b^2 x^n\right )}+\frac {3 a b^3 x^{2 n+1} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(m+2 n+1) \left (a b+b^2 x^n\right )}+\frac {a^3 (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*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2),x]

[Out]

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

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 276

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]

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 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2} \, 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 )^3 \, dx}{b^2 \left (a b+b^2 x^n\right )}\\ &=\frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int \left (a^3 b^3 (d x)^m+3 a^2 b^4 x^n (d x)^m+3 a b^5 x^{2 n} (d x)^m+b^6 x^{3 n} (d x)^m\right ) \, dx}{b^2 \left (a b+b^2 x^n\right )}\\ &=\frac {a^3 (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 (3 a^2 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 {\left (3 a b^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}\right ) \int x^{2 n} (d x)^m \, dx}{a b+b^2 x^n}+\frac {\left (b^4 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}\right ) \int x^{3 n} (d x)^m \, dx}{a b+b^2 x^n}\\ &=\frac {a^3 (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 (3 a^2 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 {\left (3 a b^3 x^{-m} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}\right ) \int x^{m+2 n} \, dx}{a b+b^2 x^n}+\frac {\left (b^4 x^{-m} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}\right ) \int x^{m+3 n} \, dx}{a b+b^2 x^n}\\ &=\frac {a^3 (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 {3 a^2 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 )}+\frac {3 a b^3 x^{1+2 n} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1+m+2 n) \left (a b+b^2 x^n\right )}+\frac {b^4 x^{1+3 n} (d x)^m \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1+m+3 n) \left (a b+b^2 x^n\right )}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 90, normalized size = 0.38 \begin {gather*} \frac {x (d x)^m \left (\left (a+b x^n\right )^2\right )^{3/2} \left (\frac {a^3}{1+m}+\frac {3 a^2 b x^n}{1+m+n}+\frac {3 a b^2 x^{2 n}}{1+m+2 n}+\frac {b^3 x^{3 n}}{1+m+3 n}\right )}{\left (a+b x^n\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.05, size = 532, normalized size = 2.24

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, x \left (9 a^{2} b \,m^{2} x^{n}+a^{3}+3 a^{2} b \,m^{3} x^{n}+b^{3} x^{3 n}+3 x^{2 n} a \,b^{2}+b^{3} m^{3} x^{3 n}+3 b^{3} m^{2} x^{3 n}+2 b^{3} n^{2} x^{3 n}+3 m \,b^{3} x^{3 n}+3 b^{3} x^{3 n} n +9 m \,a^{2} b \,x^{n}+15 a^{2} b n \,x^{n}+a^{3} m^{3}+3 a^{3} m^{2}+11 a^{3} n^{2}+6 a^{3} n +9 a \,b^{2} n^{2} x^{2 n}+3 b^{3} m^{2} n \,x^{3 n}+2 b^{3} m \,n^{2} x^{3 n}+3 a \,b^{2} m^{3} x^{2 n}+6 b^{3} m n \,x^{3 n}+3 m \,a^{3}+6 a^{3} m^{2} n +11 a^{3} m \,n^{2}+12 a^{3} m n +12 a \,b^{2} m^{2} n \,x^{2 n}+3 a^{2} b \,x^{n}+18 a^{2} b \,n^{2} x^{n}+6 a^{3} n^{3}+12 a \,b^{2} x^{2 n} n +9 a \,b^{2} m^{2} x^{2 n}+9 m a \,b^{2} x^{2 n}+15 a^{2} b \,m^{2} n \,x^{n}+18 a^{2} b m \,n^{2} x^{n}+30 a^{2} b m n \,x^{n}+9 a \,b^{2} m \,n^{2} x^{2 n}+24 a \,b^{2} m n \,x^{2 n}\right ) {\mathrm e}^{\frac {m \left (i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i d x \right )^{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i d x \right ) \mathrm {csgn}\left (i d \right )-i \pi \mathrm {csgn}\left (i d x \right )^{3}+i \pi \mathrm {csgn}\left (i d x \right )^{2} \mathrm {csgn}\left (i d \right )+2 \ln \left (x \right )+2 \ln \left (d \right )\right )}{2}}}{\left (a +b \,x^{n}\right ) \left (1+m \right ) \left (1+m +n \right ) \left (1+m +2 n \right ) \left (1+m +3 n \right )}\)	\(532\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

((a+b*x^n)^2)^(1/2)/(a+b*x^n)*x*(b^3*(x^n)^3+9*a*b^2*n^2*(x^n)^2+9*a^2*b*m^2*x^n+a^3+3*b^3*m^2*n*(x^n)^3+2*b^3
*m*n^2*(x^n)^3+3*a*b^2*m^3*(x^n)^2+6*b^3*m*n*(x^n)^3+3*a^2*b*m^3*x^n+9*m*a^2*b*x^n+15*a^2*b*n*x^n+a^3*m^3+3*a^
3*m^2+11*a^3*n^2+6*a^3*n+12*a*b^2*m^2*n*(x^n)^2+12*a*b^2*(x^n)^2*n+3*m*a^3+9*a*b^2*m^2*(x^n)^2+6*a^3*m^2*n+11*
a^3*m*n^2+12*a^3*m*n+3*(x^n)^2*a*b^2+b^3*m^3*(x^n)^3+3*b^3*m^2*(x^n)^3+2*b^3*n^2*(x^n)^3+3*m*b^3*(x^n)^3+3*b^3
*(x^n)^3*n+3*a^2*b*x^n+18*a^2*b*n^2*x^n+9*m*a*b^2*(x^n)^2+6*a^3*n^3+9*a*b^2*m*n^2*(x^n)^2+15*a^2*b*m^2*n*x^n+1
8*a^2*b*m*n^2*x^n+24*a*b^2*m*n*(x^n)^2+30*a^2*b*m*n*x^n)/(1+m)/(1+m+n)/(1+m+2*n)/(1+m+3*n)*exp(1/2*m*(I*Pi*csg
n(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 = 0.30, size = 276, normalized size = 1.16 \begin {gather*} \frac {{\left (m^{3} + 3 \, m^{2} {\left (2 \, n + 1\right )} + 6 \, n^{3} + {\left (11 \, n^{2} + 12 \, n + 3\right )} m + 11 \, n^{2} + 6 \, n + 1\right )} a^{3} d^{m} x x^{m} + {\left (m^{3} + 3 \, m^{2} {\left (n + 1\right )} + {\left (2 \, n^{2} + 6 \, n + 3\right )} m + 2 \, n^{2} + 3 \, n + 1\right )} b^{3} d^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )} + 3 \, {\left (m^{3} + m^{2} {\left (4 \, n + 3\right )} + {\left (3 \, n^{2} + 8 \, n + 3\right )} m + 3 \, n^{2} + 4 \, n + 1\right )} a b^{2} d^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )} + 3 \, {\left (m^{3} + m^{2} {\left (5 \, n + 3\right )} + {\left (6 \, n^{2} + 10 \, n + 3\right )} m + 6 \, n^{2} + 5 \, n + 1\right )} a^{2} b d^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m^{4} + 2 \, m^{3} {\left (3 \, n + 2\right )} + {\left (11 \, n^{2} + 18 \, n + 6\right )} m^{2} + 6 \, n^{3} + 2 \, {\left (3 \, n^{3} + 11 \, n^{2} + 9 \, n + 2\right )} m + 11 \, n^{2} + 6 \, 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))^(3/2),x, algorithm="maxima")

[Out]

((m^3 + 3*m^2*(2*n + 1) + 6*n^3 + (11*n^2 + 12*n + 3)*m + 11*n^2 + 6*n + 1)*a^3*d^m*x*x^m + (m^3 + 3*m^2*(n +
1) + (2*n^2 + 6*n + 3)*m + 2*n^2 + 3*n + 1)*b^3*d^m*x*e^(m*log(x) + 3*n*log(x)) + 3*(m^3 + m^2*(4*n + 3) + (3*
n^2 + 8*n + 3)*m + 3*n^2 + 4*n + 1)*a*b^2*d^m*x*e^(m*log(x) + 2*n*log(x)) + 3*(m^3 + m^2*(5*n + 3) + (6*n^2 +
10*n + 3)*m + 6*n^2 + 5*n + 1)*a^2*b*d^m*x*e^(m*log(x) + n*log(x)))/(m^4 + 2*m^3*(3*n + 2) + (11*n^2 + 18*n +
6)*m^2 + 6*n^3 + 2*(3*n^3 + 11*n^2 + 9*n + 2)*m + 11*n^2 + 6*n + 1)

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Fricas [A]
time = 0.39, size = 390, normalized size = 1.64 \begin {gather*} \frac {{\left (b^{3} m^{3} + 3 \, b^{3} m^{2} + 3 \, b^{3} m + b^{3} + 2 \, {\left (b^{3} m + b^{3}\right )} n^{2} + 3 \, {\left (b^{3} m^{2} + 2 \, b^{3} m + b^{3}\right )} n\right )} x x^{3 \, n} e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} + 3 \, {\left (a b^{2} m^{3} + 3 \, a b^{2} m^{2} + 3 \, a b^{2} m + a b^{2} + 3 \, {\left (a b^{2} m + a b^{2}\right )} n^{2} + 4 \, {\left (a b^{2} m^{2} + 2 \, a b^{2} m + a b^{2}\right )} n\right )} x x^{2 \, n} e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} + 3 \, {\left (a^{2} b m^{3} + 3 \, a^{2} b m^{2} + 3 \, a^{2} b m + a^{2} b + 6 \, {\left (a^{2} b m + a^{2} b\right )} n^{2} + 5 \, {\left (a^{2} b m^{2} + 2 \, a^{2} b m + a^{2} b\right )} n\right )} x x^{n} e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )} + {\left (a^{3} m^{3} + 6 \, a^{3} n^{3} + 3 \, a^{3} m^{2} + 3 \, a^{3} m + a^{3} + 11 \, {\left (a^{3} m + a^{3}\right )} n^{2} + 6 \, {\left (a^{3} m^{2} + 2 \, a^{3} m + a^{3}\right )} n\right )} x e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )}}{m^{4} + 6 \, {\left (m + 1\right )} n^{3} + 4 \, m^{3} + 11 \, {\left (m^{2} + 2 \, m + 1\right )} n^{2} + 6 \, m^{2} + 6 \, {\left (m^{3} + 3 \, m^{2} + 3 \, m + 1\right )} n + 4 \, 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))^(3/2),x, algorithm="fricas")

[Out]

((b^3*m^3 + 3*b^3*m^2 + 3*b^3*m + b^3 + 2*(b^3*m + b^3)*n^2 + 3*(b^3*m^2 + 2*b^3*m + b^3)*n)*x*x^(3*n)*e^(m*lo
g(d) + m*log(x)) + 3*(a*b^2*m^3 + 3*a*b^2*m^2 + 3*a*b^2*m + a*b^2 + 3*(a*b^2*m + a*b^2)*n^2 + 4*(a*b^2*m^2 + 2
*a*b^2*m + a*b^2)*n)*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 3*(a^2*b*m^3 + 3*a^2*b*m^2 + 3*a^2*b*m + a^2*b + 6*(a
^2*b*m + a^2*b)*n^2 + 5*(a^2*b*m^2 + 2*a^2*b*m + a^2*b)*n)*x*x^n*e^(m*log(d) + m*log(x)) + (a^3*m^3 + 6*a^3*n^
3 + 3*a^3*m^2 + 3*a^3*m + a^3 + 11*(a^3*m + a^3)*n^2 + 6*(a^3*m^2 + 2*a^3*m + a^3)*n)*x*e^(m*log(d) + m*log(x)
))/(m^4 + 6*(m + 1)*n^3 + 4*m^3 + 11*(m^2 + 2*m + 1)*n^2 + 6*m^2 + 6*(m^3 + 3*m^2 + 3*m + 1)*n + 4*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} \left (\left (a + b x^{n}\right )^{2}\right )^{\frac {3}{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))**(3/2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 2719 vs. \(2 (230) = 460\).
time = 5.53, size = 2719, normalized size = 11.42 \begin {gather*} \text {Too large to display} \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))^(3/2),x, algorithm="giac")

[Out]

(b^3*m^3*x*x^(3*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m^2*n*x*x^(3*n)*e^(m*log(d) + m*log(x))*sgn(
b*x^n + a) + 2*b^3*m*n^2*x*x^(3*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*a*b^2*m^3*x*x^(2*n)*e^(m*log(d)
+ m*log(x))*sgn(b*x^n + a) + b^3*m^3*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 12*a*b^2*m^2*n*x*x^(2*
n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m^2*n*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a
*b^2*m*n^2*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 2*b^3*m*n^2*x*x^(2*n)*e^(m*log(d) + m*log(x))*sg
n(b*x^n + a) + 3*a^2*b*m^3*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*a*b^2*m^3*x*x^n*e^(m*log(d) + m*lo
g(x))*sgn(b*x^n + a) + b^3*m^3*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 15*a^2*b*m^2*n*x*x^n*e^(m*log(d)
+ m*log(x))*sgn(b*x^n + a) + 12*a*b^2*m^2*n*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m^2*n*x*x^n*
e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 18*a^2*b*m*n^2*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a*b^2
*m*n^2*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 2*b^3*m*n^2*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a)
+ a^3*m^3*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*a^2*b*m^3*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3
*a*b^2*m^3*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + b^3*m^3*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 6*a^3
*m^2*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 15*a^2*b*m^2*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 12
*a*b^2*m^2*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m^2*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) +
11*a^3*m*n^2*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 18*a^2*b*m*n^2*x*e^(m*log(d) + m*log(x))*sgn(b*x^n +
a) + 9*a*b^2*m*n^2*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 2*b^3*m*n^2*x*e^(m*log(d) + m*log(x))*sgn(b*x^n
+ a) + 6*a^3*n^3*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m^2*x*x^(3*n)*e^(m*log(d) + m*log(x))*sgn(b*
x^n + a) + 6*b^3*m*n*x*x^(3*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 2*b^3*n^2*x*x^(3*n)*e^(m*log(d) + m*lo
g(x))*sgn(b*x^n + a) + 9*a*b^2*m^2*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m^2*x*x^(2*n)*e^(m
*log(d) + m*log(x))*sgn(b*x^n + a) + 24*a*b^2*m*n*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 6*b^3*m*n
*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a*b^2*n^2*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n +
a) + 2*b^3*n^2*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a^2*b*m^2*x*x^n*e^(m*log(d) + m*log(x))*sg
n(b*x^n + a) + 9*a*b^2*m^2*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m^2*x*x^n*e^(m*log(d) + m*log(
x))*sgn(b*x^n + a) + 30*a^2*b*m*n*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 24*a*b^2*m*n*x*x^n*e^(m*log(d
) + m*log(x))*sgn(b*x^n + a) + 6*b^3*m*n*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 18*a^2*b*n^2*x*x^n*e^(
m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a*b^2*n^2*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 2*b^3*n^2*x*x
^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*a^3*m^2*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a^2*b*m^2*x
*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a*b^2*m^2*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m^2*x*e
^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 12*a^3*m*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 30*a^2*b*m*n*x*e
^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 24*a*b^2*m*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 6*b^3*m*n*x*e^
(m*log(d) + m*log(x))*sgn(b*x^n + a) + 11*a^3*n^2*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 18*a^2*b*n^2*x*e^
(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a*b^2*n^2*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 2*b^3*n^2*x*e^(m
*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m*x*x^(3*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*n*x*x^(3
*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a*b^2*m*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^
3*m*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 12*a*b^2*n*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n
+ a) + 3*b^3*n*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a^2*b*m*x*x^n*e^(m*log(d) + m*log(x))*sgn(
b*x^n + a) + 9*a*b^2*m*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*b^3*m*x*x^n*e^(m*log(d) + m*log(x))*sg
n(b*x^n + a) + 15*a^2*b*n*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 12*a*b^2*n*x*x^n*e^(m*log(d) + m*log(
x))*sgn(b*x^n + a) + 3*b^3*n*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*a^3*m*x*e^(m*log(d) + m*log(x))*
sgn(b*x^n + a) + 9*a^2*b*m*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 9*a*b^2*m*x*e^(m*log(d) + m*log(x))*sgn(
b*x^n + a) + 3*b^3*m*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 6*a^3*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n +
a) + 15*a^2*b*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 12*a*b^2*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a)
+ 3*b^3*n*x*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + b^3*x*x^(3*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3
*a*b^2*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + b^3*x*x^(2*n)*e^(m*log(d) + m*log(x))*sgn(b*x^n + a)
+ 3*a^2*b*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a) + 3*a*b^2*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n + a)
+ b^3*x*x^n*e^(m*log(d) + m*log(x))*sgn(b*x^n ...

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (d\,x\right )}^m\,{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2} \,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)^(3/2),x)

[Out]

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

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