\(\int \sec ^2(c+d x) (a+a \sin (c+d x))^m \, dx\) [352]

Optimal result

Integrand size = 21, antiderivative size = 73 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {2^{-\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec (c+d x) (1+\sin (c+d x))^{\frac {1}{2}-m} (a+a \sin (c+d x))^m}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2768, 72, 71} \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {2^{m-\frac {1}{2}} \sec (c+d x) (\sin (c+d x)+1)^{\frac {1}{2}-m} (a \sin (c+d x)+a)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{d} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {2^{-\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec (c+d x) (1+\sin (c+d x))^{\frac {1}{2}-m} (a (1+\sin (c+d x)))^m}{d} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (\sec ^{2}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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