Integrand size = 23, antiderivative size = 140 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {28 \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}-\frac {2 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d} \]
-arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a ^(1/2)+28/15*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/5*sec(d*x+c)^2*tan(d*x+ c)/d/(a+a*sec(d*x+c))^(1/2)-2/15*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a/d
Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (-15 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+2 \sqrt {1-\sec (c+d x)} \left (13-\sec (c+d x)+3 \sec ^2(c+d x)\right )\right ) \tan (c+d x)}{15 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
((-15*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + 2*Sqrt[1 - Sec[c + d*x]]*(13 - Sec[c + d*x] + 3*Sec[c + d*x]^2))*Tan[c + d*x])/(15*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.79 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4309, 3042, 4498, 27, 3042, 4489, 3042, 4282, 216}
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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{\sqrt {a \sec (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4}{\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 4309 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (4 a-a \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}+\frac {2 \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}+\frac {2 \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle \frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^2-14 a^2 \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {\sec (c+d x) \left (a^2-14 a^2 \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^2-14 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {-\frac {15 a^2 \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {28 a^2 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {15 a^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {28 a^2 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {-\frac {-\frac {30 a^2 \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {28 a^2 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {-\frac {\frac {15 \sqrt {2} a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {28 a^2 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\) |
(2*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) + ((-2*Sqrt [a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d) - ((15*Sqrt[2]*a^(3/2)*ArcTan[(Sq rt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d - (28*a^2*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(5*a)
3.2.19.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*d^2*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 2)/( f*(2*n - 3)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[d^2/(b*(2*n - 3)) Int[( d*Csc[e + f*x])^(n - 2)*((2*b*(n - 2) - a*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B , e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b *(m + 1), 0] && !LtQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* ((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int [Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) *Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a *B, 0] && !LtQ[m, -1]
Time = 1.22 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.38
| method | result | size |
| default | \(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (15 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}-34 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+40 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-30 \csc \left (d x +c \right )+30 \cot \left (d x +c \right )\right )}{15 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\) | \(193\) |
-1/15/d/a*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(15*ln(csc(d*x+c) -cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc (d*x+c)^2-1)^(5/2)-34*(1-cos(d*x+c))^5*csc(d*x+c)^5+40*(1-cos(d*x+c))^3*cs c(d*x+c)^3-30*csc(d*x+c)+30*cot(d*x+c))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^ 2
Time = 0.31 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.48 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {15 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (13 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{30 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}, \frac {2 \, {\left (13 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {15 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{15 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
[1/30*(15*sqrt(2)*(a*cos(d*x + c)^3 + a*cos(d*x + c)^2)*sqrt(-1/a)*log((2* sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*si n(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(cos(d*x + c)^2 + 2*co s(d*x + c) + 1)) + 4*(13*cos(d*x + c)^2 - cos(d*x + c) + 3)*sqrt((a*cos(d* x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2), 1/15*(2*(13*cos(d*x + c)^2 - cos(d*x + c) + 3)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) + 15*sqrt(2)*(a*cos(d*x + c)^3 + a*cos( d*x + c)^2)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)]
\[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
Time = 1.04 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} {\left (\frac {15 \, \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a}} + \frac {2 \, {\left ({\left (17 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 20 \, a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )}}{15 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \]
1/15*sqrt(2)*(15*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2* d*x + 1/2*c)^2 + a)))/sqrt(-a) + 2*((17*a^2*tan(1/2*d*x + 1/2*c)^2 - 20*a^ 2)*tan(1/2*d*x + 1/2*c)^2 + 15*a^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(d*sgn(cos(d*x + c) ))
Timed out. \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]