\(\int \frac {\tan ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\) [179]
Optimal result
Integrand size = 23, antiderivative size = 63 \[
\int \frac {\tan ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}
\]
[Out]
-2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1/2)+2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
Rubi [A] (verified)
Time = 0.09 (sec) , antiderivative size = 63, 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 = {3972, 327, 209}
\[
\int \frac {\tan ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}
\]
[In]
Int[Tan[c + d*x]^2/Sqrt[a + a*Sec[c + d*x]],x]
[Out]
(-2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) + (2*Tan[c + d*x])/(d*Sqrt[a + a*Sec[
c + d*x]])
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 327
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
Rule 3972
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[-2*(a^(m/2 +
n + 1/2)/d), Subst[Int[x^m*((2 + a*x^2)^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]
Rubi steps \begin{align*}
\text {integral}& = -\frac {(2 a) \text {Subst}\left (\int \frac {x^2}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = \frac {2 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = -\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}} \\
\end{align*}
Mathematica [A] (warning: unable to verify)
Time = 0.59 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.89
\[
\int \frac {\tan ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {16 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (\frac {1}{1+\sec (c+d x)}\right )^{5/2} \left (-\arcsin \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {1}{1+\cos (c+d x)}}}\right ) \cos (c+d x)+\sqrt {\cos (c+d x)} \sqrt {\frac {1}{1+\cos (c+d x)}} \sin (c+d x)\right )}{d \sqrt {a (1+\sec (c+d x))}}
\]
[In]
Integrate[Tan[c + d*x]^2/Sqrt[a + a*Sec[c + d*x]],x]
[Out]
(16*Cos[(c + d*x)/2]^6*Sec[c + d*x]^4*((1 + Sec[c + d*x])^(-1))^(5/2)*(-(ArcSin[Tan[(c + d*x)/2]/Sqrt[(1 + Cos
[c + d*x])^(-1)]]*Cos[c + d*x]) + Sqrt[Cos[c + d*x]]*Sqrt[(1 + Cos[c + d*x])^(-1)]*Sin[c + d*x]))/(d*Sqrt[a*(1
+ Sec[c + d*x])])
Maple [A] (verified)
Time = 3.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.51
| | |
| method | result | size |
| | |
| default |
\(-\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d a}\) |
\(95\) |
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[In]
int(tan(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
[Out]
-2/d/a*(a*(1+sec(d*x+c)))^(1/2)*((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*
x+c)/(cos(d*x+c)+1))^(1/2))+cot(d*x+c)-csc(d*x+c))
Fricas [A] (verification not implemented)
none
Time = 0.27 (sec) , antiderivative size = 235, normalized size of antiderivative = 3.73
\[
\int \frac {\tan ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [-\frac {\sqrt {-a} {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d}, \frac {2 \, {\left (\sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{a d \cos \left (d x + c\right ) + a d}\right ]
\]
[In]
integrate(tan(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
[-(sqrt(-a)*(cos(d*x + c) + 1)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*co
s(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*
sin(d*x + c))/(a*d*cos(d*x + c) + a*d), 2*(sqrt(a)*(cos(d*x + c) + 1)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d
*x + c) + a*d)]
Sympy [F]
\[
\int \frac {\tan ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\tan ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx
\]
[In]
integrate(tan(d*x+c)**2/(a+a*sec(d*x+c))**(1/2),x)
[Out]
Integral(tan(c + d*x)**2/sqrt(a*(sec(c + d*x) + 1)), x)
Maxima [F]
\[
\int \frac {\tan ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{2}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x }
\]
[In]
integrate(tan(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
1/2*((2*a*d*integrate(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((cos(6*d*x +
6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x
+ 2*c) + 2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*
x + 2*c))) + (cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2
*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1
/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + ((cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)
*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*cos(3/2*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c))) - (cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c)
+ cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x +
2*c)^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*
c) + 1)))/(a*cos(6*d*x + 6*c)^2 + 4*a*cos(4*d*x + 4*c)^2 + 4*a*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + a*cos(2*d*x
+ 2*c)^2 + a*sin(6*d*x + 6*c)^2 + 4*a*sin(4*d*x + 4*c)^2 + 4*a*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + a*sin(2*d*
x + 2*c)^2 + 2*(2*a*cos(4*d*x + 4*c) + a*cos(2*d*x + 2*c))*cos(6*d*x + 6*c) + 2*(2*a*sin(4*d*x + 4*c) + a*sin(
2*d*x + 2*c))*sin(6*d*x + 6*c)), x) - 2*a*d*integrate(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
+ 2*c) + 1)^(1/4)*(((cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c
)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(1/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x
+ 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + ((cos(2*d*x + 2*c)*sin(6
*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin
(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 2*c
os(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c)
*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(1/2*arctan2(
sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))/(a*cos(6*d*x + 6*c)^2 + 4*a*cos(4*d*x + 4*c)^2 + 4*a*cos(4*d*x + 4*c
)*cos(2*d*x + 2*c) + a*cos(2*d*x + 2*c)^2 + a*sin(6*d*x + 6*c)^2 + 4*a*sin(4*d*x + 4*c)^2 + 4*a*sin(4*d*x + 4*
c)*sin(2*d*x + 2*c) + a*sin(2*d*x + 2*c)^2 + 2*(2*a*cos(4*d*x + 4*c) + a*cos(2*d*x + 2*c))*cos(6*d*x + 6*c) +
2*(2*a*sin(4*d*x + 4*c) + a*sin(2*d*x + 2*c))*sin(6*d*x + 6*c)), x) + arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x
+ 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x
+ 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*
c) + 1)) + 1) - arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1
)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^
2 + 2*cos(2*d*x + 2*c) + 1)^(1/4) + 4*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))/((cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a)*d)
Giac [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (55) = 110\).
Time = 1.14 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.97
\[
\int \frac {\tan ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} {\left (\frac {\sqrt {2} \sqrt {-a} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} - \frac {4 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}\right )}}{2 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}
\]
[In]
integrate(tan(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
1/2*sqrt(2)*(sqrt(2)*sqrt(-a)*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^
2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*
sqrt(2)*abs(a) - 6*a))/abs(a) - 4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/
2*c)^2 - a))/(d*sgn(cos(d*x + c)))
Mupad [F(-1)]
Timed out. \[
\int \frac {\tan ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x
\]
[In]
int(tan(c + d*x)^2/(a + a/cos(c + d*x))^(1/2),x)
[Out]
int(tan(c + d*x)^2/(a + a/cos(c + d*x))^(1/2), x)