Integrand size = 23, antiderivative size = 162 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {35 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{64 \sqrt {2} \sqrt {a} d}-\frac {35 a \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}-\frac {7 a \sec (c+d x)}{24 d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{48 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}} \]
-35/64*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-7/24*a*sec(d*x+c)/d/(a+a*sin( d*x+c))^(3/2)-35/128*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c ))^(1/2))*2^(1/2)/d/a^(1/2)+35/48*sec(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+1/3* sec(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.38 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},3,-\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^3(c+d x) (1+\sin (c+d x))^2}{12 d \sqrt {a (1+\sin (c+d x))}} \]
(Hypergeometric2F1[-3/2, 3, -1/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^3*(1 + Sin[c + d*x])^2)/(12*d*Sqrt[a*(1 + Sin[c + d*x])])
Time = 0.72 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3166, 3042, 3160, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{\sqrt {a \sin (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^4 \sqrt {a \sin (c+d x)+a}}dx\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {7}{6} a \int \frac {\sec ^2(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{6} a \int \frac {1}{\cos (c+d x)^2 (\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {7}{6} a \left (\frac {5 \int \frac {\sec ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{6} a \left (\frac {5 \int \frac {1}{\cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}}dx}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \left (-\frac {\int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{2 a d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {7}{6} a \left (\frac {5 \left (\frac {3}{2} a \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec ^3(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}\) |
Sec[c + d*x]^3/(3*d*Sqrt[a + a*Sin[c + d*x]]) + (7*a*(-1/4*Sec[c + d*x]/(d *(a + a*Sin[c + d*x])^(3/2)) + (5*(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*Sin[c + d*x]])]/(Sqrt[2]*a^(3/2)*d) - Cos[c + d*x]/(2*d*(a + a*Sin[c + d*x])^(3/2 ))))/2))/(8*a)))/6
3.2.66.3.1 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*(2*m + p + 1))), x] + Simp[(m + p + 1)/(a*(2*m + p + 1)) Int[ (g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] & & IntegersQ[2*m, 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 = 0.73 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.49
| method | result | size |
| default | \(\frac {-210 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{\frac {7}{2}}-105 \sqrt {2}\, \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (\cos ^{2}\left (d x +c \right )\right )-70 \left (\cos ^{2}\left (d x +c \right )\right ) a^{\frac {7}{2}}+210 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (d x +c \right ) a^{2}-112 \sin \left (d x +c \right ) a^{\frac {7}{2}}+210 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-16 a^{\frac {7}{2}}}{384 a^{\frac {7}{2}} \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(241\) |
1/384*(-210*cos(d*x+c)^2*sin(d*x+c)*a^(7/2)-105*2^(1/2)*(a-a*sin(d*x+c))^( 3/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2*cos(d*x+c)^2- 70*cos(d*x+c)^2*a^(7/2)+210*(a-a*sin(d*x+c))^(3/2)*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^2-112*sin(d*x+c)*a^(7/2) +210*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2*(a-a* sin(d*x+c))^(3/2)-16*a^(7/2))/a^(7/2)/(sin(d*x+c)-1)/(1+sin(d*x+c))/cos(d* x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.31 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {105 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (35 \, \cos \left (d x + c\right )^{2} + 7 \, {\left (15 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{768 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \]
1/768*(105*sqrt(2)*(cos(d*x + c)^3*sin(d*x + c) + cos(d*x + c)^3)*sqrt(a)* log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d *x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*si n(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*(35*cos(d*x + c)^2 + 7*(15*cos(d*x + c)^2 + 8)*sin(d*x + c) + 8)*sqrt(a*sin(d*x + c) + a))/(a*d*cos(d*x + c)^3*sin(d*x + c) + a*d *cos(d*x + c)^3)
\[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]
Time = 0.38 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left (11 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 13 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {8 \, \sqrt {2} {\left (9 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sqrt {a}\right )}}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{384 \, d} \]
-1/384*(3*sqrt(2)*(11*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 13*sqrt(a )*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^ 2*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 8*sqrt(2)*(9*sqrt(a)*sin(-1/4*p i + 1/2*d*x + 1/2*c)^2 + sqrt(a))/(a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*s in(-1/4*pi + 1/2*d*x + 1/2*c)^3))/d
Timed out. \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]