\(\int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [134]

Optimal result

Integrand size = 23, antiderivative size = 69 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} d}+\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 69, 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 = {2755, 2746, 65, 212} \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {a \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{d}-\frac {a^{5/2} \text {arctanh}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} d} \]

[In]

Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

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

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])

Rule 2755

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

Rubi steps \begin{align*} \text {integral}& = \frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac {1}{2} a^2 \int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac {a^3 \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{2 d} \\ & = \frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{d}-\frac {a^3 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{d} \\ & = -\frac {a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} d}+\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {a^2 \left (-\sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a (1+\sin (c+d x))}}{\sqrt {2} \sqrt {a}}\right )-\frac {2 \sqrt {a (1+\sin (c+d x))}}{-1+\sin (c+d x)}\right )}{2 d} \]

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 31.83 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {a^{3} \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}}{a \sin \left (d x +c \right )-a}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}\right )}{d}\)	\(66\)

[In]

int(sec(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-a^3*((a+a*sin(d*x+c))^(1/2)/(a*sin(d*x+c)-a)+1/2*2^(1/2)/a^(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 = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.48 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {\sqrt {2} {\left (a^{2} \sin \left (d x + c\right ) - a^{2}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{2}}{4 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**3*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.36 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {\sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) - \frac {4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{4}}{a \sin \left (d x + c\right ) - a}}{4 \, a d} \]

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/4*(sqrt(2)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c)
+ a))) - 4*sqrt(a*sin(d*x + c) + a)*a^4/(a*sin(d*x + c) - a))/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.32 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} a^{\frac {5}{2}} {\left (\frac {2 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{4 \, d} \]

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/4*sqrt(2)*a^(5/2)*(2*cos(-1/4*pi + 1/2*d*x + 1/2*c)/(cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1) + log(cos(-1/4*p
i + 1/2*d*x + 1/2*c) + 1) - log(-cos(-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. \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]

[In]

int((a + a*sin(c + d*x))^(5/2)/cos(c + d*x)^3,x)

[Out]

int((a + a*sin(c + d*x))^(5/2)/cos(c + d*x)^3, x)