\(\int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\) [126]

Optimal result

Integrand size = 23, antiderivative size = 183 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {15 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {31 \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {9 \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {13 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d} \]

[Out]

-15/4*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*sec(d*x+c)^3*tan(d*x
+c)/d/(a+a*sec(d*x+c))^(3/2)+31/5*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+9/10*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*
sec(d*x+c))^(1/2)-13/10*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3901, 4106, 4095, 4086, 3880, 209} \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {15 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {13 \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{10 a^2 d}-\frac {\tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac {9 \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt {a \sec (c+d x)+a}}+\frac {31 \tan (c+d x)}{5 a d \sqrt {a \sec (c+d x)+a}} \]

[In]

Int[Sec[c + d*x]^5/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(-15*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - (Sec[c + d*x]^
3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + (31*Tan[c + d*x])/(5*a*d*Sqrt[a + a*Sec[c + d*x]]) + (9*Sec
[c + d*x]^2*Tan[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) - (13*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(10*a
^2*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 3901

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

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)]

Rule 4095

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)),
Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4106

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {\int \frac {\sec ^3(c+d x) \left (3 a-\frac {9}{2} a \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {\sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {9 \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {\int \frac {\sec ^2(c+d x) \left (-9 a^2+\frac {39}{4} a^2 \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{5 a^3} \\ & = -\frac {\sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {9 \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {13 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}-\frac {2 \int \frac {\sec (c+d x) \left (\frac {39 a^3}{8}-\frac {93}{4} a^3 \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{15 a^4} \\ & = -\frac {\sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {31 \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {9 \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {13 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}-\frac {15 \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a} \\ & = -\frac {\sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {31 \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {9 \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {13 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}+\frac {15 \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 {15 \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 {\sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {31 \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {9 \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {13 \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.68 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-75 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) (1+\sec (c+d x))+2 \sqrt {1-\sec (c+d x)} \left (49+36 \sec (c+d x)-4 \sec ^2(c+d x)+4 \sec ^3(c+d x)\right )\right ) \tan (c+d x)}{20 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]

[In]

Integrate[Sec[c + d*x]^5/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

((-75*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*(1 + Sec[c + d*x]) + 2*Sqrt[1 - Sec[c + d*x]]*(49 + 36*S
ec[c + d*x] - 4*Sec[c + d*x]^2 + 4*Sec[c + d*x]^3))*Tan[c + d*x])/(20*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c +
d*x]))^(3/2))

Maple [A] (warning: unable to verify)

Time = 1.19 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.17

method	result	size
default	\(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (5 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+75 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}-127 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+175 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-85 \csc \left (d x +c \right )+85 \cot \left (d x +c \right )\right )}{20 d \,a^{2} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\)	\(215\)

[In]

int(sec(d*x+c)^5/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/20/d/a^2*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(5*(1-cos(d*x+c))^7*csc(d*x+c)^7+75*ln(csc(d*x+c)-c
ot(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)^(5/2)-127*(1-cos(d*x+c))^
5*csc(d*x+c)^5+175*(1-cos(d*x+c))^3*csc(d*x+c)^3-85*csc(d*x+c)+85*cot(d*x+c))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1
)^2

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.26 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (49 \, \cos \left (d x + c\right )^{3} + 36 \, \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right ) + 4\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}, \frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left (49 \, \cos \left (d x + c\right )^{3} + 36 \, \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right ) + 4\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{20 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[-1/40*(75*sqrt(2)*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 + cos(d*x + c)^2)*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)/(co
s(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(49*cos(d*x + c)^3 + 36*cos(d*x + c)^2 - 4*cos(d*x + c) + 4)*sqrt((a*c
os(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x +
c)^2), 1/20*(75*sqrt(2)*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2)*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*(49*cos(d*x + c)^3 + 36*cos(d*x + c)^2
- 4*cos(d*x + c) + 4)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*co
s(d*x + c)^3 + a^2*d*cos(d*x + c)^2)]

Sympy [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**5/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**5/(a*(sec(c + d*x) + 1))**(3/2), x)

Maxima [F]

\[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{5}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^5/(a*sec(d*x + c) + a)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 1.00 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {{\left ({\left ({\left (\frac {5 \, \sqrt {2} a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {127 \, \sqrt {2} a}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {175 \, \sqrt {2} a}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {85 \, \sqrt {2} a}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {75 \, \sqrt {2} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{20 \, d} \]

[In]

integrate(sec(d*x+c)^5/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/20*((((5*sqrt(2)*a*tan(1/2*d*x + 1/2*c)^2/sgn(cos(d*x + c)) - 127*sqrt(2)*a/sgn(cos(d*x + c)))*tan(1/2*d*x
+ 1/2*c)^2 + 175*sqrt(2)*a/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 - 85*sqrt(2)*a/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)) - 75*sqrt(2)*log(abs(-sqr
t(-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))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int(1/(cos(c + d*x)^5*(a + a/cos(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^5*(a + a/cos(c + d*x))^(3/2)), x)