3.361 \(\int (d \cos (a+b x))^n \sin ^4(a+b x) \, dx\)

Optimal. Leaf size=69 \[ -\frac{\sin (a+b x) (d \cos (a+b x))^{n+1} \, _2F_1\left (-\frac{3}{2},\frac{n+1}{2};\frac{n+3}{2};\cos ^2(a+b x)\right )}{b d (n+1) \sqrt{\sin ^2(a+b x)}} \]

[Out]

-(((d*Cos[a + b*x])^(1 + n)*Hypergeometric2F1[-3/2, (1 + n)/2, (3 + n)/2, Cos[a + b*x]^2]*Sin[a + b*x])/(b*d*(
1 + n)*Sqrt[Sin[a + b*x]^2]))

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Rubi [A]  time = 0.0391226, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {2576} \[ -\frac{\sin (a+b x) (d \cos (a+b x))^{n+1} \, _2F_1\left (-\frac{3}{2},\frac{n+1}{2};\frac{n+3}{2};\cos ^2(a+b x)\right )}{b d (n+1) \sqrt{\sin ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[a + b*x])^n*Sin[a + b*x]^4,x]

[Out]

-(((d*Cos[a + b*x])^(1 + n)*Hypergeometric2F1[-3/2, (1 + n)/2, (3 + n)/2, Cos[a + b*x]^2]*Sin[a + b*x])/(b*d*(
1 + n)*Sqrt[Sin[a + b*x]^2]))

Rule 2576

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^(2*IntPar
t[(n - 1)/2] + 1)*(b*Sin[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Cos[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/
2, (1 - n)/2, (3 + m)/2, Cos[e + f*x]^2])/(a*f*(m + 1)*(Sin[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a,
b, e, f, m, n}, x] && SimplerQ[n, m]

Rubi steps

\begin{align*} \int (d \cos (a+b x))^n \sin ^4(a+b x) \, dx &=-\frac{(d \cos (a+b x))^{1+n} \, _2F_1\left (-\frac{3}{2},\frac{1+n}{2};\frac{3+n}{2};\cos ^2(a+b x)\right ) \sin (a+b x)}{b d (1+n) \sqrt{\sin ^2(a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.112452, size = 68, normalized size = 0.99 \[ -\frac{\sin (2 (a+b x)) (d \cos (a+b x))^n \, _2F_1\left (-\frac{3}{2},\frac{n+1}{2};\frac{n+3}{2};\cos ^2(a+b x)\right )}{2 b (n+1) \sqrt{\sin ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[a + b*x])^n*Sin[a + b*x]^4,x]

[Out]

-((d*Cos[a + b*x])^n*Hypergeometric2F1[-3/2, (1 + n)/2, (3 + n)/2, Cos[a + b*x]^2]*Sin[2*(a + b*x)])/(2*b*(1 +
 n)*Sqrt[Sin[a + b*x]^2])

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Maple [F]  time = 0.655, size = 0, normalized size = 0. \begin{align*} \int \left ( d\cos \left ( bx+a \right ) \right ) ^{n} \left ( \sin \left ( bx+a \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(b*x+a))^n*sin(b*x+a)^4,x)

[Out]

int((d*cos(b*x+a))^n*sin(b*x+a)^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \cos \left (b x + a\right )\right )^{n} \sin \left (b x + a\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^n*sin(b*x+a)^4,x, algorithm="maxima")

[Out]

integrate((d*cos(b*x + a))^n*sin(b*x + a)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \left (d \cos \left (b x + a\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^n*sin(b*x+a)^4,x, algorithm="fricas")

[Out]

integral((cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*(d*cos(b*x + a))^n, 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((d*cos(b*x+a))**n*sin(b*x+a)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \cos \left (b x + a\right )\right )^{n} \sin \left (b x + a\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(b*x+a))^n*sin(b*x+a)^4,x, algorithm="giac")

[Out]

integrate((d*cos(b*x + a))^n*sin(b*x + a)^4, x)