\(\int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\) [91]

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\frac {14 a \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {4 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 a d} \] Output:

14/15*a*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/15*(a+a*sec(d*x+c))^(1/2)*ta 
n(d*x+c)/d+2/5*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.56 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\frac {2 a \left (8+4 \sec (c+d x)+3 \sec ^2(c+d x)\right ) \tan (c+d x)}{15 d \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]],x]
 

Output:

(2*a*(8 + 4*Sec[c + d*x] + 3*Sec[c + d*x]^2)*Tan[c + d*x])/(15*d*Sqrt[a*(1 
 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4287, 27, 3042, 4489, 3042, 4279}

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.

\(\displaystyle \int \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\)

\(\Big \downarrow \) 4287

\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\)

Input:

Int[Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]],x]
 

Output:

(2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d) + ((14*a^2*Tan[c + d*x 
])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + 
d*x])/(3*d))/(5*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 4279

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free 
Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4287

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 
))), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( 
m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - 
b^2, 0] &&  !LtQ[m, -2^(-1)]
 

rule 4489

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

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {\left (16 \sin \left (d x +c \right )+8 \tan \left (d x +c \right )+6 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right )}\)	\(60\)

Input:

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

Output:

1/d*(16*sin(d*x+c)+8*tan(d*x+c)+6*sec(d*x+c)*tan(d*x+c))/(15*cos(d*x+c)+15 
)*(a*(1+sec(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\frac {2 \, {\left (8 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \] Input:

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

Output:

2/15*(8*cos(d*x + c)^2 + 4*cos(d*x + c) + 3)*sqrt((a*cos(d*x + c) + a)/cos 
(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)
 

Sympy [F]

\[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3} \,d x } \] Input:

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

Output:

8/15*(15*(d*cos(2*d*x + 2*c)^2 + d*sin(2*d*x + 2*c)^2 + 2*d*cos(2*d*x + 2* 
c) + d)*(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)*integrate((((cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 3*cos(6*d* 
x + 6*c)*cos(2*d*x + 2*c) + 3*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d* 
x + 2*c)^2 + sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 3*sin(6*d*x + 6*c)*sin(2* 
d*x + 2*c) + 3*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( 
8*d*x + 8*c) + 3*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 3*cos(2*d*x + 2*c)*si 
n(4*d*x + 4*c) - cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 3*cos(6*d*x + 6*c)*si 
n(2*d*x + 2*c) - 3*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(8*d*x + 8*c) + 3*cos(2*d*x + 2*c 
)*sin(6*d*x + 6*c) + 3*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(8*d*x + 8*c 
)*sin(2*d*x + 2*c) - 3*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 3*cos(4*d*x + 4 
*c)*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) 
 - (cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 3*cos(6*d*x + 6*c)*cos(2*d*x + 2*c 
) + 3*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(8*d*x + 
 8*c)*sin(2*d*x + 2*c) + 3*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 3*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 ...
 

Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\frac {2 \, \sqrt {2} {\left (15 \, a^{3} + {\left (7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:

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

Output:

2/15*sqrt(2)*(15*a^3 + (7*a^3*tan(1/2*d*x + 1/2*c)^2 - 10*a^3)*tan(1/2*d*x 
 + 1/2*c)^2)*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2* 
c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)
 

Mupad [B] (verification not implemented)

Time = 13.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.34 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\frac {8\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,5{}\mathrm {i}-{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,5{}\mathrm {i}-{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,2{}\mathrm {i}+2{}\mathrm {i}\right )}{15\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \] Input:

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

Output:

(8*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*2i 
 + d*x*2i)*5i - exp(c*3i + d*x*3i)*5i - exp(c*5i + d*x*5i)*2i + 2i))/(15*d 
*(exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2)
 

Reduce [F]

\[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\sqrt {a}\, \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) \] Input:

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

Output:

sqrt(a)*int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x)