\(\int \sec (c+d x) (a+b \sin (c+d x))^m \, dx\) [634]

Optimal result

Integrand size = 19, antiderivative size = 115 \[ \int \sec (c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a-b}\right ) (a+b \sin (c+d x))^{1+m}}{2 (a-b) d (1+m)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a+b}\right ) (a+b \sin (c+d x))^{1+m}}{2 (a+b) d (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2747, 726, 70} \[ \int \sec (c+d x) (a+b \sin (c+d x))^m \, dx=\frac {(a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \sin (c+d x)}{a+b}\right )}{2 d (m+1) (a+b)}-\frac {(a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \sin (c+d x)}{a-b}\right )}{2 d (m+1) (a-b)} \]

[In]

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

[Out]

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

Rule 70

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

Rule 726

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b \text {Subst}\left (\int \frac {(a+x)^m}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b \text {Subst}\left (\int \left (\frac {(a+x)^m}{2 b (b-x)}+\frac {(a+x)^m}{2 b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {(a+x)^m}{b-x} \, dx,x,b \sin (c+d x)\right )}{2 d}+\frac {\text {Subst}\left (\int \frac {(a+x)^m}{b+x} \, dx,x,b \sin (c+d x)\right )}{2 d} \\ & = -\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a-b}\right ) (a+b \sin (c+d x))^{1+m}}{2 (a-b) d (1+m)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a+b}\right ) (a+b \sin (c+d x))^{1+m}}{2 (a+b) d (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86 \[ \int \sec (c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {\left ((a+b) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a-b}\right )+(-a+b) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \sin (c+d x)}{a+b}\right )\right ) (a+b \sin (c+d x))^{1+m}}{2 (a-b) (a+b) d (1+m)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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