\(\int (a-a \sec ^2(c+d x))^{5/2} \, dx\) [221]

Optimal result

Integrand size = 17, antiderivative size = 101 \[ \int \left (a-a \sec ^2(c+d x)\right )^{5/2} \, dx=-\frac {a^2 \cot (c+d x) \log (\cos (c+d x)) \sqrt {-a \tan ^2(c+d x)}}{d}-\frac {a^2 \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}+\frac {a^2 \tan ^3(c+d x) \sqrt {-a \tan ^2(c+d x)}}{4 d} \] Output:

-a^2*cot(d*x+c)*ln(cos(d*x+c))*(-a*tan(d*x+c)^2)^(1/2)/d-1/2*a^2*tan(d*x+c 
)*(-a*tan(d*x+c)^2)^(1/2)/d+1/4*a^2*tan(d*x+c)^3*(-a*tan(d*x+c)^2)^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int \left (a-a \sec ^2(c+d x)\right )^{5/2} \, dx=\frac {a \csc ^3(c+d x) (2+3 \log (\cos (c+d x))+\cos (4 (c+d x)) \log (\cos (c+d x))+4 \cos (2 (c+d x)) (1+\log (\cos (c+d x)))) \sec (c+d x) \left (-a \tan ^2(c+d x)\right )^{3/2}}{8 d} \] Input:

Integrate[(a - a*Sec[c + d*x]^2)^(5/2),x]
 

Output:

(a*Csc[c + d*x]^3*(2 + 3*Log[Cos[c + d*x]] + Cos[4*(c + d*x)]*Log[Cos[c + 
d*x]] + 4*Cos[2*(c + d*x)]*(1 + Log[Cos[c + d*x]]))*Sec[c + d*x]*(-(a*Tan[ 
c + d*x]^2))^(3/2))/(8*d)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.588, Rules used = {3042, 4609, 3042, 4141, 3042, 3954, 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.

\(\displaystyle \int \left (a-a \sec ^2(c+d x)\right )^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a-a \sec (c+d x)^2\right )^{5/2}dx\)

\(\Big \downarrow \) 4609

\(\displaystyle \int \left (-a \tan ^2(c+d x)\right )^{5/2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (-a \tan (c+d x)^2\right )^{5/2}dx\)

\(\Big \downarrow \) 4141

\(\displaystyle a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \int \tan ^5(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \int \tan (c+d x)^5dx\)

\(\Big \downarrow \) 3954

\(\displaystyle a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \left (\frac {\tan ^4(c+d x)}{4 d}-\int \tan ^3(c+d x)dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \left (\frac {\tan ^4(c+d x)}{4 d}-\int \tan (c+d x)^3dx\right )\)

\(\Big \downarrow \) 3954

\(\displaystyle a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \left (\int \tan (c+d x)dx+\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \left (\int \tan (c+d x)dx+\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}\right )\)

\(\Big \downarrow \) 3956

\(\displaystyle a^2 \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \left (\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}-\frac {\log (\cos (c+d x))}{d}\right )\)

Input:

Int[(a - a*Sec[c + d*x]^2)^(5/2),x]
 

Output:

a^2*Cot[c + d*x]*Sqrt[-(a*Tan[c + d*x]^2)]*(-(Log[Cos[c + d*x]]/d) - Tan[c 
 + d*x]^2/(2*d) + Tan[c + d*x]^4/(4*d))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3954

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]
 

rule 3956

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

rule 4141

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]])
 

rule 4609

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]
 

Maple [A] (verified)

Time = 4.03 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32

method	result	size
default	\(\frac {\left (-\cos \left (d x +c \right )^{4} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\cos \left (d x +c \right )^{4} \ln \left (\frac {2}{\cos \left (d x +c \right )+1}\right )-\cos \left (d x +c \right )^{4} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\frac {3 \cos \left (d x +c \right )^{4}}{4}-\cos \left (d x +c \right )^{2}+\frac {1}{4}\right ) a^{2} \sqrt {-a \tan \left (d x +c \right )^{2}}\, \sec \left (d x +c \right )^{3} \csc \left (d x +c \right )}{d}\)	\(133\)
risch	\(\frac {a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, x}{{\mathrm e}^{2 i \left (d x +c \right )}-1}-\frac {2 a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left (d x +c \right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}-\frac {4 i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{6 i \left (d x +c \right )}+{\mathrm e}^{4 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}-\frac {i a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) d}\)	\(296\)

Input:

int((a-sec(d*x+c)^2*a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(-cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)+1)+cos(d*x+c)^4*ln(2/(cos(d*x 
+c)+1))-cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)-1)+3/4*cos(d*x+c)^4-cos(d*x 
+c)^2+1/4)*a^2*(-a*tan(d*x+c)^2)^(1/2)*sec(d*x+c)^3*csc(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.86 \[ \int \left (a-a \sec ^2(c+d x)\right )^{5/2} \, dx=-\frac {{\left (4 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) + 4 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right )^{2} - a}{\cos \left (d x + c\right )^{2}}}}{4 \, d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )} \] Input:

integrate((a-a*sec(d*x+c)^2)^(5/2),x, algorithm="fricas")
 

Output:

-1/4*(4*a^2*cos(d*x + c)^4*log(-cos(d*x + c)) + 4*a^2*cos(d*x + c)^2 - a^2 
)*sqrt((a*cos(d*x + c)^2 - a)/cos(d*x + c)^2)/(d*cos(d*x + c)^3*sin(d*x + 
c))
 

Sympy [F]

\[ \int \left (a-a \sec ^2(c+d x)\right )^{5/2} \, dx=\int \left (- a \sec ^{2}{\left (c + d x \right )} + a\right )^{\frac {5}{2}}\, dx \] Input:

integrate((a-a*sec(d*x+c)**2)**(5/2),x)
 

Output:

Integral((-a*sec(c + d*x)**2 + a)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.61 \[ \int \left (a-a \sec ^2(c+d x)\right )^{5/2} \, dx=\frac {\sqrt {-a} a^{2} \tan \left (d x + c\right )^{4} - 2 \, \sqrt {-a} a^{2} \tan \left (d x + c\right )^{2} + 2 \, \sqrt {-a} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{4 \, d} \] Input:

integrate((a-a*sec(d*x+c)^2)^(5/2),x, algorithm="maxima")
 

Output:

1/4*(sqrt(-a)*a^2*tan(d*x + c)^4 - 2*sqrt(-a)*a^2*tan(d*x + c)^2 + 2*sqrt( 
-a)*a^2*log(tan(d*x + c)^2 + 1))/d
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.80 \[ \int \left (a-a \sec ^2(c+d x)\right )^{5/2} \, dx=-\frac {2 \, \sqrt {-a} a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + 2\right ) - 2 \, \sqrt {-a} a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - 2\right ) + \frac {3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}\right )}^{2} \sqrt {-a} a^{2} - 20 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}\right )} \sqrt {-a} a^{2} + 44 \, \sqrt {-a} a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - 2\right )}^{2}}}{4 \, d} \] Input:

integrate((a-a*sec(d*x+c)^2)^(5/2),x, algorithm="giac")
 

Output:

-1/4*(2*sqrt(-a)*a^2*log(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 
 + 2) - 2*sqrt(-a)*a^2*log(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c) 
^2 - 2) + (3*(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2)^2*sqrt(-a 
)*a^2 - 20*(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2)*sqrt(-a)*a^ 
2 + 44*sqrt(-a)*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 - 
2)^2)/d
 

Mupad [F(-1)]

Timed out. \[ \int \left (a-a \sec ^2(c+d x)\right )^{5/2} \, dx=\int {\left (a-\frac {a}{{\cos \left (c+d\,x\right )}^2}\right )}^{5/2} \,d x \] Input:

int((a - a/cos(c + d*x)^2)^(5/2),x)
 

Output:

int((a - a/cos(c + d*x)^2)^(5/2), x)
 

Reduce [F]

\[ \int \left (a-a \sec ^2(c+d x)\right )^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {-\sec \left (d x +c \right )^{2}+1}d x +\int \sqrt {-\sec \left (d x +c \right )^{2}+1}\, \sec \left (d x +c \right )^{4}d x -2 \left (\int \sqrt {-\sec \left (d x +c \right )^{2}+1}\, \sec \left (d x +c \right )^{2}d x \right )\right ) \] Input:

int((a-a*sec(d*x+c)^2)^(5/2),x)
 

Output:

sqrt(a)*a**2*(int(sqrt( - sec(c + d*x)**2 + 1),x) + int(sqrt( - sec(c + d* 
x)**2 + 1)*sec(c + d*x)**4,x) - 2*int(sqrt( - sec(c + d*x)**2 + 1)*sec(c + 
 d*x)**2,x))