\(\int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [122]

Optimal result

Integrand size = 23, antiderivative size = 26 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 a \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2752} \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 a \sec (c+d x) \sqrt {a \sin (c+d x)+a}}{d} \]

[In]

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

[Out]

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

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 a \sec (c+d x) \sqrt {a (1+\sin (c+d x))}}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42

method	result	size
default	\(\frac {2 a^{2} \left (1+\sin \left (d x +c \right )\right )}{\cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, \sqrt {a \sin \left (d x + c\right ) + a} a}{d \cos \left (d x + c\right )} \]

[In]

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

[Out]

2*sqrt(a*sin(d*x + c) + a)*a/(d*cos(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (24) = 48\).

Time = 0.33 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.77 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (a^{\frac {3}{2}} + \frac {2 \, a^{\frac {3}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{\frac {3}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )}}{d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {\sqrt {2} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.79 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {4\,a\,\cos \left (c+d\,x\right )\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}}{d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

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

[Out]

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