\(\int (e \cos (c+d x))^{1-m} (a+a \sin (c+d x))^m \, dx\) [365]

Optimal result

Integrand size = 27, antiderivative size = 97 \[ \int (e \cos (c+d x))^{1-m} (a+a \sin (c+d x))^m \, dx=\frac {2^{1-\frac {m}{2}} (e \cos (c+d x))^{2-m} \operatorname {Hypergeometric2F1}\left (\frac {m}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (1-\sin (c+d x))^{-1+\frac {m}{2}} (a+a \sin (c+d x))^m}{d e (2+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2768, 72, 71} \[ \int (e \cos (c+d x))^{1-m} (a+a \sin (c+d x))^m \, dx=\frac {2^{1-\frac {m}{2}} (1-\sin (c+d x))^{\frac {m}{2}-1} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{2-m} \operatorname {Hypergeometric2F1}\left (\frac {m}{2},\frac {m+2}{2},\frac {m+4}{2},\frac {1}{2} (\sin (c+d x)+1)\right )}{d e (m+2)} \]

[In]

Int[(e*Cos[c + d*x])^(1 - m)*(a + a*Sin[c + d*x])^m,x]

[Out]

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

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))^{2-m} (a-a \sin (c+d x))^{\frac {1}{2} (-2+m)} (a+a \sin (c+d x))^{\frac {1}{2} (-2+m)}\right ) \text {Subst}\left (\int (a-a x)^{-m/2} (a+a x)^{m/2} \, dx,x,\sin (c+d x)\right )}{d e} \\ & = \frac {\left (2^{-m/2} a^2 (e \cos (c+d x))^{2-m} (a-a \sin (c+d x))^{\frac {1}{2} (-2+m)-\frac {m}{2}} \left (\frac {a-a \sin (c+d x)}{a}\right )^{m/2} (a+a \sin (c+d x))^{\frac {1}{2} (-2+m)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-m/2} (a+a x)^{m/2} \, dx,x,\sin (c+d x)\right )}{d e} \\ & = \frac {2^{1-\frac {m}{2}} (e \cos (c+d x))^{2-m} \operatorname {Hypergeometric2F1}\left (\frac {m}{2},\frac {2+m}{2},\frac {4+m}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (1-\sin (c+d x))^{-1+\frac {m}{2}} (a+a \sin (c+d x))^m}{d e (2+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00 \[ \int (e \cos (c+d x))^{1-m} (a+a \sin (c+d x))^m \, dx=\frac {2^{1+\frac {m}{2}} (e \cos (c+d x))^{2-m} \operatorname {Hypergeometric2F1}\left (1-\frac {m}{2},-\frac {m}{2},2-\frac {m}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-1-\frac {m}{2}} (a (1+\sin (c+d x)))^m}{d e (-2+m)} \]

[In]

Integrate[(e*Cos[c + d*x])^(1 - m)*(a + a*Sin[c + d*x])^m,x]

[Out]

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

Maple [F]

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

[In]

int((e*cos(d*x+c))^(1-m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(1-m)*(a+a*sin(d*x+c))^m,x)

Fricas [F]

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

[In]

integrate((e*cos(d*x+c))^(1-m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((e*cos(d*x + c))^(-m + 1)*(a*sin(d*x + c) + a)^m, x)

Sympy [F]

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

[In]

integrate((e*cos(d*x+c))**(1-m)*(a+a*sin(d*x+c))**m,x)

[Out]

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

Maxima [F]

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

[In]

integrate((e*cos(d*x+c))^(1-m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(-m + 1)*(a*sin(d*x + c) + a)^m, x)

Giac [F]

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

[In]

integrate((e*cos(d*x+c))^(1-m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-m + 1)*(a*sin(d*x + c) + a)^m, x)

Mupad [F(-1)]

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

[In]

int((e*cos(c + d*x))^(1 - m)*(a + a*sin(c + d*x))^m,x)

[Out]

int((e*cos(c + d*x))^(1 - m)*(a + a*sin(c + d*x))^m, x)