3.351 \(\int \cos ^2(c+d x) (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=81 \[ -\frac{a 2^{m+\frac{3}{2}} \cos ^3(c+d x) (\sin (c+d x)+1)^{-m-\frac{1}{2}} (a \sin (c+d x)+a)^{m-1} \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0744746, antiderivative size = 81, normalized size of antiderivative = 1., 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} \[ -\frac{a 2^{m+\frac{3}{2}} \cos ^3(c+d x) (\sin (c+d x)+1)^{-m-\frac{1}{2}} (a \sin (c+d x)+a)^{m-1} \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^m,x]

[Out]

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

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]

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 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]))

Rubi steps

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

Mathematica [A]  time = 0.0942731, size = 78, normalized size = 0.96 \[ -\frac{2^{m+\frac{3}{2}} \cos ^3(c+d x) (\sin (c+d x)+1)^{-m-\frac{3}{2}} (a (\sin (c+d x)+1))^m \, _2F_1\left (\frac{3}{2},-m-\frac{1}{2};\frac{5}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sin[c + d*x])^m,x]

[Out]

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

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Maple [F]  time = 0.749, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sin(d*x+c))^m,x)

[Out]

int(cos(d*x+c)^2*(a+a*sin(d*x+c))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^m*cos(d*x + c)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((a*sin(d*x + c) + a)^m*cos(d*x + c)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (c + d x \right )} + 1\right )\right )^{m} \cos ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*sin(d*x+c))**m,x)

[Out]

Integral((a*(sin(c + d*x) + 1))**m*cos(c + d*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^m*cos(d*x + c)^2, x)