3.1.30 \(\int \frac {(e x)^m (a+b x^n)^2 (A+B x^n)}{(c+d x^n)^2} \, dx\) [30]

Optimal. Leaf size=267 \[ -\frac {b^2 (A d (1+m+n)-B c (1+m+2 n)) x^{1+n} (e x)^m}{c d^2 n (1+m+n)}-\frac {b (2 a d (A d (1+m)-B c (1+m+n))-b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m}}{c d^3 e (1+m) n}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac {(b c-a d) (a d (B c (1+m)-A d (1+m-n))+b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c^2 d^3 e (1+m) n} \]

[Out]

-b^2*(A*d*(1+m+n)-B*c*(1+m+2*n))*x^(1+n)*(e*x)^m/c/d^2/n/(1+m+n)-b*(2*a*d*(A*d*(1+m)-B*c*(1+m+n))-b*c*(A*d*(1+
m+n)-B*c*(1+m+2*n)))*(e*x)^(1+m)/c/d^3/e/(1+m)/n-(-A*d+B*c)*(e*x)^(1+m)*(a+b*x^n)^2/c/d/e/n/(c+d*x^n)-(-a*d+b*
c)*(a*d*(B*c*(1+m)-A*d*(1+m-n))+b*c*(A*d*(1+m+n)-B*c*(1+m+2*n)))*(e*x)^(1+m)*hypergeom([1, (1+m)/n],[(1+m+n)/n
],-d*x^n/c)/c^2/d^3/e/(1+m)/n

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Rubi [A]
time = 0.45, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {608, 584, 20, 30, 371} \begin {gather*} -\frac {(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right ) (a d (B c (m+1)-A d (m-n+1))+b c (A d (m+n+1)-B c (m+2 n+1)))}{c^2 d^3 e (m+1) n}-\frac {b (e x)^{m+1} (2 a d (A d (m+1)-B c (m+n+1))-b c (A d (m+n+1)-B c (m+2 n+1)))}{c d^3 e (m+1) n}-\frac {(e x)^{m+1} \left (a+b x^n\right )^2 (B c-A d)}{c d e n \left (c+d x^n\right )}-\frac {b^2 x^{n+1} (e x)^m (A d (m+n+1)-B c (m+2 n+1))}{c d^2 n (m+n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*(A*d*(1 + m + n) - B*c*(1 + m + 2*n))*x^(1 + n)*(e*x)^m)/(c*d^2*n*(1 + m + n))) - (b*(2*a*d*(A*d*(1 + m
) - B*c*(1 + m + n)) - b*c*(A*d*(1 + m + n) - B*c*(1 + m + 2*n)))*(e*x)^(1 + m))/(c*d^3*e*(1 + m)*n) - ((B*c -
A*d)*(e*x)^(1 + m)*(a + b*x^n)^2)/(c*d*e*n*(c + d*x^n)) - ((b*c - a*d)*(a*d*(B*c*(1 + m) - A*d*(1 + m - n)) +
b*c*(A*d*(1 + m + n) - B*c*(1 + m + 2*n)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*
x^n)/c)])/(c^2*d^3*e*(1 + m)*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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 584

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]

Rule 608

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*g*n*(p + 1))), x] + Dis
t[1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(
m + 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n},
x] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rubi steps

\begin {align*} \int \frac {(e x)^m \left (a+b x^n\right )^2 \left (A+B x^n\right )}{\left (c+d x^n\right )^2} \, dx &=-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac {\int \frac {(e x)^m \left (a+b x^n\right ) \left (-a (B c (1+m)-A d (1+m-n))+b (A d (1+m+n)-B c (1+m+2 n)) x^n\right )}{c+d x^n} \, dx}{c d n}\\ &=-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac {\int \left (\frac {b (2 a d (A d (1+m)-B c (1+m+n))-b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^m}{d^2}+\frac {b^2 (A d (1+m+n)-B c (1+m+2 n)) x^n (e x)^m}{d}+\frac {(b c-a d) (a d (B c (1+m)-A d (1+m-n))+b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^m}{d^2 \left (c+d x^n\right )}\right ) \, dx}{c d n}\\ &=-\frac {b (2 a d (A d (1+m)-B c (1+m+n))-b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m}}{c d^3 e (1+m) n}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac {\left (b^2 (A d (1+m+n)-B c (1+m+2 n))\right ) \int x^n (e x)^m \, dx}{c d^2 n}-\frac {((b c-a d) (a d (B c (1+m)-A d (1+m-n))+b c (A d (1+m+n)-B c (1+m+2 n)))) \int \frac {(e x)^m}{c+d x^n} \, dx}{c d^3 n}\\ &=-\frac {b (2 a d (A d (1+m)-B c (1+m+n))-b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m}}{c d^3 e (1+m) n}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac {(b c-a d) (a d (B c (1+m)-A d (1+m-n))+b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c^2 d^3 e (1+m) n}-\frac {\left (b^2 (A d (1+m+n)-B c (1+m+2 n)) x^{-m} (e x)^m\right ) \int x^{m+n} \, dx}{c d^2 n}\\ &=-\frac {b^2 (A d (1+m+n)-B c (1+m+2 n)) x^{1+n} (e x)^m}{c d^2 n (1+m+n)}-\frac {b (2 a d (A d (1+m)-B c (1+m+n))-b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m}}{c d^3 e (1+m) n}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^n\right )^2}{c d e n \left (c+d x^n\right )}-\frac {(b c-a d) (a d (B c (1+m)-A d (1+m-n))+b c (A d (1+m+n)-B c (1+m+2 n))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c^2 d^3 e (1+m) n}\\ \end {align*}

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Mathematica [A]
time = 0.82, size = 249, normalized size = 0.93 \begin {gather*} x (e x)^m \left (\frac {2 a b \left (-A d (1+m)+B c (1+m+n)+B d n x^n\right )}{d^2 (1+m) n \left (c+d x^n\right )}-\frac {a^2 B c-a^2 A d}{c^2 d n+c d^2 n x^n}+\frac {b^2 \left (A d \left (\frac {1}{1+m}+\frac {c}{c n+d n x^n}\right )+B \left (-\frac {2 c}{1+m}+\frac {d x^n}{1+m+n}-\frac {c^2}{c n+d n x^n}\right )\right )}{d^3}+\frac {(b c-a d) (a d (-B c (1+m)+A d (1+m-n))+b c (-A d (1+m+n)+B c (1+m+2 n))) \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c^2 d^3 (1+m) n}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*(e*x)^m*((2*a*b*(-(A*d*(1 + m)) + B*c*(1 + m + n) + B*d*n*x^n))/(d^2*(1 + m)*n*(c + d*x^n)) - (a^2*B*c - a^2
*A*d)/(c^2*d*n + c*d^2*n*x^n) + (b^2*(A*d*((1 + m)^(-1) + c/(c*n + d*n*x^n)) + B*((-2*c)/(1 + m) + (d*x^n)/(1
+ m + n) - c^2/(c*n + d*n*x^n))))/d^3 + ((b*c - a*d)*(a*d*(-(B*c*(1 + m)) + A*d*(1 + m - n)) + b*c*(-(A*d*(1 +
m + n)) + B*c*(1 + m + 2*n)))*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c^2*d^3*(1 + m)*
n))

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (a +b \,x^{n}\right )^{2} \left (A +B \,x^{n}\right )}{\left (c +d \,x^{n}\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((m*e^m + (n + 1)*e^m)*b^2*c^2*d - 2*(m*e^m + e^m)*a*b*c*d^2 + (m*e^m - (n - 1)*e^m)*a^2*d^3)*A - ((m*e^m +
(2*n + 1)*e^m)*b^2*c^3 - 2*(m*e^m + (n + 1)*e^m)*a*b*c^2*d + (m*e^m + e^m)*a^2*c*d^2)*B)*integrate(x^m/(c*d^4*
n*x^n + c^2*d^3*n), x) + ((m*n*e^m + n*e^m)*B*b^2*c*d^2*x*e^(m*log(x) + 2*n*log(x)) + (((m^2*e^m + 2*m*(n + 1)
*e^m + (n^2 + 2*n + 1)*e^m)*b^2*c^2*d - 2*(m^2*e^m + m*(n + 2)*e^m + (n + 1)*e^m)*a*b*c*d^2 + (m^2*e^m + m*(n
+ 2)*e^m + (n + 1)*e^m)*a^2*d^3)*A - ((m^2*e^m + m*(3*n + 2)*e^m + (2*n^2 + 3*n + 1)*e^m)*b^2*c^3 - 2*(m^2*e^m
+ 2*m*(n + 1)*e^m + (n^2 + 2*n + 1)*e^m)*a*b*c^2*d + (m^2*e^m + m*(n + 2)*e^m + (n + 1)*e^m)*a^2*c*d^2)*B)*x*
x^m + ((m*n*e^m + (n^2 + n)*e^m)*A*b^2*c*d^2 - ((m*n*e^m + (2*n^2 + n)*e^m)*b^2*c^2*d - 2*(m*n*e^m + (n^2 + n)
*e^m)*a*b*c*d^2)*B)*x*e^(m*log(x) + n*log(x)))/((m^2*n + (n^2 + 2*n)*m + n^2 + n)*c*d^4*x^n + (m^2*n + (n^2 +
2*n)*m + n^2 + n)*c^2*d^3)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^2*x^(3*n) + A*a^2 + (2*B*a*b + A*b^2)*x^(2*n) + (B*a^2 + 2*A*a*b)*x^n)*(x*e)^m/(d^2*x^(2*n) + 2*
c*d*x^n + c^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*x)**m*(A + B*x**n)*(a + b*x**n)**2/(c + d*x**n)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^n + A)*(b*x^n + a)^2*(x*e)^m/(d*x^n + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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