\(\int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [108]

Optimal result

Integrand size = 21, antiderivative size = 40 \[ \int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \]

[Out]

arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 40, 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 = {2746, 65, 212} \[ \int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{d} \]

[In]

Int[Sec[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2746

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \]

[In]

Integrate[Sec[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, \sqrt {a}}{d}\)	\(32\)

[In]

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

[Out]

arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.30 \[ \int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\left [\frac {\sqrt {2} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right )}{2 \, d}, -\frac {\sqrt {2} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a}}{\sqrt {a \sin \left (d x + c\right ) + a}}\right )}{d}\right ] \]

[In]

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

[Out]

[1/2*sqrt(2)*sqrt(a)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) -
1))/d, -sqrt(2)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)/sqrt(a*sin(d*x + c) + a))/d]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.45 \[ \int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right )}{2 \, d} \]

[In]

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

[Out]

-1/2*sqrt(2)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c)
+ a)))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.48 \[ \int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} \sqrt {a} {\left (\log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{2 \, d} \]

[In]

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

[Out]

1/2*sqrt(2)*sqrt(a)*(log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) - log(-cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1))*sgn(c
os(-1/4*pi + 1/2*d*x + 1/2*c))/d

Mupad [F(-1)]

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

[In]

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

[Out]

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