\(\int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx\) [583]

Optimal result

Integrand size = 34, antiderivative size = 34 \[ \int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx=\frac {b (b x)^{1-n} \sin ^{-1+n}(a x)}{a^2 \left (a c^2 x \cos (a x)-c^2 \sin (a x)\right )}+\frac {b^2 (1-n) \text {Int}\left ((b x)^{-n} \sin ^{-2+n}(a x),x\right )}{a^2 c^2} \]

[Out]

b*(b*x)^(1-n)*sin(a*x)^(-1+n)/a^2/(a*c^2*x*cos(a*x)-c^2*sin(a*x))+b^2*(1-n)*Unintegrable(sin(a*x)^(-2+n)/((b*x
)^n),x)/a^2/c^2

Rubi [N/A]

Not integrable

Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx=\int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx \]

[In]

Int[((b*x)^(2 - n)*Sin[a*x]^n)/(a*c*x*Cos[a*x] - c*Sin[a*x])^2,x]

[Out]

(b*(b*x)^(1 - n)*Sin[a*x]^(-1 + n))/(a^2*(a*c^2*x*Cos[a*x] - c^2*Sin[a*x])) + (b^2*(1 - n)*Defer[Int][Sin[a*x]
^(-2 + n)/(b*x)^n, x])/(a^2*c^2)

Rubi steps \begin{align*} \text {integral}& = \frac {b (b x)^{1-n} \sin ^{-1+n}(a x)}{a^2 \left (a c^2 x \cos (a x)-c^2 \sin (a x)\right )}+\frac {\left (b^2 (1-n)\right ) \int (b x)^{-n} \sin ^{-2+n}(a x) \, dx}{a^2 c^2} \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 8.96 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx=\int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx \]

[In]

Integrate[((b*x)^(2 - n)*Sin[a*x]^n)/(a*c*x*Cos[a*x] - c*Sin[a*x])^2,x]

[Out]

Integrate[((b*x)^(2 - n)*Sin[a*x]^n)/(a*c*x*Cos[a*x] - c*Sin[a*x])^2, x]

Maple [N/A] (verified)

Not integrable

Time = 0.96 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00

\[\int \frac {\left (x b \right )^{2-n} \sin \left (a x \right )^{n}}{\left (a c x \cos \left (a x \right )-c \sin \left (a x \right )\right )^{2}}d x\]

[In]

int((x*b)^(2-n)*sin(a*x)^n/(a*c*x*cos(a*x)-c*sin(a*x))^2,x)

[Out]

int((x*b)^(2-n)*sin(a*x)^n/(a*c*x*cos(a*x)-c*sin(a*x))^2,x)

Fricas [N/A]

Not integrable

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.94 \[ \int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx=\int { \frac {\left (b x\right )^{-n + 2} \sin \left (a x\right )^{n}}{{\left (a c x \cos \left (a x\right ) - c \sin \left (a x\right )\right )}^{2}} \,d x } \]

[In]

integrate((b*x)^(2-n)*sin(a*x)^n/(a*c*x*cos(a*x)-c*sin(a*x))^2,x, algorithm="fricas")

[Out]

integral(-(b*x)^(-n + 2)*sin(a*x)^n/(2*a*c^2*x*cos(a*x)*sin(a*x) - (a^2*c^2*x^2 - c^2)*cos(a*x)^2 - c^2), x)

Sympy [N/A]

Not integrable

Time = 82.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx=\frac {\int \frac {\left (b x\right )^{2 - n} \sin ^{n}{\left (a x \right )}}{a^{2} x^{2} \cos ^{2}{\left (a x \right )} - 2 a x \sin {\left (a x \right )} \cos {\left (a x \right )} + \sin ^{2}{\left (a x \right )}}\, dx}{c^{2}} \]

[In]

integrate((b*x)**(2-n)*sin(a*x)**n/(a*c*x*cos(a*x)-c*sin(a*x))**2,x)

[Out]

Integral((b*x)**(2 - n)*sin(a*x)**n/(a**2*x**2*cos(a*x)**2 - 2*a*x*sin(a*x)*cos(a*x) + sin(a*x)**2), x)/c**2

Maxima [N/A]

Not integrable

Time = 2.80 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx=\int { \frac {\left (b x\right )^{-n + 2} \sin \left (a x\right )^{n}}{{\left (a c x \cos \left (a x\right ) - c \sin \left (a x\right )\right )}^{2}} \,d x } \]

[In]

integrate((b*x)^(2-n)*sin(a*x)^n/(a*c*x*cos(a*x)-c*sin(a*x))^2,x, algorithm="maxima")

[Out]

integrate((b*x)^(-n + 2)*sin(a*x)^n/(a*c*x*cos(a*x) - c*sin(a*x))^2, x)

Giac [N/A]

Not integrable

Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx=\int { \frac {\left (b x\right )^{-n + 2} \sin \left (a x\right )^{n}}{{\left (a c x \cos \left (a x\right ) - c \sin \left (a x\right )\right )}^{2}} \,d x } \]

[In]

integrate((b*x)^(2-n)*sin(a*x)^n/(a*c*x*cos(a*x)-c*sin(a*x))^2,x, algorithm="giac")

[Out]

integrate((b*x)^(-n + 2)*sin(a*x)^n/(a*c*x*cos(a*x) - c*sin(a*x))^2, x)

Mupad [N/A]

Not integrable

Time = 28.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(b x)^{2-n} \sin ^n(a x)}{(a c x \cos (a x)-c \sin (a x))^2} \, dx=\int \frac {{\sin \left (a\,x\right )}^n\,{\left (b\,x\right )}^{2-n}}{{\left (c\,\sin \left (a\,x\right )-a\,c\,x\,\cos \left (a\,x\right )\right )}^2} \,d x \]

[In]

int((sin(a*x)^n*(b*x)^(2 - n))/(c*sin(a*x) - a*c*x*cos(a*x))^2,x)

[Out]

int((sin(a*x)^n*(b*x)^(2 - n))/(c*sin(a*x) - a*c*x*cos(a*x))^2, x)