\(\int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\) [129]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {3 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}} \]

[Out]

3/4*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*tan(d*x+c)/d/(a+a*sec(
d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3882, 3880, 209} \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {3 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

[In]

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

[Out]

(3*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - Tan[c + d*x]/(2*
d*(a + a*Sec[c + d*x])^(3/2))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3880

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-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
]

Rule 3882

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(
(a + b*Csc[e + f*x])^m/(f*(2*m + 1))), x] + Dist[m/(b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {3 \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a} \\ & = -\frac {\tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {3 \text {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{2 a d} \\ & = \frac {3 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-2 \sqrt {1-\sec (c+d x)}+3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) (1+\sec (c+d x))\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]

[In]

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

[Out]

((-2*Sqrt[1 - Sec[c + d*x]] + 3*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*(1 + Sec[c + d*x]))*Tan[c + d*
x])/(4*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(148\) vs. \(2(62)=124\).

Time = 0.85 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.94

method result size
default \(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{4 d \,a^{2}}\) \(149\)

[In]

int(sec(d*x+c)^2/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (62) = 124\).

Time = 0.31 (sec) , antiderivative size = 329, normalized size of antiderivative = 4.27 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \]

[In]

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

[Out]

[-1/8*(3*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) +
 a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*c
os(d*x + c) + 1)) + 4*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2
 + 2*a^2*d*cos(d*x + c) + a^2*d), -1/4*(3*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*sqrt((a*cos(d*x + c) + a)/co
s(d*x + c))*cos(d*x + c)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]

Sympy [F]

\[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{4 \, d} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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