Integrand size = 23, antiderivative size = 91 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {3 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {3 a^3 \sqrt {a+a \sin (c+d x)}}{d}+\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d} \]
a*sec(d*x+c)^2*(a+a*sin(d*x+c))^(5/2)/d-3*a^(7/2)*arctanh(1/2*(a+a*sin(d*x +c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d+3*a^3*(a+a*sin(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.46 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {a \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (a+a \sin (c+d x))^{5/2}}{10 d} \]
Time = 0.37 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3155, 3042, 3146, 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.
\(\displaystyle \int \sec ^3(c+d x) (a \sin (c+d x)+a)^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^{7/2}}{\cos (c+d x)^3}dx\) |
\(\Big \downarrow \) 3155 |
\(\displaystyle \frac {a \sec ^2(c+d x) (a \sin (c+d x)+a)^{5/2}}{d}-\frac {3}{2} a^2 \int \sec (c+d x) (\sin (c+d x) a+a)^{3/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a \sec ^2(c+d x) (a \sin (c+d x)+a)^{5/2}}{d}-\frac {3}{2} a^2 \int \frac {(\sin (c+d x) a+a)^{3/2}}{\cos (c+d x)}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {a \sec ^2(c+d x) (a \sin (c+d x)+a)^{5/2}}{d}-\frac {3 a^3 \int \frac {\sqrt {\sin (c+d x) a+a}}{a-a \sin (c+d x)}d(a \sin (c+d x))}{2 d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a \sec ^2(c+d x) (a \sin (c+d x)+a)^{5/2}}{d}-\frac {3 a^3 \left (2 a \int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))-2 \sqrt {a \sin (c+d x)+a}\right )}{2 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {a \sec ^2(c+d x) (a \sin (c+d x)+a)^{5/2}}{d}-\frac {3 a^3 \left (4 a \int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}-2 \sqrt {a \sin (c+d x)+a}\right )}{2 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a \sec ^2(c+d x) (a \sin (c+d x)+a)^{5/2}}{d}-\frac {3 a^3 \left (2 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )-2 \sqrt {a \sin (c+d x)+a}\right )}{2 d}\) |
(a*Sec[c + d*x]^2*(a + a*Sin[c + d*x])^(5/2))/d - (3*a^3*(2*Sqrt[2]*Sqrt[a ]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[2]] - 2*Sqrt[a + a*Sin[c + d*x]]))/( 2*d)
3.2.48.3.1 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])
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] + Simp[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] && Int egersQ[2*m, 2*p]
Time = 29.89 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.91
method | result | size |
default | \(\frac {2 a^{3} \left (\sqrt {a +a \sin \left (d x +c \right )}+4 a \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 \left (a \sin \left (d x +c \right )-a \right )}-\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )}{d}\) | \(83\) |
2*a^3*((a+a*sin(d*x+c))^(1/2)+4*a*(-1/4*(a+a*sin(d*x+c))^(1/2)/(a*sin(d*x+ c)-a)-3/8*2^(1/2)/a^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/ 2))))/d
Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.27 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {3 \, \sqrt {2} {\left (a^{3} \sin \left (d x + c\right ) - a^{3}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, {\left (a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]
1/2*(3*sqrt(2)*(a^3*sin(d*x + c) - a^3)*sqrt(a)*log(-(a*sin(d*x + c) - 2*s qrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) - 1)) + 4*(a^ 3*sin(d*x + c) - 2*a^3)*sqrt(a*sin(d*x + c) + a))/(d*sin(d*x + c) - d)
Timed out. \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.23 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {3 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + 4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{4} - \frac {4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{5}}{a \sin \left (d x + c\right ) - a}}{2 \, a d} \]
1/2*(3*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*sqrt(a*sin(d*x + c) + a)* a^4 - 4*sqrt(a*sin(d*x + c) + a)*a^5/(a*sin(d*x + c) - a))/(a*d)
Time = 0.32 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.18 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {\sqrt {2} a^{\frac {7}{2}} {\left (\frac {2 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 4 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 3 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{2 \, d} \]
-1/2*sqrt(2)*a^(7/2)*(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) - 4*cos(-1/4*pi + 1/2*d*x + 1/2*c) + 3*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) - 3*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
Timed out. \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]