\(\int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [150]

Optimal result

Integrand size = 23, antiderivative size = 106 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} d}-\frac {a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{8 d}+\frac {a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d} \]

[Out]

-1/8*a^2*sec(d*x+c)^2*(a+a*sin(d*x+c))^(3/2)/d+1/2*a*sec(d*x+c)^4*(a+a*sin(d*x+c))^(5/2)/d-1/16*a^(7/2)*arctan
h(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 106, 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 = {2755, 2754, 2746, 65, 212} \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {a^{7/2} \text {arctanh}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} d}-\frac {a^2 \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{8 d}+\frac {a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{2 d} \]

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

-1/8*(a^(7/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) - (a^2*Sec[c + d*x]^2*(a + a*Si
n[c + d*x])^(3/2))/(8*d) + (a*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2))/(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 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 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 ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d}-\frac {1}{4} a^2 \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx \\ & = -\frac {a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{8 d}+\frac {a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d}-\frac {1}{16} a^3 \int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{8 d}+\frac {a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d}-\frac {a^4 \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{16 d} \\ & = -\frac {a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{8 d}+\frac {a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d}-\frac {a^4 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{8 d} \\ & = -\frac {a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} d}-\frac {a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{8 d}+\frac {a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {-\sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a (1+\sin (c+d x))}}{\sqrt {2} \sqrt {a}}\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4+2 a^3 \sqrt {a (1+\sin (c+d x))} (3+\sin (c+d x))}{16 d (-1+\sin (c+d x))^2} \]

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(-(Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])]*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
^4) + 2*a^3*Sqrt[a*(1 + Sin[c + d*x])]*(3 + Sin[c + d*x]))/(16*d*(-1 + Sin[c + d*x])^2)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71

\[-\frac {2 a^{5} \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \left (3+\sin \left (d x +c \right )\right )}{16 \left (a \sin \left (d x +c \right )-a \right )^{2}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{32 a^{\frac {3}{2}}}\right )}{d}\]

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x)

[Out]

-2*a^5*(-1/16*(a+a*sin(d*x+c))^(1/2)*(3+sin(d*x+c))/(a*sin(d*x+c)-a)^2+1/32/a^(3/2)*2^(1/2)*arctanh(1/2*(a+a*s
in(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.37 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {{\left (\sqrt {2} a^{3} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} a^{3} \sin \left (d x + c\right ) - 2 \, \sqrt {2} a^{3}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (a^{3} \sin \left (d x + c\right ) + 3 \, a^{3}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{32 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/32*((sqrt(2)*a^3*cos(d*x + c)^2 + 2*sqrt(2)*a^3*sin(d*x + c) - 2*sqrt(2)*a^3)*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*(a^3*sin(d*x + c) + 3*a^3)*sqrt(a*s
in(d*x + c) + a))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.25 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {\sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left ({\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5} + 2 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{6}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )} a + 4 \, a^{2}}}{32 \, a d} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

1/32*(sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c)
+ a))) + 4*((a*sin(d*x + c) + a)^(3/2)*a^5 + 2*sqrt(a*sin(d*x + c) + a)*a^6)/((a*sin(d*x + c) + a)^2 - 4*(a*s
in(d*x + c) + a)*a + 4*a^2))/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.41 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {\sqrt {2} a^{\frac {7}{2}} {\left (\frac {4 \, {\left (\frac {1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\frac {1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2} - 4} - \log \left ({\left | \frac {1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) + \log \left ({\left | \frac {1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{64 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

1/64*sqrt(2)*a^(7/2)*(4*(1/cos(-1/4*pi + 1/2*d*x + 1/2*c) + cos(-1/4*pi + 1/2*d*x + 1/2*c))/((1/cos(-1/4*pi +
1/2*d*x + 1/2*c) + cos(-1/4*pi + 1/2*d*x + 1/2*c))^2 - 4) - log(abs(1/cos(-1/4*pi + 1/2*d*x + 1/2*c) + cos(-1/
4*pi + 1/2*d*x + 1/2*c) + 2)) + log(abs(1/cos(-1/4*pi + 1/2*d*x + 1/2*c) + cos(-1/4*pi + 1/2*d*x + 1/2*c) - 2)
))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d

Mupad [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \]

[In]

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^5,x)

[Out]

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^5, x)