3.345 \(\int \cos ^4(a+b x) (c \sin (a+b x))^m \, dx\)

Optimal. Leaf size=68 \[ \frac{\cos (a+b x) (c \sin (a+b x))^{m+1} \, _2F_1\left (-\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b c (m+1) \sqrt{\cos ^2(a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.0409665, antiderivative size = 68, 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 = {2577} \[ \frac{\cos (a+b x) (c \sin (a+b x))^{m+1} \, _2F_1\left (-\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b c (m+1) \sqrt{\cos ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^4*(c*Sin[a + b*x])^m,x]

[Out]

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

Rule 2577

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

Rubi steps

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

Mathematica [A]  time = 0.0533877, size = 63, normalized size = 0.93 \[ \frac{\sqrt{\cos ^2(a+b x)} \tan (a+b x) (c \sin (a+b x))^m \, _2F_1\left (-\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(a+b x)\right )}{b (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^4*(c*Sin[a + b*x])^m,x]

[Out]

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

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Maple [F]  time = 0.885, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( bx+a \right ) \right ) ^{4} \left ( c\sin \left ( bx+a \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^4*(c*sin(b*x+a))^m,x)

[Out]

int(cos(b*x+a)^4*(c*sin(b*x+a))^m,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sin(b*x + a))^m*cos(b*x + a)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (c \sin \left (b x + a\right )\right )^{m} \cos \left (b x + a\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*sin(b*x + a))^m*cos(b*x + a)^4, 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(cos(b*x+a)**4*(c*sin(b*x+a))**m,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sin(b*x + a))^m*cos(b*x + a)^4, x)