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(b*x+a)*hypergeom([-3/2, 1/2+1/2*m],[3/2+1/2*m],sin(b*x+a)^2)*(c*sin(b*x+a))^(1+m)/b/c/(1+m)/(cos(b*x+a)^2)
^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.05, 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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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (c \sin \left (b x + a\right )\right )^{m} \cos \left (b x + a\right )^{4}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin \left (b x + a\right )\right )^{m} \cos \left (b x + a\right )^{4}\,{d x} \]

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)

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maple [F]  time = 1.26, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{4}\left (b x +a \right )\right ) \left (c \sin \left (b x +a \right )\right )^{m}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin \left (b x + a\right )\right )^{m} \cos \left (b x + a\right )^{4}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (a+b\,x\right )}^4\,{\left (c\,\sin \left (a+b\,x\right )\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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