Integrand size = 23, antiderivative size = 169 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {7 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} d}-\frac {7 a^3 \cos (c+d x)}{16 d (a+a \sin (c+d x))^{3/2}}+\frac {7 a^2 \sec (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{30 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{5 d} \]
-7/16*a^3*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)+1/5*sec(d*x+c)^5*(a+a*sin(d* x+c))^(3/2)/d-7/32*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)+7/12*a^2*sec(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+7/3 0*a*sec(d*x+c)^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 = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.32 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},2,-\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^5(c+d x) (a (1+\sin (c+d x)))^{5/2}}{10 a d} \]
(Hypergeometric2F1[-5/2, 2, -3/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^5*(a* (1 + Sin[c + d*x]))^(5/2))/(10*a*d)
Time = 0.73 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3154, 3042, 3154, 3042, 3166, 3042, 3129, 3042, 3128, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^{3/2}}{\cos (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {7}{10} a \int \sec ^4(c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{10} a \int \frac {\sqrt {\sin (c+d x) a+a}}{\cos (c+d x)^4}dx+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \int \frac {\sec ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \int \frac {1}{\cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3}{2} a \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3}{2} a \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3}{2} a \left (-\frac {\int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{2 a d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3}{2} a \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d}\) |
(Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(5*d) + (7*a*((Sec[c + d*x]^3* Sqrt[a + a*Sin[c + d*x]])/(3*d) + (5*a*(Sec[c + d*x]/(d*Sqrt[a + a*Sin[c + d*x]]) + (3*a*(-1/2*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Si n[c + d*x]])]/(Sqrt[2]*a^(3/2)*d) - Cos[c + d*x]/(2*d*(a + a*Sin[c + d*x]) ^(3/2))))/2))/6))/10
3.2.26.3.1 Defintions of rubi rules used
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[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
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] + Simp[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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_. )*(x_)]], x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*S qrt[a + b*Sin[e + f*x]])), x] + Simp[a*((2*p + 1)/(2*g^2*(p + 1))) Int[(g *Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e , f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*p]
Time = 0.50 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.04
| method | result | size |
| default | \(-\frac {210 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{\frac {7}{2}}+105 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (d x +c \right ) a -350 \left (\cos ^{2}\left (d x +c \right )\right ) a^{\frac {7}{2}}+105 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -168 \sin \left (d x +c \right ) a^{\frac {7}{2}}+72 a^{\frac {7}{2}}}{480 a^{\frac {3}{2}} \left (\sin \left (d x +c \right )-1\right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(176\) |
-1/480/a^(3/2)*(210*cos(d*x+c)^2*sin(d*x+c)*a^(7/2)+105*(a-a*sin(d*x+c))^( 5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+c )*a-350*cos(d*x+c)^2*a^(7/2)+105*(a-a*sin(d*x+c))^(5/2)*2^(1/2)*arctanh(1/ 2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a-168*sin(d*x+c)*a^(7/2)+72*a^(7 /2))/(sin(d*x+c)-1)^2/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.38 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.47 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {105 \, {\left (\sqrt {2} a \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - \sqrt {2} a \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (175 \, a \cos \left (d x + c\right )^{2} - 21 \, {\left (5 \, a \cos \left (d x + c\right )^{2} - 4 \, a\right )} \sin \left (d x + c\right ) - 36 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{960 \, {\left (d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{3}\right )}} \]
1/960*(105*(sqrt(2)*a*cos(d*x + c)^3*sin(d*x + c) - sqrt(2)*a*cos(d*x + c) ^3)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(a*sin(d*x + c) + a)*(sqrt(2)*c os(d*x + c) - sqrt(2)*sin(d*x + c) + sqrt(2))*sqrt(a) + 3*a*cos(d*x + c) - (a*cos(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*(175*a*cos(d*x + c)^2 - 21*(5 *a*cos(d*x + c)^2 - 4*a)*sin(d*x + c) - 36*a)*sqrt(a*sin(d*x + c) + a))/(d *cos(d*x + c)^3*sin(d*x + c) - d*cos(d*x + c)^3)
Timed out. \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.85 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {\sqrt {2} a^{\frac {3}{2}} {\left (\frac {30 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {4 \, {\left (45 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3\right )}}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} - 105 \, \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 105 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{960 \, d} \]
-1/960*sqrt(2)*a^(3/2)*(30*sin(-1/4*pi + 1/2*d*x + 1/2*c)/(sin(-1/4*pi + 1 /2*d*x + 1/2*c)^2 - 1) + 4*(45*sin(-1/4*pi + 1/2*d*x + 1/2*c)^4 + 10*sin(- 1/4*pi + 1/2*d*x + 1/2*c)^2 + 3)/sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 105*lo g(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + 105*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
Timed out. \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^6} \,d x \]