3.225 \(\int \frac {1}{(a-a \sec ^2(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=67 \[ \frac {\cot (c+d x)}{2 a d \sqrt {-a \tan ^2(c+d x)}}+\frac {\tan (c+d x) \log (\sin (c+d x))}{a d \sqrt {-a \tan ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 67, 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} \[ \frac {\cot (c+d x)}{2 a d \sqrt {-a \tan ^2(c+d x)}}+\frac {\tan (c+d x) \log (\sin (c+d x))}{a d \sqrt {-a \tan ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sec[c + d*x]^2)^(-3/2),x]

[Out]

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

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 \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{3/2}} \, dx &=\int \frac {1}{\left (-a \tan ^2(c+d x)\right )^{3/2}} \, dx\\ &=-\frac {\tan (c+d x) \int \cot ^3(c+d x) \, dx}{a \sqrt {-a \tan ^2(c+d x)}}\\ &=\frac {\cot (c+d x)}{2 a d \sqrt {-a \tan ^2(c+d x)}}+\frac {\tan (c+d x) \int \cot (c+d x) \, dx}{a \sqrt {-a \tan ^2(c+d x)}}\\ &=\frac {\cot (c+d x)}{2 a d \sqrt {-a \tan ^2(c+d x)}}+\frac {\log (\sin (c+d x)) \tan (c+d x)}{a d \sqrt {-a \tan ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 57, normalized size = 0.85 \[ -\frac {\tan ^3(c+d x) \left (\cot ^2(c+d x)+2 \log (\tan (c+d x))+2 \log (\cos (c+d x))\right )}{2 d \left (-a \tan ^2(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((Cot[c + d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]])*Tan[c + d*x]^3)/(d*(-(a*Tan[c + d*x]^2))^(3
/2))

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fricas [A]  time = 0.84, size = 94, normalized size = 1.40 \[ -\frac {{\left (2 \, {\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right )^{2} - a}{\cos \left (d x + c\right )^{2}}}}{2 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.85, size = 212, normalized size = 3.16 \[ \frac {\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\sqrt {-a} a \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {8 \, \sqrt {-a} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}{a^{2} \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {4 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )}{\sqrt {-a} a \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}{\sqrt {-a} a \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)^2)^(3/2),x, algorithm="giac")

[Out]

1/8*(tan(1/2*d*x + 1/2*c)^2/(sqrt(-a)*a*sgn(-tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))) + 8*sqrt(-a)*log(
tan(1/2*d*x + 1/2*c)^2 + 1)/(a^2*sgn(-tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))) + 4*log(tan(1/2*d*x + 1/
2*c)^2)/(sqrt(-a)*a*sgn(-tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))) - (4*tan(1/2*d*x + 1/2*c)^2 - 1)/(sqr
t(-a)*a*sgn(-tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c)^2))/d

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maple [B]  time = 1.66, size = 141, normalized size = 2.10 \[ \frac {\left (4 \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-4 \ln \left (\frac {2}{1+\cos \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-\left (\cos ^{2}\left (d x +c \right )\right )-4 \ln \left (-\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+4 \ln \left (\frac {2}{1+\cos \left (d x +c \right )}\right )-1\right ) \sin \left (d x +c \right )}{4 d \cos \left (d x +c \right )^{3} \left (-\frac {a \left (\sin ^{2}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{2}}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-a*sec(d*x+c)^2)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.43, size = 60, normalized size = 0.90 \[ -\frac {\frac {\log \left (\tan \left (d x + c\right )^{2} + 1\right )}{\sqrt {-a} a} - \frac {2 \, \log \left (\tan \left (d x + c\right )\right )}{\sqrt {-a} a} + \frac {\sqrt {-a}}{a^{2} \tan \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a-\frac {a}{{\cos \left (c+d\,x\right )}^2}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - a/cos(c + d*x)^2)^(3/2),x)

[Out]

int(1/(a - a/cos(c + d*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- a \sec ^{2}{\left (c + d x \right )} + a\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sec(d*x+c)**2)**(3/2),x)

[Out]

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

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