\(\int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\) [120]

Optimal result

Integrand size = 23, antiderivative size = 104 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 \tan (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3885, 4086, 3880, 209} \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 a d}-\frac {4 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}} \]

[In]

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

[Out]

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

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 3885

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] + Dist[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 4086

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(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)]

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.83 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\left (-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\frac {2}{3} (1-\sec (c+d x))^{3/2}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.49

method	result	size
default	\(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-4 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}\right )}{3 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}\)	\(155\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 316, normalized size of antiderivative = 3.04 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}, -\frac {2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + \frac {3 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{3 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}\right ] \]

[In]

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

[Out]

[1/6*(3*sqrt(2)*(a*cos(d*x + c)^2 + a*cos(d*x + c))*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d
*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)/(cos(d*x + c)^2 + 2*cos
(d*x + c) + 1)) - 4*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 1)*sin(d*x + c))/(a*d*cos(d*x + c)
^2 + a*d*cos(d*x + c)), -1/3*(2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 1)*sin(d*x + c) + 3*sq
rt(2)*(a*cos(d*x + c)^2 + a*cos(d*x + c))*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/
(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.00 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} {\left (\frac {4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} + \frac {3 \, \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a}}\right )}}{3 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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