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

Optimal. Leaf size=185 \[ \frac{(e x)^{m+1} \left (a^2 B d^2-a b d (A d+2 B c)+b^2 c (2 A d+B c)\right )}{b^3 e (m+1)}+\frac{(e x)^{m+1} (A b-a B) (b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a b^3 e (m+1)}+\frac{d x^{n+1} (e x)^m (-a B d+A b d+2 b B c)}{b^2 (m+n+1)}+\frac{B d^2 x^{2 n+1} (e x)^m}{b (m+2 n+1)} \]

[Out]

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

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Rubi [A]  time = 0.2257, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {570, 20, 30, 364} \[ \frac{(e x)^{m+1} \left (a^2 B d^2-a b d (A d+2 B c)+b^2 c (2 A d+B c)\right )}{b^3 e (m+1)}+\frac{(e x)^{m+1} (A b-a B) (b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a b^3 e (m+1)}+\frac{d x^{n+1} (e x)^m (-a B d+A b d+2 b B c)}{b^2 (m+n+1)}+\frac{B d^2 x^{2 n+1} (e x)^m}{b (m+2 n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 570

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 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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

Mathematica [A]  time = 0.259551, size = 153, normalized size = 0.83 \[ \frac{x (e x)^m \left (\frac{a^2 B d^2-a b d (A d+2 B c)+b^2 c (2 A d+B c)}{m+1}+\frac{(A b-a B) (b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a (m+1)}+\frac{b d x^n (-a B d+A b d+2 b B c)}{m+n+1}+\frac{b^2 B d^2 x^{2 n}}{m+2 n+1}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.505, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( A+B{x}^{n} \right ) \left ( c+d{x}^{n} \right ) ^{2}}{a+b{x}^{n}}}\, dx \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/(a+b*x^n),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\left ({\left (b^{3} c^{2} e^{m} - 2 \, a b^{2} c d e^{m} + a^{2} b d^{2} e^{m}\right )} A -{\left (a b^{2} c^{2} e^{m} - 2 \, a^{2} b c d e^{m} + a^{3} d^{2} e^{m}\right )} B\right )} \int \frac{x^{m}}{b^{4} x^{n} + a b^{3}}\,{d x} + \frac{{\left (m^{2} + m{\left (n + 2\right )} + n + 1\right )} B b^{2} d^{2} e^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )} +{\left ({\left (2 \,{\left (m^{2} + m{\left (3 \, n + 2\right )} + 2 \, n^{2} + 3 \, n + 1\right )} b^{2} c d e^{m} -{\left (m^{2} + m{\left (3 \, n + 2\right )} + 2 \, n^{2} + 3 \, n + 1\right )} a b d^{2} e^{m}\right )} A +{\left ({\left (m^{2} + m{\left (3 \, n + 2\right )} + 2 \, n^{2} + 3 \, n + 1\right )} b^{2} c^{2} e^{m} - 2 \,{\left (m^{2} + m{\left (3 \, n + 2\right )} + 2 \, n^{2} + 3 \, n + 1\right )} a b c d e^{m} +{\left (m^{2} + m{\left (3 \, n + 2\right )} + 2 \, n^{2} + 3 \, n + 1\right )} a^{2} d^{2} e^{m}\right )} B\right )} x x^{m} +{\left ({\left (m^{2} + 2 \, m{\left (n + 1\right )} + 2 \, n + 1\right )} A b^{2} d^{2} e^{m} +{\left (2 \,{\left (m^{2} + 2 \, m{\left (n + 1\right )} + 2 \, n + 1\right )} b^{2} c d e^{m} -{\left (m^{2} + 2 \, m{\left (n + 1\right )} + 2 \, n + 1\right )} a b d^{2} e^{m}\right )} B\right )} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{{\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}} \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/(a+b*x^n),x, algorithm="maxima")

[Out]

((b^3*c^2*e^m - 2*a*b^2*c*d*e^m + a^2*b*d^2*e^m)*A - (a*b^2*c^2*e^m - 2*a^2*b*c*d*e^m + a^3*d^2*e^m)*B)*integr
ate(x^m/(b^4*x^n + a*b^3), x) + ((m^2 + m*(n + 2) + n + 1)*B*b^2*d^2*e^m*x*e^(m*log(x) + 2*n*log(x)) + ((2*(m^
2 + m*(3*n + 2) + 2*n^2 + 3*n + 1)*b^2*c*d*e^m - (m^2 + m*(3*n + 2) + 2*n^2 + 3*n + 1)*a*b*d^2*e^m)*A + ((m^2
+ m*(3*n + 2) + 2*n^2 + 3*n + 1)*b^2*c^2*e^m - 2*(m^2 + m*(3*n + 2) + 2*n^2 + 3*n + 1)*a*b*c*d*e^m + (m^2 + m*
(3*n + 2) + 2*n^2 + 3*n + 1)*a^2*d^2*e^m)*B)*x*x^m + ((m^2 + 2*m*(n + 1) + 2*n + 1)*A*b^2*d^2*e^m + (2*(m^2 +
2*m*(n + 1) + 2*n + 1)*b^2*c*d*e^m - (m^2 + 2*m*(n + 1) + 2*n + 1)*a*b*d^2*e^m)*B)*x*e^(m*log(x) + n*log(x)))/
((m^3 + 3*m^2*(n + 1) + (2*n^2 + 6*n + 3)*m + 2*n^2 + 3*n + 1)*b^3)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B d^{2} x^{3 \, n} + A c^{2} +{\left (2 \, B c d + A d^{2}\right )} x^{2 \, n} +{\left (B c^{2} + 2 \, A c d\right )} x^{n}\right )} \left (e x\right )^{m}}{b x^{n} + a}, x\right ) \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/(a+b*x^n),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 27.2062, size = 1085, normalized size = 5.86 \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/(a+b*x**n),x)

[Out]

A*c**2*e**m*m*x*x**m*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1/n)*gamma(m/n + 1/n)/(a*n**2*gamma(m/n + 1 +
1/n)) + A*c**2*e**m*x*x**m*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1/n)*gamma(m/n + 1/n)/(a*n**2*gamma(m/
n + 1 + 1/n)) + 2*A*c*d*e**m*m*x*x**m*x**n*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1 + 1/n)*gamma(m/n + 1
+ 1/n)/(a*n**2*gamma(m/n + 2 + 1/n)) + 2*A*c*d*e**m*x*x**m*x**n*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1
+ 1/n)*gamma(m/n + 1 + 1/n)/(a*n*gamma(m/n + 2 + 1/n)) + 2*A*c*d*e**m*x*x**m*x**n*lerchphi(b*x**n*exp_polar(I*
pi)/a, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(a*n**2*gamma(m/n + 2 + 1/n)) + A*d**2*e**m*m*x*x**m*x**(2*n)*le
rchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(a*n**2*gamma(m/n + 3 + 1/n)) + 2*A*d*
*2*e**m*x*x**m*x**(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(a*n*gamma(m
/n + 3 + 1/n)) + A*d**2*e**m*x*x**m*x**(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2 + 1/n)*gamma(m/n +
2 + 1/n)/(a*n**2*gamma(m/n + 3 + 1/n)) + B*c**2*e**m*m*x*x**m*x**n*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n +
1 + 1/n)*gamma(m/n + 1 + 1/n)/(a*n**2*gamma(m/n + 2 + 1/n)) + B*c**2*e**m*x*x**m*x**n*lerchphi(b*x**n*exp_pol
ar(I*pi)/a, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(a*n*gamma(m/n + 2 + 1/n)) + B*c**2*e**m*x*x**m*x**n*lerchp
hi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(a*n**2*gamma(m/n + 2 + 1/n)) + 2*B*c*d*e*
*m*m*x*x**m*x**(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(a*n**2*gamma(m
/n + 3 + 1/n)) + 4*B*c*d*e**m*x*x**m*x**(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2 + 1/n)*gamma(m/n +
2 + 1/n)/(a*n*gamma(m/n + 3 + 1/n)) + 2*B*c*d*e**m*x*x**m*x**(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n
+ 2 + 1/n)*gamma(m/n + 2 + 1/n)/(a*n**2*gamma(m/n + 3 + 1/n)) + B*d**2*e**m*m*x*x**m*x**(3*n)*lerchphi(b*x**n*
exp_polar(I*pi)/a, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(a*n**2*gamma(m/n + 4 + 1/n)) + 3*B*d**2*e**m*x*x**m
*x**(3*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(a*n*gamma(m/n + 4 + 1/n))
+ B*d**2*e**m*x*x**m*x**(3*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(a*n*
*2*gamma(m/n + 4 + 1/n))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{n} + A\right )}{\left (d x^{n} + c\right )}^{2} \left (e x\right )^{m}}{b x^{n} + a}\,{d x} \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/(a+b*x^n),x, algorithm="giac")

[Out]

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