Integrand size = 21, antiderivative size = 110 \[ \int \sec (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {8 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 a^3 \sqrt {a+a \sin (c+d x)}}{d}-\frac {4 a^2 (a+a \sin (c+d x))^{3/2}}{3 d}-\frac {2 a (a+a \sin (c+d x))^{5/2}}{5 d} \]
-4/3*a^2*(a+a*sin(d*x+c))^(3/2)/d-2/5*a*(a+a*sin(d*x+c))^(5/2)/d+8*a^(7/2) *arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-8*a^3*(a+a* sin(d*x+c))^(1/2)/d
Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.77 \[ \int \sec (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {120 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a (1+\sin (c+d x))}}{\sqrt {2} \sqrt {a}}\right )-2 a^3 \sqrt {a (1+\sin (c+d x))} \left (73+16 \sin (c+d x)+3 \sin ^2(c+d x)\right )}{15 d} \]
(120*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])] - 2*a^3*Sqrt[a*(1 + Sin[c + d*x])]*(73 + 16*Sin[c + d*x] + 3*Sin[c + d*x] ^2))/(15*d)
Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.87, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3146, 60, 60, 60, 73, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec (c+d x) (a \sin (c+d x)+a)^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^{7/2}}{\cos (c+d x)}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {a \int \frac {(\sin (c+d x) a+a)^{5/2}}{a-a \sin (c+d x)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \left (2 a \int \frac {(\sin (c+d x) a+a)^{3/2}}{a-a \sin (c+d x)}d(a \sin (c+d x))-\frac {2}{5} (a \sin (c+d x)+a)^{5/2}\right )}{d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \left (2 a \left (2 a \int \frac {\sqrt {\sin (c+d x) a+a}}{a-a \sin (c+d x)}d(a \sin (c+d x))-\frac {2}{3} (a \sin (c+d x)+a)^{3/2}\right )-\frac {2}{5} (a \sin (c+d x)+a)^{5/2}\right )}{d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \left (2 a \left (2 a \left (2 a \int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))-2 \sqrt {a \sin (c+d x)+a}\right )-\frac {2}{3} (a \sin (c+d x)+a)^{3/2}\right )-\frac {2}{5} (a \sin (c+d x)+a)^{5/2}\right )}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {a \left (2 a \left (2 a \left (4 a \int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}-2 \sqrt {a \sin (c+d x)+a}\right )-\frac {2}{3} (a \sin (c+d x)+a)^{3/2}\right )-\frac {2}{5} (a \sin (c+d x)+a)^{5/2}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a \left (2 a \left (2 a \left (2 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )-2 \sqrt {a \sin (c+d x)+a}\right )-\frac {2}{3} (a \sin (c+d x)+a)^{3/2}\right )-\frac {2}{5} (a \sin (c+d x)+a)^{5/2}\right )}{d}\) |
(a*((-2*(a + a*Sin[c + d*x])^(5/2))/5 + 2*a*((-2*(a + a*Sin[c + d*x])^(3/2 ))/3 + 2*a*(2*Sqrt[2]*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[2]] - 2* Sqrt[a + a*Sin[c + d*x]]))))/d
3.2.46.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[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] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 1.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.75
method | result | size |
default | \(-\frac {2 a \left (\frac {\left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a^{2} \sqrt {a +a \sin \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\) | \(83\) |
-2*a*(1/5*(a+a*sin(d*x+c))^(5/2)+2/3*a*(a+a*sin(d*x+c))^(3/2)+4*a^2*(a+a*s in(d*x+c))^(1/2)-4*a^(5/2)*2^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1 /2)/a^(1/2)))/d
Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.93 \[ \int \sec (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 \, {\left (30 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + {\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 16 \, a^{3} \sin \left (d x + c\right ) - 76 \, a^{3}\right )} \sqrt {a \sin \left (d x + c\right ) + a}\right )}}{15 \, d} \]
2/15*(30*sqrt(2)*a^(7/2)*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)) + (3*a^3*cos(d*x + c)^2 - 16*a ^3*sin(d*x + c) - 76*a^3)*sqrt(a*sin(d*x + c) + a))/d
Timed out. \[ \int \sec (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05 \[ \int \sec (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {2 \, {\left (30 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + 3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} + 10 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 60 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{4}\right )}}{15 \, a d} \]
-2/15*(30*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))) + 3*(a*sin(d*x + c) + a)^( 5/2)*a^2 + 10*(a*sin(d*x + c) + a)^(3/2)*a^3 + 60*sqrt(a*sin(d*x + c) + a) *a^4)/(a*d)
Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.97 \[ \int \sec (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {4 \, \sqrt {2} {\left (6 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 15 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} a^{\frac {7}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{15 \, d} \]
-4/15*sqrt(2)*(6*cos(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 10*cos(-1/4*pi + 1/2*d *x + 1/2*c)^3 + 30*cos(-1/4*pi + 1/2*d*x + 1/2*c) - 15*log(cos(-1/4*pi + 1 /2*d*x + 1/2*c) + 1) + 15*log(-cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1))*a^(7/2 )*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d
Timed out. \[ \int \sec (c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{\cos \left (c+d\,x\right )} \,d x \]