3.576 \(\int \frac{(c x)^m (d+e x^n+f x^{2 n}+g x^{3 n})}{a+b x^n} \, dx\)

Optimal. Leaf size=162 \[ \frac{(c x)^{m+1} \left (a^2 b f+a^3 (-g)-a b^2 e+b^3 d\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a b^3 c (m+1)}+\frac{(c x)^{m+1} \left (a^2 g-a b f+b^2 e\right )}{b^3 c (m+1)}+\frac{x^{n+1} (c x)^m (b f-a g)}{b^2 (m+n+1)}+\frac{g x^{2 n+1} (c x)^m}{b (m+2 n+1)} \]

[Out]

((b*f - a*g)*x^(1 + n)*(c*x)^m)/(b^2*(1 + m + n)) + (g*x^(1 + 2*n)*(c*x)^m)/(b*(1 + m + 2*n)) + ((b^2*e - a*b*
f + a^2*g)*(c*x)^(1 + m))/(b^3*c*(1 + m)) + ((b^3*d - a*b^2*e + a^2*b*f - a^3*g)*(c*x)^(1 + m)*Hypergeometric2
F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a*b^3*c*(1 + m))

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Rubi [A]  time = 0.167104, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1844, 20, 30, 364} \[ \frac{(c x)^{m+1} \left (a^2 b f+a^3 (-g)-a b^2 e+b^3 d\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a b^3 c (m+1)}+\frac{(c x)^{m+1} \left (a^2 g-a b f+b^2 e\right )}{b^3 c (m+1)}+\frac{x^{n+1} (c x)^m (b f-a g)}{b^2 (m+n+1)}+\frac{g x^{2 n+1} (c x)^m}{b (m+2 n+1)} \]

Antiderivative was successfully verified.

[In]

Int[((c*x)^m*(d + e*x^n + f*x^(2*n) + g*x^(3*n)))/(a + b*x^n),x]

[Out]

((b*f - a*g)*x^(1 + n)*(c*x)^m)/(b^2*(1 + m + n)) + (g*x^(1 + 2*n)*(c*x)^m)/(b*(1 + m + 2*n)) + ((b^2*e - a*b*
f + a^2*g)*(c*x)^(1 + m))/(b^3*c*(1 + m)) + ((b^3*d - a*b^2*e + a^2*b*f - a^3*g)*(c*x)^(1 + m)*Hypergeometric2
F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a*b^3*c*(1 + m))

Rule 1844

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IGtQ[m, 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{(c x)^m \left (d+e x^n+f x^{2 n}+g x^{3 n}\right )}{a+b x^n} \, dx &=\int \left (\frac{\left (b^2 e-a b f+a^2 g\right ) (c x)^m}{b^3}+\frac{(b f-a g) x^n (c x)^m}{b^2}+\frac{g x^{2 n} (c x)^m}{b}+\frac{\left (b^3 d-a b^2 e+a^2 b f-a^3 g\right ) (c x)^m}{b^3 \left (a+b x^n\right )}\right ) \, dx\\ &=\frac{\left (b^2 e-a b f+a^2 g\right ) (c x)^{1+m}}{b^3 c (1+m)}+\frac{g \int x^{2 n} (c x)^m \, dx}{b}+\frac{(b f-a g) \int x^n (c x)^m \, dx}{b^2}+\frac{\left (b^3 d-a b^2 e+a^2 b f-a^3 g\right ) \int \frac{(c x)^m}{a+b x^n} \, dx}{b^3}\\ &=\frac{\left (b^2 e-a b f+a^2 g\right ) (c x)^{1+m}}{b^3 c (1+m)}+\frac{\left (b^3 d-a b^2 e+a^2 b f-a^3 g\right ) (c 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 c (1+m)}+\frac{\left (g x^{-m} (c x)^m\right ) \int x^{m+2 n} \, dx}{b}+\frac{\left ((b f-a g) x^{-m} (c x)^m\right ) \int x^{m+n} \, dx}{b^2}\\ &=\frac{(b f-a g) x^{1+n} (c x)^m}{b^2 (1+m+n)}+\frac{g x^{1+2 n} (c x)^m}{b (1+m+2 n)}+\frac{\left (b^2 e-a b f+a^2 g\right ) (c x)^{1+m}}{b^3 c (1+m)}+\frac{\left (b^3 d-a b^2 e+a^2 b f-a^3 g\right ) (c 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 c (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.341273, size = 130, normalized size = 0.8 \[ \frac{x (c x)^m \left (\frac{\left (a^2 b f+a^3 (-g)-a b^2 e+b^3 d\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a (m+1)}+\frac{a^2 g-a b f+b^2 e}{m+1}+\frac{b x^n (b f-a g)}{m+n+1}+\frac{b^2 g x^{2 n}}{m+2 n+1}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c*x)^m*((b^2*e - a*b*f + a^2*g)/(1 + m) + (b*(b*f - a*g)*x^n)/(1 + m + n) + (b^2*g*x^(2*n))/(1 + m + 2*n)
+ ((b^3*d - a*b^2*e + a^2*b*f - a^3*g)*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 = 0.384, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( cx \right ) ^{m} \left ( d+e{x}^{n}+f{x}^{2\,n}+g{x}^{3\,n} \right ) }{a+b{x}^{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

(b^3*c^m*d - a*b^2*c^m*e + a^2*b*c^m*f - a^3*c^m*g)*integrate(x^m/(b^4*x^n + a*b^3), x) + ((m^2 + m*(n + 2) +
n + 1)*b^2*c^m*g*x*e^(m*log(x) + 2*n*log(x)) + ((m^2 + m*(3*n + 2) + 2*n^2 + 3*n + 1)*b^2*c^m*e - (m^2 + m*(3*
n + 2) + 2*n^2 + 3*n + 1)*a*b*c^m*f + (m^2 + m*(3*n + 2) + 2*n^2 + 3*n + 1)*a^2*c^m*g)*x*x^m + ((m^2 + 2*m*(n
+ 1) + 2*n + 1)*b^2*c^m*f - (m^2 + 2*m*(n + 1) + 2*n + 1)*a*b*c^m*g)*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 (g x^{3 \, n} + f x^{2 \, n} + e x^{n} + d\right )} \left (c x\right )^{m}}{b x^{n} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^m*(d+e*x^n+f*x^(2*n)+g*x^(3*n))/(a+b*x^n),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**m*(d+e*x**n+f*x**(2*n)+g*x**(3*n))/(a+b*x**n),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x^{3 \, n} + f x^{2 \, n} + e x^{n} + d\right )} \left (c x\right )^{m}}{b x^{n} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^m*(d+e*x^n+f*x^(2*n)+g*x^(3*n))/(a+b*x^n),x, algorithm="giac")

[Out]

integrate((g*x^(3*n) + f*x^(2*n) + e*x^n + d)*(c*x)^m/(b*x^n + a), x)