Integrand size = 17, antiderivative size = 64 \[ \int \left (a-a \sec ^2(c+d x)\right )^{3/2} \, dx=-\frac {a \cot (c+d x) \log (\cos (c+d x)) \sqrt {-a \tan ^2(c+d x)}}{d}-\frac {a \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d} \] Output:
-a*cot(d*x+c)*ln(cos(d*x+c))*(-a*tan(d*x+c)^2)^(1/2)/d-1/2*a*tan(d*x+c)*(- a*tan(d*x+c)^2)^(1/2)/d
Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.75 \[ \int \left (a-a \sec ^2(c+d x)\right )^{3/2} \, dx=\frac {\cot ^3(c+d x) \left (2 \log (\cos (c+d x))+\sec ^2(c+d x)\right ) \left (-a \tan ^2(c+d x)\right )^{3/2}}{2 d} \] Input:
Integrate[(a - a*Sec[c + d*x]^2)^(3/2),x]
Output:
(Cot[c + d*x]^3*(2*Log[Cos[c + d*x]] + Sec[c + d*x]^2)*(-(a*Tan[c + d*x]^2 ))^(3/2))/(2*d)
Time = 0.36 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 4609, 3042, 4141, 3042, 3954, 3042, 3956}
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.
\(\displaystyle \int \left (a-a \sec ^2(c+d x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a-a \sec (c+d x)^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 4609 |
\(\displaystyle \int \left (-a \tan ^2(c+d x)\right )^{3/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (-a \tan (c+d x)^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle -a \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \int \tan ^3(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \int \tan (c+d x)^3dx\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -a \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \left (\frac {\tan ^2(c+d x)}{2 d}-\int \tan (c+d x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \left (\frac {\tan ^2(c+d x)}{2 d}-\int \tan (c+d x)dx\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -a \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \left (\frac {\tan ^2(c+d x)}{2 d}+\frac {\log (\cos (c+d x))}{d}\right )\) |
Input:
Int[(a - a*Sec[c + d*x]^2)^(3/2),x]
Output:
-(a*Cot[c + d*x]*Sqrt[-(a*Tan[c + d*x]^2)]*(Log[Cos[c + d*x]]/d + Tan[c + d*x]^2/(2*d)))
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Ta n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(b*tan[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a + b, 0]
Time = 1.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.56
method | result | size |
default | \(\frac {a \sqrt {-a \tan \left (d x +c \right )^{2}}\, \left (2 \ln \left (\frac {2}{\cos \left (d x +c \right )+1}\right ) \cot \left (d x +c \right )-2 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cot \left (d x +c \right )-2 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cot \left (d x +c \right )-\tan \left (d x +c \right )\right )}{2 d}\) | \(100\) |
risch | \(-\frac {a \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left (i {\mathrm e}^{4 i \left (d x +c \right )} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )+{\mathrm e}^{4 i \left (d x +c \right )} d x +2 \,{\mathrm e}^{4 i \left (d x +c \right )} c +2 i {\mathrm e}^{2 i \left (d x +c \right )} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )+2 \,{\mathrm e}^{2 i \left (d x +c \right )} d x +2 i {\mathrm e}^{2 i \left (d x +c \right )}+4 \,{\mathrm e}^{2 i \left (d x +c \right )} c +i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )+d x +2 c \right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\) | \(194\) |
Input:
int((a-sec(d*x+c)^2*a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2/d*a*(-a*tan(d*x+c)^2)^(1/2)*(2*ln(2/(cos(d*x+c)+1))*cot(d*x+c)-2*ln(-c ot(d*x+c)+csc(d*x+c)+1)*cot(d*x+c)-2*ln(-cot(d*x+c)+csc(d*x+c)-1)*cot(d*x+ c)-tan(d*x+c))
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.06 \[ \int \left (a-a \sec ^2(c+d x)\right )^{3/2} \, dx=-\frac {{\left (2 \, a \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) + a\right )} \sqrt {\frac {a \cos \left (d x + c\right )^{2} - a}{\cos \left (d x + c\right )^{2}}}}{2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \] Input:
integrate((a-a*sec(d*x+c)^2)^(3/2),x, algorithm="fricas")
Output:
-1/2*(2*a*cos(d*x + c)^2*log(-cos(d*x + c)) + a)*sqrt((a*cos(d*x + c)^2 - a)/cos(d*x + c)^2)/(d*cos(d*x + c)*sin(d*x + c))
\[ \int \left (a-a \sec ^2(c+d x)\right )^{3/2} \, dx=\int \left (- a \sec ^{2}{\left (c + d x \right )} + a\right )^{\frac {3}{2}}\, dx \] Input:
integrate((a-a*sec(d*x+c)**2)**(3/2),x)
Output:
Integral((-a*sec(c + d*x)**2 + a)**(3/2), x)
Time = 0.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.62 \[ \int \left (a-a \sec ^2(c+d x)\right )^{3/2} \, dx=-\frac {\sqrt {-a} a \tan \left (d x + c\right )^{2} - \sqrt {-a} a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \] Input:
integrate((a-a*sec(d*x+c)^2)^(3/2),x, algorithm="maxima")
Output:
-1/2*(sqrt(-a)*a*tan(d*x + c)^2 - sqrt(-a)*a*log(tan(d*x + c)^2 + 1))/d
Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (58) = 116\).
Time = 0.23 (sec) , antiderivative size = 137, normalized size of antiderivative = 2.14 \[ \int \left (a-a \sec ^2(c+d x)\right )^{3/2} \, dx=-\frac {\sqrt {-a} a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + 2\right ) - \sqrt {-a} a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - 2\right ) + \frac {{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}\right )} \sqrt {-a} a - 6 \, \sqrt {-a} a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - 2}}{2 \, d} \] Input:
integrate((a-a*sec(d*x+c)^2)^(3/2),x, algorithm="giac")
Output:
-1/2*(sqrt(-a)*a*log(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 + 2 ) - sqrt(-a)*a*log(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 - 2) + ((tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2)*sqrt(-a)*a - 6*sqrt (-a)*a)/(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 - 2))/d
Timed out. \[ \int \left (a-a \sec ^2(c+d x)\right )^{3/2} \, dx=\int {\left (a-\frac {a}{{\cos \left (c+d\,x\right )}^2}\right )}^{3/2} \,d x \] Input:
int((a - a/cos(c + d*x)^2)^(3/2),x)
Output:
int((a - a/cos(c + d*x)^2)^(3/2), x)
\[ \int \left (a-a \sec ^2(c+d x)\right )^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {-\sec \left (d x +c \right )^{2}+1}d x -\left (\int \sqrt {-\sec \left (d x +c \right )^{2}+1}\, \sec \left (d x +c \right )^{2}d x \right )\right ) \] Input:
int((a-a*sec(d*x+c)^2)^(3/2),x)
Output:
sqrt(a)*a*(int(sqrt( - sec(c + d*x)**2 + 1),x) - int(sqrt( - sec(c + d*x)* *2 + 1)*sec(c + d*x)**2,x))