\(\int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [176]

Optimal result

Integrand size = 21, antiderivative size = 89 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {1}{3 d (a+a \sin (c+d x))^{3/2}}-\frac {1}{2 a d \sqrt {a+a \sin (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2746, 53, 65, 212} \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {1}{2 a d \sqrt {a \sin (c+d x)+a}}-\frac {1}{3 d (a \sin (c+d x)+a)^{3/2}} \]

[In]

Int[Sec[c + d*x]/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(2*Sqrt[2]*a^(3/2)*d) - 1/(3*d*(a + a*Sin[c + d*x])^(3/2))
- 1/(2*a*d*Sqrt[a + a*Sin[c + d*x]])

Rule 53

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

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) (a+x)^{5/2}} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {1}{3 d (a+a \sin (c+d x))^{3/2}}+\frac {\text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{2 d} \\ & = -\frac {1}{3 d (a+a \sin (c+d x))^{3/2}}-\frac {1}{2 a d \sqrt {a+a \sin (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{4 a d} \\ & = -\frac {1}{3 d (a+a \sin (c+d x))^{3/2}}-\frac {1}{2 a d \sqrt {a+a \sin (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{2 a d} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {1}{3 d (a+a \sin (c+d x))^{3/2}}-\frac {1}{2 a 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 = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.46 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right )}{3 d (a+a \sin (c+d x))^{3/2}} \]

[In]

Integrate[Sec[c + d*x]/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {a \left (\frac {1}{2 a^{2} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {1}{3 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 a^{\frac {5}{2}}}\right )}{d}\)	\(71\)

[In]

int(sec(d*x+c)/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-a*(1/2/a^2/(a+a*sin(d*x+c))^(1/2)+1/3/a/(a+a*sin(d*x+c))^(3/2)-1/4/a^(5/2)*2^(1/2)*arctanh(1/2*(a+a*sin(d*x+c
))^(1/2)*2^(1/2)/a^(1/2)))/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.48 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (3 \, \sin \left (d x + c\right ) + 5\right )}}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - 2 \, a^{2} d \sin \left (d x + c\right ) - 2 \, a^{2} d\right )}} \]

[In]

integrate(sec(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/24*(3*sqrt(2)*(cos(d*x + c)^2 - 2*sin(d*x + c) - 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)) + 4*sqrt(a*sin(d*x + c) + a)*(3*sin(d*x + c) + 5))/(a^2*d*cos(d*x
+ c)^2 - 2*a^2*d*sin(d*x + c) - 2*a^2*d)

Sympy [F]

\[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)/(a*(sin(c + d*x) + 1))**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.02 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (3 \, a \sin \left (d x + c\right ) + 5 \, a\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}}{24 \, a d} \]

[In]

integrate(sec(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-1/24*(3*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a)
))/sqrt(a) + 4*(3*a*sin(d*x + c) + 5*a)/(a*sin(d*x + c) + a)^(3/2))/(a*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (70) = 140\).

Time = 0.40 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.60 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {a} {\left (\frac {3 \, \sqrt {2} \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {3 \, \sqrt {2} \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (3 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{24 \, d} \]

[In]

integrate(sec(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/24*sqrt(a)*(3*sqrt(2)*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 3*
sqrt(2)*log(-cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 2*sqrt(2)*(3*cos(
-1/4*pi + 1/2*d*x + 1/2*c)^2 + 1)/(a^2*cos(-1/4*pi + 1/2*d*x + 1/2*c)^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/
d

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(1/(cos(c + d*x)*(a + a*sin(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)*(a + a*sin(c + d*x))^(3/2)), x)