3.350 \(\int \cos ^4(c+d x) (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=83 \[ -\frac {a^2 2^{m+\frac {5}{2}} \cos ^5(c+d x) (\sin (c+d x)+1)^{-m-\frac {1}{2}} (a \sin (c+d x)+a)^{m-2} \, _2F_1\left (\frac {5}{2},-m-\frac {3}{2};\frac {7}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{5 d} \]

[Out]

-1/5*2^(5/2+m)*a^2*cos(d*x+c)^5*hypergeom([5/2, -3/2-m],[7/2],1/2-1/2*sin(d*x+c))*(1+sin(d*x+c))^(-1/2-m)*(a+a
*sin(d*x+c))^(-2+m)/d

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Rubi [A]  time = 0.08, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2689, 70, 69} \[ -\frac {a^2 2^{m+\frac {5}{2}} \cos ^5(c+d x) (\sin (c+d x)+1)^{-m-\frac {1}{2}} (a \sin (c+d x)+a)^{m-2} \, _2F_1\left (\frac {5}{2},-m-\frac {3}{2};\frac {7}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2^(5/2 + m)*a^2*Cos[c + d*x]^5*Hypergeometric2F1[5/2, -3/2 - m, 7/2, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x]
)^(-1/2 - m)*(a + a*Sin[c + d*x])^(-2 + m))/(5*d)

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 2689

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+a \sin (c+d x))^m \, dx &=\frac {\left (a^2 \cos ^5(c+d x)\right ) \operatorname {Subst}\left (\int (a-a x)^{3/2} (a+a x)^{\frac {3}{2}+m} \, dx,x,\sin (c+d x)\right )}{d (a-a \sin (c+d x))^{5/2} (a+a \sin (c+d x))^{5/2}}\\ &=\frac {\left (2^{\frac {3}{2}+m} a^3 \cos ^5(c+d x) (a+a \sin (c+d x))^{-2+m} \left (\frac {a+a \sin (c+d x)}{a}\right )^{-\frac {1}{2}-m}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{\frac {3}{2}+m} (a-a x)^{3/2} \, dx,x,\sin (c+d x)\right )}{d (a-a \sin (c+d x))^{5/2}}\\ &=-\frac {2^{\frac {5}{2}+m} a^2 \cos ^5(c+d x) \, _2F_1\left (\frac {5}{2},-\frac {3}{2}-m;\frac {7}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-\frac {1}{2}-m} (a+a \sin (c+d x))^{-2+m}}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 78, normalized size = 0.94 \[ -\frac {2^{m+\frac {5}{2}} \cos ^5(c+d x) (\sin (c+d x)+1)^{-m-\frac {5}{2}} (a (\sin (c+d x)+1))^m \, _2F_1\left (\frac {5}{2},-m-\frac {3}{2};\frac {7}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(2^(5/2 + m)*Cos[c + d*x]^5*Hypergeometric2F1[5/2, -3/2 - m, 7/2, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x]
)^(-5/2 - m)*(a*(1 + Sin[c + d*x]))^m)/d

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.44, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{4}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \cos ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(c + d*x) + 1))**m*cos(c + d*x)**4, x)

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