\(\int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx\) [1193]

Optimal result

Integrand size = 33, antiderivative size = 33 \[ \int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx=\text {Int}\left (\cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p,x\right ) \]

[Out]

Unintegrable(cos(d*x+c)^4*sin(d*x+c)^(-3-p)*(a+b*sin(d*x+c))^p,x)

Rubi [N/A]

Not integrable

Time = 0.07 (sec) , antiderivative size = 33, 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 = {} \[ \int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx=\int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx \]

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^(-3 - p)*(a + b*Sin[c + d*x])^p,x]

[Out]

Defer[Int][Cos[c + d*x]^4*Sin[c + d*x]^(-3 - p)*(a + b*Sin[c + d*x])^p, x]

Rubi steps \begin{align*} \text {integral}& = \int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 4.83 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx=\int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx \]

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^(-3 - p)*(a + b*Sin[c + d*x])^p,x]

[Out]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^(-3 - p)*(a + b*Sin[c + d*x])^p, x]

Maple [N/A] (verified)

Not integrable

Time = 0.93 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00

\[\int \left (\cos ^{4}\left (d x +c \right )\right ) \left (\sin ^{-3-p}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{p}d x\]

[In]

int(cos(d*x+c)^4*sin(d*x+c)^(-3-p)*(a+b*sin(d*x+c))^p,x)

[Out]

int(cos(d*x+c)^4*sin(d*x+c)^(-3-p)*(a+b*sin(d*x+c))^p,x)

Fricas [N/A]

Not integrable

Time = 0.50 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{-p - 3} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^(-3-p)*(a+b*sin(d*x+c))^p,x, algorithm="fricas")

[Out]

integral((b*sin(d*x + c) + a)^p*sin(d*x + c)^(-p - 3)*cos(d*x + c)^4, x)

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**(-3-p)*(a+b*sin(d*x+c))**p,x)

[Out]

Timed out

Maxima [N/A]

Not integrable

Time = 5.55 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{-p - 3} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^(-3-p)*(a+b*sin(d*x+c))^p,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^p*sin(d*x + c)^(-p - 3)*cos(d*x + c)^4, x)

Giac [N/A]

Not integrable

Time = 1.93 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{-p - 3} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^(-3-p)*(a+b*sin(d*x+c))^p,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^p*sin(d*x + c)^(-p - 3)*cos(d*x + c)^4, x)

Mupad [N/A]

Not integrable

Time = 16.68 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \cos ^4(c+d x) \sin ^{-3-p}(c+d x) (a+b \sin (c+d x))^p \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^p}{{\sin \left (c+d\,x\right )}^{p+3}} \,d x \]

[In]

int((cos(c + d*x)^4*(a + b*sin(c + d*x))^p)/sin(c + d*x)^(p + 3),x)

[Out]

int((cos(c + d*x)^4*(a + b*sin(c + d*x))^p)/sin(c + d*x)^(p + 3), x)