\(\int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [111]

Optimal result

Integrand size = 23, antiderivative size = 137 \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {5 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{8 \sqrt {2} d}-\frac {5 a^2 \cos (c+d x)}{8 d (a+a \sin (c+d x))^{3/2}}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \]

[Out]

-5/8*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-5/16*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2
))*a^(1/2)/d*2^(1/2)+5/6*a*sec(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+1/3*sec(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2754, 2766, 2729, 2728, 212} \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {5 a^2 \cos (c+d x)}{8 d (a \sin (c+d x)+a)^{3/2}}-\frac {5 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{8 \sqrt {2} d}+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-5*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(8*Sqrt[2]*d) - (5*a^2*Cos[c +
d*x])/(8*d*(a + a*Sin[c + d*x])^(3/2)) + (5*a*Sec[c + d*x])/(6*d*Sqrt[a + a*Sin[c + d*x]]) + (Sec[c + d*x]^3*
Sqrt[a + a*Sin[c + d*x]])/(3*d)

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 2754

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

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 ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {1}{6} (5 a) \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = \frac {5 a \sec (c+d x)}{6 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {1}{4} \left (5 a^2\right ) \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx \\ & = -\frac {5 a^2 \cos (c+d x)}{8 d (a+a \sin (c+d x))^{3/2}}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {1}{16} (5 a) \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\frac {5 a^2 \cos (c+d x)}{8 d (a+a \sin (c+d x))^{3/2}}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {(5 a) \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 )}{8 d} \\ & = -\frac {5 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{8 \sqrt {2} d}-\frac {5 a^2 \cos (c+d x)}{8 d (a+a \sin (c+d x))^{3/2}}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.39 \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^3(c+d x) (a (1+\sin (c+d x)))^{3/2}}{6 a d} \]

[In]

Integrate[Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Hypergeometric2F1[-3/2, 2, -1/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^3*(a*(1 + Sin[c + d*x]))^(3/2))/(6*a*d)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {15 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (d x +c \right ) a -30 a^{\frac {5}{2}} \left (\cos ^{2}\left (d x +c \right )\right )+15 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -20 a^{\frac {5}{2}} \sin \left (d x +c \right )+4 a^{\frac {5}{2}}}{48 a^{\frac {3}{2}} \left (\sin \left (d x +c \right )-1\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(157\)

[In]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/48/a^(3/2)*(15*(a-a*sin(d*x+c))^(3/2)*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-30*a^(5/2)*cos(d*x+c)^2+15*(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2
))*a-20*a^(5/2)*sin(d*x+c)+4*a^(5/2))/(sin(d*x+c)-1)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.37 \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {15 \, \sqrt {2} \sqrt {a} \cos \left (d x + c\right )^{3} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (15 \, \cos \left (d x + c\right )^{2} + 10 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{96 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/96*(15*sqrt(2)*sqrt(a)*cos(d*x + c)^3*log(-(a*cos(d*x + c)^2 - 2*sqrt(a*sin(d*x + c) + a)*(sqrt(2)*cos(d*x +
c) - sqrt(2)*sin(d*x + c) + sqrt(2))*sqrt(a) + 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*(15*cos(d*x + c)^2 + 10*sin(d*x + c
) - 2)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^3)

Sympy [F]

\[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sec ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**4*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*sec(c + d*x)**4, x)

Maxima [F]

\[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*sec(d*x + c)^4, x)

Giac [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.30 \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (15 \, \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 15 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {6 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {4 \, {\left (6 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )}}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}\right )} \sqrt {a}}{96 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/96*sqrt(2)*(15*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 15*log(-sin(-1/
4*pi + 1/2*d*x + 1/2*c) + 1)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 6*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-
1/4*pi + 1/2*d*x + 1/2*c)/(sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1) - 4*(6*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*si
n(-1/4*pi + 1/2*d*x + 1/2*c)^2 + sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))/sin(-1/4*pi + 1/2*d*x + 1/2*c)^3)*sqrt(a
)/d

Mupad [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^4} \,d x \]

[In]

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

[Out]

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