\(\int (a+a \sec (c+d x))^n \sqrt {\sin (c+d x)} \, dx\) [157]

Optimal result

Integrand size = 23, antiderivative size = 105 \[ \int (a+a \sec (c+d x))^n \sqrt {\sin (c+d x)} \, dx=-\frac {\operatorname {AppellF1}\left (1-n,\frac {1}{4},\frac {1}{4}-n,2-n,\cos (c+d x),-\cos (c+d x)\right ) \sqrt [4]{1-\cos (c+d x)} \cos (c+d x) (1+\cos (c+d x))^{\frac {1}{4}-n} (a+a \sec (c+d x))^n}{d (1-n) \sqrt {\sin (c+d x)}} \]

[Out]

-AppellF1(1-n,1/4-n,1/4,2-n,-cos(d*x+c),cos(d*x+c))*(1-cos(d*x+c))^(1/4)*cos(d*x+c)*(1+cos(d*x+c))^(1/4-n)*(a+
a*sec(d*x+c))^n/d/(1-n)/sin(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3961, 2965, 140, 138} \[ \int (a+a \sec (c+d x))^n \sqrt {\sin (c+d x)} \, dx=-\frac {\sqrt [4]{1-\cos (c+d x)} \cos (c+d x) (\cos (c+d x)+1)^{\frac {1}{4}-n} (a \sec (c+d x)+a)^n \operatorname {AppellF1}\left (1-n,\frac {1}{4},\frac {1}{4}-n,2-n,\cos (c+d x),-\cos (c+d x)\right )}{d (1-n) \sqrt {\sin (c+d x)}} \]

[In]

Int[(a + a*Sec[c + d*x])^n*Sqrt[Sin[c + d*x]],x]

[Out]

-((AppellF1[1 - n, 1/4, 1/4 - n, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 - Cos[c + d*x])^(1/4)*Cos[c + d*x]*(1
+ Cos[c + d*x])^(1/4 - n)*(a + a*Sec[c + d*x])^n)/(d*(1 - n)*Sqrt[Sin[c + d*x]]))

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 140

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[c^IntPart[n]*((c +
d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 2965

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[g*((g*Cos[e + f*x])^(p - 1)/(f*(a + b*Sin[e + f*x])^((p - 1)/2)*(a - b*Sin[e +
f*x])^((p - 1)/2))), Subst[Int[(d*x)^n*(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]],
x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rule 3961

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Sin[e
+ f*x]^FracPart[m]*((a + b*Csc[e + f*x])^FracPart[m]/(b + a*Sin[e + f*x])^FracPart[m]), Int[(g*Cos[e + f*x])^
p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && (EqQ[a^2 - b^2, 0] ||
IntegersQ[2*m, p])

Rubi steps \begin{align*} \text {integral}& = \left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int (-\cos (c+d x))^{-n} (-a-a \cos (c+d x))^n \sqrt {\sin (c+d x)} \, dx \\ & = -\frac {\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{\frac {1}{4}-n} \sqrt [4]{-a+a \cos (c+d x)} (a+a \sec (c+d x))^n\right ) \text {Subst}\left (\int \frac {(-x)^{-n} (-a-a x)^{-\frac {1}{4}+n}}{\sqrt [4]{-a+a x}} \, dx,x,\cos (c+d x)\right )}{d \sqrt {\sin (c+d x)}} \\ & = -\frac {\left ((-\cos (c+d x))^n (1+\cos (c+d x))^{\frac {1}{4}-n} \sqrt [4]{-a+a \cos (c+d x)} (a+a \sec (c+d x))^n\right ) \text {Subst}\left (\int \frac {(-x)^{-n} (1+x)^{-\frac {1}{4}+n}}{\sqrt [4]{-a+a x}} \, dx,x,\cos (c+d x)\right )}{d \sqrt {\sin (c+d x)}} \\ & = -\frac {\left (\sqrt [4]{1-\cos (c+d x)} (-\cos (c+d x))^n (1+\cos (c+d x))^{\frac {1}{4}-n} (a+a \sec (c+d x))^n\right ) \text {Subst}\left (\int \frac {(-x)^{-n} (1+x)^{-\frac {1}{4}+n}}{\sqrt [4]{1-x}} \, dx,x,\cos (c+d x)\right )}{d \sqrt {\sin (c+d x)}} \\ & = -\frac {\operatorname {AppellF1}\left (1-n,\frac {1}{4},\frac {1}{4}-n,2-n,\cos (c+d x),-\cos (c+d x)\right ) \sqrt [4]{1-\cos (c+d x)} \cos (c+d x) (1+\cos (c+d x))^{\frac {1}{4}-n} (a+a \sec (c+d x))^n}{d (1-n) \sqrt {\sin (c+d x)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(214\) vs. \(2(105)=210\).

Time = 3.04 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.04 \[ \int (a+a \sec (c+d x))^n \sqrt {\sin (c+d x)} \, dx=\frac {14 \operatorname {AppellF1}\left (\frac {3}{4},n,\frac {3}{2},\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x)) (a (1+\sec (c+d x)))^n \sin ^{\frac {3}{2}}(c+d x)}{d \left (6 \left (3 \operatorname {AppellF1}\left (\frac {7}{4},n,\frac {5}{2},\frac {11}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 n \operatorname {AppellF1}\left (\frac {7}{4},1+n,\frac {3}{2},\frac {11}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+21 \operatorname {AppellF1}\left (\frac {3}{4},n,\frac {3}{2},\frac {7}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \]

[In]

Integrate[(a + a*Sec[c + d*x])^n*Sqrt[Sin[c + d*x]],x]

[Out]

(14*AppellF1[3/4, n, 3/2, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])*(a*(1 + Sec[c + d*x
]))^n*Sin[c + d*x]^(3/2))/(d*(6*(3*AppellF1[7/4, n, 5/2, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*n*
AppellF1[7/4, 1 + n, 3/2, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + 21*AppellF1[3/
4, n, 3/2, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])))

Maple [F]

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

[In]

int((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x)

[Out]

int((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x)

Fricas [F]

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

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*sqrt(sin(d*x + c)), x)

Sympy [F]

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

[In]

integrate((a+a*sec(d*x+c))**n*sin(d*x+c)**(1/2),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**n*sqrt(sin(c + d*x)), x)

Maxima [F]

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

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*sqrt(sin(d*x + c)), x)

Giac [F]

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

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*sqrt(sin(d*x + c)), x)

Mupad [F(-1)]

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

[In]

int(sin(c + d*x)^(1/2)*(a + a/cos(c + d*x))^n,x)

[Out]

int(sin(c + d*x)^(1/2)*(a + a/cos(c + d*x))^n, x)