\(\int (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^m \, dx\) [372]

Optimal result

Integrand size = 27, antiderivative size = 89 \[ \int (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^m \, dx=\frac {2^{\frac {5}{2}-m} (e \cos (c+d x))^{5-2 m} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2} (-3+2 m),\frac {7}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (1-\sin (c+d x))^{-\frac {5}{2}+m} (a+a \sin (c+d x))^m}{5 d e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2768, 7, 72, 71} \[ \int (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^m \, dx=\frac {2^{\frac {5}{2}-m} (1-\sin (c+d x))^{m-\frac {5}{2}} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{5-2 m} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2} (2 m-3),\frac {7}{2},\frac {1}{2} (\sin (c+d x)+1)\right )}{5 d e} \]

[In]

Int[(e*Cos[c + d*x])^(4 - 2*m)*(a + a*Sin[c + d*x])^m,x]

[Out]

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

Rule 7

Int[(u_.)*(Px_)^(p_), x_Symbol] :> Int[u*Px^Simplify[p], x] /; PolyQ[Px, x] &&  !RationalQ[p] && FreeQ[p, x] &
& RationalQ[Simplify[p]]

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 2768

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*} \text {integral}& = \frac {\left (a^2 (e \cos (c+d x))^{5-2 m} (a-a \sin (c+d x))^{\frac {1}{2} (-5+2 m)} (a+a \sin (c+d x))^{\frac {1}{2} (-5+2 m)}\right ) \text {Subst}\left (\int (a-a x)^{\frac {1}{2} (3-2 m)} (a+a x)^{\frac {1}{2} (3-2 m)+m} \, dx,x,\sin (c+d x)\right )}{d e} \\ & = \frac {\left (a^2 (e \cos (c+d x))^{5-2 m} (a-a \sin (c+d x))^{\frac {1}{2} (-5+2 m)} (a+a \sin (c+d x))^{\frac {1}{2} (-5+2 m)}\right ) \text {Subst}\left (\int (a-a x)^{\frac {1}{2} (3-2 m)} (a+a x)^{3/2} \, dx,x,\sin (c+d x)\right )}{d e} \\ & = \frac {\left (2^{\frac {3}{2}-m} a^3 (e \cos (c+d x))^{5-2 m} (a-a \sin (c+d x))^{\frac {1}{2}-m+\frac {1}{2} (-5+2 m)} \left (\frac {a-a \sin (c+d x)}{a}\right )^{-\frac {1}{2}+m} (a+a \sin (c+d x))^{\frac {1}{2} (-5+2 m)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{\frac {1}{2} (3-2 m)} (a+a x)^{3/2} \, dx,x,\sin (c+d x)\right )}{d e} \\ & = \frac {2^{\frac {5}{2}-m} (e \cos (c+d x))^{5-2 m} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2} (-3+2 m),\frac {7}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (1-\sin (c+d x))^{-\frac {5}{2}+m} (a+a \sin (c+d x))^m}{5 d e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.08 \[ \int (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^m \, dx=\frac {4 \sqrt {2} e^4 \cos ^5(c+d x) (e \cos (c+d x))^{-2 m} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-m,\frac {7}{2}-m,\frac {1}{2} (1-\sin (c+d x))\right ) (a (1+\sin (c+d x)))^m}{d (-5+2 m) (1+\sin (c+d x))^{5/2}} \]

[In]

Integrate[(e*Cos[c + d*x])^(4 - 2*m)*(a + a*Sin[c + d*x])^m,x]

[Out]

(4*Sqrt[2]*e^4*Cos[c + d*x]^5*Hypergeometric2F1[-3/2, 5/2 - m, 7/2 - m, (1 - Sin[c + d*x])/2]*(a*(1 + Sin[c +
d*x]))^m)/(d*(-5 + 2*m)*(e*Cos[c + d*x])^(2*m)*(1 + Sin[c + d*x])^(5/2))

Maple [F]

\[\int \left (e \cos \left (d x +c \right )\right )^{4-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]

[In]

int((e*cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c))^m,x)

Fricas [F]

\[ \int (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-2 \, m + 4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((e*cos(d*x + c))^(-2*m + 4)*(a*sin(d*x + c) + a)^m, x)

Sympy [F]

\[ \int (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^m \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{4 - 2 m}\, dx \]

[In]

integrate((e*cos(d*x+c))**(4-2*m)*(a+a*sin(d*x+c))**m,x)

[Out]

Integral((a*(sin(c + d*x) + 1))**m*(e*cos(c + d*x))**(4 - 2*m), x)

Maxima [F]

\[ \int (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-2 \, m + 4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(-2*m + 4)*(a*sin(d*x + c) + a)^m, x)

Giac [F]

\[ \int (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-2 \, m + 4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-2*m + 4)*(a*sin(d*x + c) + a)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^m \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^{4-2\,m}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m \,d x \]

[In]

int((e*cos(c + d*x))^(4 - 2*m)*(a + a*sin(c + d*x))^m,x)

[Out]

int((e*cos(c + d*x))^(4 - 2*m)*(a + a*sin(c + d*x))^m, x)