\(\int \sqrt {a+a \sec (c+d x)} \tan ^2(c+d x) \, dx\) [143]

Optimal result

Integrand size = 23, antiderivative size = 96 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^2(c+d x) \, dx=-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.69 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.35 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^2(c+d x) \, dx=\frac {8 \sqrt {2} \left (\frac {1}{1+\sec (c+d x)}\right )^{7/2} \sqrt {a (1+\sec (c+d x))} \left (-\frac {\cos (c+d x) (7+3 \cos (c+d x)) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-3 \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \cos (c+d x)+(-1+4 \cos (c+d x)) \sqrt {1-\sec (c+d x)}\right )}{24 \sqrt {1-\sec (c+d x)}}-\frac {4}{7} \operatorname {Hypergeometric2F1}\left (2,\frac {7}{2},\frac {9}{2},-2 \sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^3(c+d x)}{3 d \left (1-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{5/2}} \] Input:

Integrate[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^2,x]
 

Output:

(8*Sqrt[2]*((1 + Sec[c + d*x])^(-1))^(7/2)*Sqrt[a*(1 + Sec[c + d*x])]*(-1/ 
24*(Cos[c + d*x]*(7 + 3*Cos[c + d*x])*Csc[(c + d*x)/2]^4*Sec[(c + d*x)/2]^ 
2*(-3*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Cos[c + d*x] + (-1 + 4*Cos[c + d*x]) 
*Sqrt[1 - Sec[c + d*x]]))/Sqrt[1 - Sec[c + d*x]] - (4*Hypergeometric2F1[2, 
 7/2, 9/2, -2*Sec[c + d*x]*Sin[(c + d*x)/2]^2]*Sec[c + d*x]*Tan[(c + d*x)/ 
2]^2)/7)*Tan[c + d*x]^3)/(3*d*(1 - Tan[(c + d*x)/2]^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4375, 363, 262, 216}

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.

\(\displaystyle \int \tan ^2(c+d x) \sqrt {a \sec (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4375

\(\displaystyle -\frac {2 a^2 \int \frac {\tan ^2(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}{(\sec (c+d x) a+a) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\)

\(\Big \downarrow \) 363

\(\displaystyle -\frac {2 a^2 \left (\int \frac {\tan ^2(c+d x)}{(\sec (c+d x) a+a) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )-\frac {\tan ^3(c+d x)}{3 (a \sec (c+d x)+a)^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 262

\(\displaystyle -\frac {2 a^2 \left (-\frac {\int \frac {1}{\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{a}-\frac {\tan ^3(c+d x)}{3 (a \sec (c+d x)+a)^{3/2}}-\frac {\tan (c+d x)}{a \sqrt {a \sec (c+d x)+a}}\right )}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 a^2 \left (\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2}}-\frac {\tan ^3(c+d x)}{3 (a \sec (c+d x)+a)^{3/2}}-\frac {\tan (c+d x)}{a \sqrt {a \sec (c+d x)+a}}\right )}{d}\)

Input:

Int[Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x]^2,x]
 

Output:

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

Defintions of rubi rules used

rule 216

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

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 3042

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

rule 4375

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_.), x_Symbol] :> Simp[-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] && I 
ntegerQ[n - 1/2]
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.34

method	result	size
default	\(-\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-2 \sin \left (d x +c \right )-\tan \left (d x +c \right )+3 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )\right )}{3 d \left (1+\cos \left (d x +c \right )\right )}\)	\(129\)

Input:

int((a+a*sec(d*x+c))^(1/2)*tan(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.95 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^2(c+d x) \, dx=\left [\frac {3 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {2 \, {\left (3 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )\right )}}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(sqrt(a*(sec(c + d*x) + 1))*tan(c + d*x)**2, x)
 

Maxima [F]

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

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

Output:

-1/6*(3*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) 
^(3/4)*(2*d*integrate((((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(5/2* 
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(5/2*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(5/2*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*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(5/2*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)))/ 
(((2*(2*cos(4*d*x + 4*c) + cos(2*d*x + 2*c))*cos(6*d*x + 6*c) + cos(6*d*x 
+ 6*c)^2 + 4*cos(4*d*x + 4*c)^2 + 4*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + co 
s(2*d*x + 2*c)^2 + 2*(2*sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6 
*c) + sin(6*d*x + 6*c)^2 + 4*sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2 
*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (84) = 168\).

Time = 0.37 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.17 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^2(c+d x) \, dx=\frac {\sqrt {2} {\left (\frac {3 \, \sqrt {2} \sqrt {-a} a \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} + \frac {4 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{6 \, d} \] Input:

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

Output:

1/6*sqrt(2)*(3*sqrt(2)*sqrt(-a)*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*(a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^2)*t 
an(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 
1/2*c)^2 + a)))*sgn(cos(d*x + c))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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