3.222 \(\int (a-a \sec ^2(c+d x))^{3/2} \, dx\)

Optimal. Leaf size=64 \[ -\frac {a \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}-\frac {a \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \log (\cos (c+d x))}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {4121, 3658, 3473, 3475} \[ -\frac {a \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}-\frac {a \cot (c+d x) \sqrt {-a \tan ^2(c+d x)} \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sec[c + d*x]^2)^(3/2),x]

[Out]

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

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

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

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[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 4121

Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(b*tan[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rubi steps

\begin {align*} \int \left (a-a \sec ^2(c+d x)\right )^{3/2} \, dx &=\int \left (-a \tan ^2(c+d x)\right )^{3/2} \, dx\\ &=-\left (\left (a \cot (c+d x) \sqrt {-a \tan ^2(c+d x)}\right ) \int \tan ^3(c+d x) \, dx\right )\\ &=-\frac {a \tan (c+d x) \sqrt {-a \tan ^2(c+d x)}}{2 d}+\left (a \cot (c+d x) \sqrt {-a \tan ^2(c+d x)}\right ) \int \tan (c+d x) \, 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}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 48, normalized size = 0.75 \[ \frac {\cot ^3(c+d x) \left (-a \tan ^2(c+d x)\right )^{3/2} \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sec[c + d*x]^2)^(3/2),x]

[Out]

(Cot[c + d*x]^3*(-(a*Tan[c + d*x]^2))^(3/2)*(2*Log[Cos[c + d*x]] + Tan[c + d*x]^2))/(2*d)

________________________________________________________________________________________

fricas [A]  time = 1.33, size = 68, normalized size = 1.06 \[ -\frac {{\left (2 \, a \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) + a\right )} \sqrt {\frac {a \cos \left (d x + c\right )^{2} - a}{\cos \left (d x + c\right )^{2}}}}{2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sec(d*x+c)^2)^(3/2),x, algorithm="fricas")

[Out]

-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)*s
in(d*x + c))

________________________________________________________________________________________

giac [B]  time = 1.15, size = 137, normalized size = 2.14 \[ -\frac {\sqrt {-a} a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + 2\right ) - \sqrt {-a} a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - 2\right ) + \frac {{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}\right )} \sqrt {-a} a - 6 \, \sqrt {-a} a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {1}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - 2}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sec(d*x+c)^2)^(3/2),x, algorithm="giac")

[Out]

-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*sq
rt(-a)*a)/(tan(1/2*d*x + 1/2*c)^2 + 1/tan(1/2*d*x + 1/2*c)^2 - 2))/d

________________________________________________________________________________________

maple [B]  time = 2.34, size = 148, normalized size = 2.31 \[ \frac {\left (2 \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\frac {-\sin \left (d x +c \right )-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+2 \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-2 \ln \left (\frac {2}{1+\cos \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-\left (\cos ^{2}\left (d x +c \right )\right )+1\right ) \cos \left (d x +c \right ) \left (-\frac {a \left (\sin ^{2}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{2}}\right )^{\frac {3}{2}}}{2 d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-a*sec(d*x+c)^2)^(3/2),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 40, normalized size = 0.62 \[ -\frac {\sqrt {-a} a \tan \left (d x + c\right )^{2} - \sqrt {-a} a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sec(d*x+c)^2)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (a-\frac {a}{{\cos \left (c+d\,x\right )}^2}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - a/cos(c + d*x)^2)^(3/2),x)

[Out]

int((a - a/cos(c + d*x)^2)^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (- a \sec ^{2}{\left (c + d x \right )} + a\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sec(d*x+c)**2)**(3/2),x)

[Out]

Integral((-a*sec(c + d*x)**2 + a)**(3/2), x)

________________________________________________________________________________________