\(\int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [124]

Optimal result

Integrand size = 23, antiderivative size = 107 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{2 \sqrt {2} d}+\frac {a \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{2 d}+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2754, 2728, 212} \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} d}+\frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}+\frac {a \sec (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d} \]

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

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

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

Mathematica [C] (verified)

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

Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.48 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^3(c+d x) (a (1+\sin (c+d x)))^{3/2}}{3 d} \]

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\left (1+\sin \left (d x +c \right )\right ) \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-10 a^{\frac {7}{2}}+6 \sin \left (d x +c \right ) a^{\frac {7}{2}}\right )}{12 a^{\frac {3}{2}} \left (\sin \left (d x +c \right )-1\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(107\)

[In]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/12/a^(3/2)*(1+sin(d*x+c))/(sin(d*x+c)-1)*(3*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2*
(a-a*sin(d*x+c))^(3/2)-10*a^(7/2)+6*sin(d*x+c)*a^(7/2))/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. 215 vs. \(2 (88) = 176\).

Time = 0.31 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.01 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {3 \, {\left (\sqrt {2} a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \sqrt {2} a \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (3 \, a \sin \left (d x + c\right ) - 5 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{24 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**4*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {\sqrt {2} a^{\frac {3}{2}} {\left (\frac {2 \, {\left (3 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - 3 \, \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 3 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{24 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \]

[In]

int((a + a*sin(c + d*x))^(3/2)/cos(c + d*x)^4,x)

[Out]

int((a + a*sin(c + d*x))^(3/2)/cos(c + d*x)^4, x)