\(\int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [164]

Optimal result

Integrand size = 23, antiderivative size = 102 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{4 \sqrt {2} \sqrt {a} d}-\frac {3 a \cos (c+d x)}{4 d (a+a \sin (c+d x))^{3/2}}+\frac {\sec (c+d x)}{d \sqrt {a+a \sin (c+d x)}} \]

[Out]

-3/4*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-3/8*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*
2^(1/2)/d/a^(1/2)+sec(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2766, 2729, 2728, 212} \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{4 \sqrt {2} \sqrt {a} d}-\frac {3 a \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[Sec[c + d*x]^2/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-3*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(4*Sqrt[2]*Sqrt[a]*d) - (3*a*Cos[c + d
*x])/(4*d*(a + a*Sin[c + d*x])^(3/2)) + Sec[c + d*x]/(d*Sqrt[a + a*Sin[c + d*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 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2766

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*((
g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[a*((2*p + 1)/(2*g^2*(p + 1))), In
t[(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0
] && LtQ[p, -1] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {\sec (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {1}{2} (3 a) \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx \\ & = -\frac {3 a \cos (c+d x)}{4 d (a+a \sin (c+d x))^{3/2}}+\frac {\sec (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {3}{8} \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\frac {3 a \cos (c+d x)}{4 d (a+a \sin (c+d x))^{3/2}}+\frac {\sec (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {3 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 d} \\ & = -\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{4 \sqrt {2} \sqrt {a} d}-\frac {3 a \cos (c+d x)}{4 d (a+a \sin (c+d x))^{3/2}}+\frac {\sec (c+d x)}{d \sqrt {a+a \sin (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.51 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec (c+d x) \sqrt {a (1+\sin (c+d x))}}{2 a d} \]

[In]

Integrate[Sec[c + d*x]^2/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.31

method	result	size
default	\(-\frac {3 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (d x +c \right ) a +3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sqrt {a -a \sin \left (d x +c \right )}-6 a^{\frac {3}{2}} \sin \left (d x +c \right )-2 a^{\frac {3}{2}}}{8 a^{\frac {3}{2}} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(134\)

[In]

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

[Out]

-1/8*(3*(a-a*sin(d*x+c))^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+c)*a+3*2^(1
/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a*(a-a*sin(d*x+c))^(1/2)-6*a^(3/2)*sin(d*x+c)-2*a^(3/2
))/a^(3/2)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (85) = 170\).

Time = 0.30 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.96 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right ) \sin \left (d x + c\right ) + \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (3 \, \sin \left (d x + c\right ) + 1\right )}}{16 \, {\left (a d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/16*(3*sqrt(2)*(cos(d*x + c)*sin(d*x + c) + cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*s
in(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x
+ c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 4*sqrt(a*sin(d*x + c) +
a)*(3*sin(d*x + c) + 1))/(a*d*cos(d*x + c)*sin(d*x + c) + a*d*cos(d*x + c))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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