3.11 \(\int (e x)^m (A+B x^n) (c+d x^n)^2 \, dx\)

Optimal. Leaf size=102 \[ \frac{c x^{n+1} (e x)^m (2 A d+B c)}{m+n+1}+\frac{d x^{2 n+1} (e x)^m (A d+2 B c)}{m+2 n+1}+\frac{A c^2 (e x)^{m+1}}{e (m+1)}+\frac{B d^2 x^{3 n+1} (e x)^m}{m+3 n+1} \]

[Out]

(c*(B*c + 2*A*d)*x^(1 + n)*(e*x)^m)/(1 + m + n) + (d*(2*B*c + A*d)*x^(1 + 2*n)*(e*x)^m)/(1 + m + 2*n) + (B*d^2
*x^(1 + 3*n)*(e*x)^m)/(1 + m + 3*n) + (A*c^2*(e*x)^(1 + m))/(e*(1 + m))

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Rubi [A]  time = 0.0759361, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {448, 20, 30} \[ \frac{c x^{n+1} (e x)^m (2 A d+B c)}{m+n+1}+\frac{d x^{2 n+1} (e x)^m (A d+2 B c)}{m+2 n+1}+\frac{A c^2 (e x)^{m+1}}{e (m+1)}+\frac{B d^2 x^{3 n+1} (e x)^m}{m+3 n+1} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*(A + B*x^n)*(c + d*x^n)^2,x]

[Out]

(c*(B*c + 2*A*d)*x^(1 + n)*(e*x)^m)/(1 + m + n) + (d*(2*B*c + A*d)*x^(1 + 2*n)*(e*x)^m)/(1 + m + 2*n) + (B*d^2
*x^(1 + 3*n)*(e*x)^m)/(1 + m + 3*n) + (A*c^2*(e*x)^(1 + m))/(e*(1 + m))

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

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 (e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^2 \, dx &=\int \left (A c^2 (e x)^m+c (B c+2 A d) x^n (e x)^m+d (2 B c+A d) x^{2 n} (e x)^m+B d^2 x^{3 n} (e x)^m\right ) \, dx\\ &=\frac{A c^2 (e x)^{1+m}}{e (1+m)}+\left (B d^2\right ) \int x^{3 n} (e x)^m \, dx+(d (2 B c+A d)) \int x^{2 n} (e x)^m \, dx+(c (B c+2 A d)) \int x^n (e x)^m \, dx\\ &=\frac{A c^2 (e x)^{1+m}}{e (1+m)}+\left (B d^2 x^{-m} (e x)^m\right ) \int x^{m+3 n} \, dx+\left (d (2 B c+A d) x^{-m} (e x)^m\right ) \int x^{m+2 n} \, dx+\left (c (B c+2 A d) x^{-m} (e x)^m\right ) \int x^{m+n} \, dx\\ &=\frac{c (B c+2 A d) x^{1+n} (e x)^m}{1+m+n}+\frac{d (2 B c+A d) x^{1+2 n} (e x)^m}{1+m+2 n}+\frac{B d^2 x^{1+3 n} (e x)^m}{1+m+3 n}+\frac{A c^2 (e x)^{1+m}}{e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.110322, size = 78, normalized size = 0.76 \[ x (e x)^m \left (\frac{c x^n (2 A d+B c)}{m+n+1}+\frac{d x^{2 n} (A d+2 B c)}{m+2 n+1}+\frac{A c^2}{m+1}+\frac{B d^2 x^{3 n}}{m+3 n+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*(A + B*x^n)*(c + d*x^n)^2,x]

[Out]

x*(e*x)^m*((A*c^2)/(1 + m) + (c*(B*c + 2*A*d)*x^n)/(1 + m + n) + (d*(2*B*c + A*d)*x^(2*n))/(1 + m + 2*n) + (B*
d^2*x^(3*n))/(1 + m + 3*n))

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Maple [C]  time = 0.052, size = 732, normalized size = 7.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(A+B*x^n)*(c+d*x^n)^2,x)

[Out]

x*(12*A*c*d*m*n^2*x^n+16*B*c*d*m*n*(x^n)^2+20*A*c*d*m*n*x^n+A*c^2+2*(x^n)^2*B*c*d+3*B*c^2*m^2*x^n+6*B*c^2*n^2*
x^n+3*B*c^2*x^n*m+5*B*c^2*x^n*n+6*A*c^2*n^3+3*A*c^2*m^2+11*A*c^2*n^2+A*c^2*m^3+B*c^2*m^3*x^n+3*m*B*d^2*(x^n)^3
+3*B*d^2*(x^n)^3*n+3*A*d^2*(x^n)^2*m+4*A*d^2*(x^n)^2*n+2*B*d^2*n^2*(x^n)^3+3*A*d^2*m^2*(x^n)^2+3*A*d^2*n^2*(x^
n)^2+B*d^2*m^3*(x^n)^3+A*d^2*m^3*(x^n)^2+3*B*d^2*m^2*(x^n)^3+10*A*c*d*m^2*n*x^n+8*B*c*d*m^2*n*(x^n)^2+6*B*c*d*
m*n^2*(x^n)^2+6*A*c*d*x^n*m+10*A*c*d*x^n*n+3*B*d^2*m^2*n*(x^n)^3+2*B*d^2*m*n^2*(x^n)^3+4*A*d^2*m^2*n*(x^n)^2+3
*A*d^2*m*n^2*(x^n)^2+2*B*c*d*m^3*(x^n)^2+6*B*d^2*m*n*(x^n)^3+2*A*c*d*m^3*x^n+8*A*d^2*m*n*(x^n)^2+5*B*c^2*m^2*n
*x^n+6*B*c^2*m*n^2*x^n+6*B*c*d*m^2*(x^n)^2+6*B*c*d*n^2*(x^n)^2+6*A*c*d*m^2*x^n+12*A*c*d*n^2*x^n+10*B*c^2*m*n*x
^n+6*B*c*d*(x^n)^2*m+8*B*c*d*(x^n)^2*n+3*A*c^2*m+6*A*c^2*n+B*c^2*x^n+B*d^2*(x^n)^3+A*d^2*(x^n)^2+6*A*c^2*m^2*n
+11*A*c^2*m*n^2+12*A*c^2*m*n+2*x^n*A*c*d)/(1+m)/(m+n+1)/(1+m+2*n)/(1+m+3*n)*exp(1/2*m*(-I*Pi*csgn(I*e*x)^3+I*P
i*csgn(I*e*x)^2*csgn(I*e)+I*Pi*csgn(I*e*x)^2*csgn(I*x)-I*Pi*csgn(I*e*x)*csgn(I*e)*csgn(I*x)+2*ln(e)+2*ln(x)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(A+B*x^n)*(c+d*x^n)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.11974, size = 1208, normalized size = 11.84 \begin{align*} \frac{{\left (B d^{2} m^{3} + 3 \, B d^{2} m^{2} + 3 \, B d^{2} m + B d^{2} + 2 \,{\left (B d^{2} m + B d^{2}\right )} n^{2} + 3 \,{\left (B d^{2} m^{2} + 2 \, B d^{2} m + B d^{2}\right )} n\right )} x x^{3 \, n} e^{\left (m \log \left (e\right ) + m \log \left (x\right )\right )} +{\left ({\left (2 \, B c d + A d^{2}\right )} m^{3} + 2 \, B c d + A d^{2} + 3 \,{\left (2 \, B c d + A d^{2}\right )} m^{2} + 3 \,{\left (2 \, B c d + A d^{2} +{\left (2 \, B c d + A d^{2}\right )} m\right )} n^{2} + 3 \,{\left (2 \, B c d + A d^{2}\right )} m + 4 \,{\left (2 \, B c d + A d^{2} +{\left (2 \, B c d + A d^{2}\right )} m^{2} + 2 \,{\left (2 \, B c d + A d^{2}\right )} m\right )} n\right )} x x^{2 \, n} e^{\left (m \log \left (e\right ) + m \log \left (x\right )\right )} +{\left ({\left (B c^{2} + 2 \, A c d\right )} m^{3} + B c^{2} + 2 \, A c d + 3 \,{\left (B c^{2} + 2 \, A c d\right )} m^{2} + 6 \,{\left (B c^{2} + 2 \, A c d +{\left (B c^{2} + 2 \, A c d\right )} m\right )} n^{2} + 3 \,{\left (B c^{2} + 2 \, A c d\right )} m + 5 \,{\left (B c^{2} + 2 \, A c d +{\left (B c^{2} + 2 \, A c d\right )} m^{2} + 2 \,{\left (B c^{2} + 2 \, A c d\right )} m\right )} n\right )} x x^{n} e^{\left (m \log \left (e\right ) + m \log \left (x\right )\right )} +{\left (A c^{2} m^{3} + 6 \, A c^{2} n^{3} + 3 \, A c^{2} m^{2} + 3 \, A c^{2} m + A c^{2} + 11 \,{\left (A c^{2} m + A c^{2}\right )} n^{2} + 6 \,{\left (A c^{2} m^{2} + 2 \, A c^{2} m + A c^{2}\right )} n\right )} x e^{\left (m \log \left (e\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(A+B*x^n)*(c+d*x^n)^2,x, algorithm="fricas")

[Out]

((B*d^2*m^3 + 3*B*d^2*m^2 + 3*B*d^2*m + B*d^2 + 2*(B*d^2*m + B*d^2)*n^2 + 3*(B*d^2*m^2 + 2*B*d^2*m + B*d^2)*n)
*x*x^(3*n)*e^(m*log(e) + m*log(x)) + ((2*B*c*d + A*d^2)*m^3 + 2*B*c*d + A*d^2 + 3*(2*B*c*d + A*d^2)*m^2 + 3*(2
*B*c*d + A*d^2 + (2*B*c*d + A*d^2)*m)*n^2 + 3*(2*B*c*d + A*d^2)*m + 4*(2*B*c*d + A*d^2 + (2*B*c*d + A*d^2)*m^2
+ 2*(2*B*c*d + A*d^2)*m)*n)*x*x^(2*n)*e^(m*log(e) + m*log(x)) + ((B*c^2 + 2*A*c*d)*m^3 + B*c^2 + 2*A*c*d + 3*
(B*c^2 + 2*A*c*d)*m^2 + 6*(B*c^2 + 2*A*c*d + (B*c^2 + 2*A*c*d)*m)*n^2 + 3*(B*c^2 + 2*A*c*d)*m + 5*(B*c^2 + 2*A
*c*d + (B*c^2 + 2*A*c*d)*m^2 + 2*(B*c^2 + 2*A*c*d)*m)*n)*x*x^n*e^(m*log(e) + m*log(x)) + (A*c^2*m^3 + 6*A*c^2*
n^3 + 3*A*c^2*m^2 + 3*A*c^2*m + A*c^2 + 11*(A*c^2*m + A*c^2)*n^2 + 6*(A*c^2*m^2 + 2*A*c^2*m + A*c^2)*n)*x*e^(m
*log(e) + 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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(A+B*x**n)*(c+d*x**n)**2,x)

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.09045, size = 1381, normalized size = 13.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(A+B*x^n)*(c+d*x^n)^2,x, algorithm="giac")

[Out]

(B*d^2*m^3*x*x^m*x^(3*n)*e^m + 3*B*d^2*m^2*n*x*x^m*x^(3*n)*e^m + 2*B*d^2*m*n^2*x*x^m*x^(3*n)*e^m + 2*B*c*d*m^3
*x*x^m*x^(2*n)*e^m + A*d^2*m^3*x*x^m*x^(2*n)*e^m + 8*B*c*d*m^2*n*x*x^m*x^(2*n)*e^m + 4*A*d^2*m^2*n*x*x^m*x^(2*
n)*e^m + 6*B*c*d*m*n^2*x*x^m*x^(2*n)*e^m + 3*A*d^2*m*n^2*x*x^m*x^(2*n)*e^m + B*c^2*m^3*x*x^m*x^n*e^m + 2*A*c*d
*m^3*x*x^m*x^n*e^m + 5*B*c^2*m^2*n*x*x^m*x^n*e^m + 10*A*c*d*m^2*n*x*x^m*x^n*e^m + 6*B*c^2*m*n^2*x*x^m*x^n*e^m
+ 12*A*c*d*m*n^2*x*x^m*x^n*e^m + A*c^2*m^3*x*x^m*e^m + 6*A*c^2*m^2*n*x*x^m*e^m + 11*A*c^2*m*n^2*x*x^m*e^m + 6*
A*c^2*n^3*x*x^m*e^m + 3*B*d^2*m^2*x*x^m*x^(3*n)*e^m + 6*B*d^2*m*n*x*x^m*x^(3*n)*e^m + 2*B*d^2*n^2*x*x^m*x^(3*n
)*e^m + 6*B*c*d*m^2*x*x^m*x^(2*n)*e^m + 3*A*d^2*m^2*x*x^m*x^(2*n)*e^m + 16*B*c*d*m*n*x*x^m*x^(2*n)*e^m + 8*A*d
^2*m*n*x*x^m*x^(2*n)*e^m + 6*B*c*d*n^2*x*x^m*x^(2*n)*e^m + 3*A*d^2*n^2*x*x^m*x^(2*n)*e^m + 3*B*c^2*m^2*x*x^m*x
^n*e^m + 6*A*c*d*m^2*x*x^m*x^n*e^m + 10*B*c^2*m*n*x*x^m*x^n*e^m + 20*A*c*d*m*n*x*x^m*x^n*e^m + 6*B*c^2*n^2*x*x
^m*x^n*e^m + 12*A*c*d*n^2*x*x^m*x^n*e^m + 3*A*c^2*m^2*x*x^m*e^m + 12*A*c^2*m*n*x*x^m*e^m + 11*A*c^2*n^2*x*x^m*
e^m + 3*B*d^2*m*x*x^m*x^(3*n)*e^m + 3*B*d^2*n*x*x^m*x^(3*n)*e^m + 6*B*c*d*m*x*x^m*x^(2*n)*e^m + 3*A*d^2*m*x*x^
m*x^(2*n)*e^m + 8*B*c*d*n*x*x^m*x^(2*n)*e^m + 4*A*d^2*n*x*x^m*x^(2*n)*e^m + 3*B*c^2*m*x*x^m*x^n*e^m + 6*A*c*d*
m*x*x^m*x^n*e^m + 5*B*c^2*n*x*x^m*x^n*e^m + 10*A*c*d*n*x*x^m*x^n*e^m + 3*A*c^2*m*x*x^m*e^m + 6*A*c^2*n*x*x^m*e
^m + B*d^2*x*x^m*x^(3*n)*e^m + 2*B*c*d*x*x^m*x^(2*n)*e^m + A*d^2*x*x^m*x^(2*n)*e^m + B*c^2*x*x^m*x^n*e^m + 2*A
*c*d*x*x^m*x^n*e^m + A*c^2*x*x^m*e^m)/(m^4 + 6*m^3*n + 11*m^2*n^2 + 6*m*n^3 + 4*m^3 + 18*m^2*n + 22*m*n^2 + 6*
n^3 + 6*m^2 + 18*m*n + 11*n^2 + 4*m + 6*n + 1)