3.619 \(\int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx\)

Optimal. Leaf size=175 \[ \frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{15 b c}+\frac {14 \tan (2 a+2 b x)}{15 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b \sqrt {c}} \]

[Out]

-1/2*arctanh(1/2*c^(1/2)*tan(2*b*x+2*a)*2^(1/2)/(-c+c*sec(2*b*x+2*a))^(1/2))/b*2^(1/2)/c^(1/2)+14/15*tan(2*b*x
+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/5*sec(2*b*x+2*a)^2*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/15*(-c
+c*sec(2*b*x+2*a))^(1/2)*tan(2*b*x+2*a)/b/c

________________________________________________________________________________________

Rubi [A]  time = 0.60, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4397, 3822, 4010, 4001, 3795, 207} \[ \frac {\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{5 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {\tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{15 b c}+\frac {14 \tan (2 a+2 b x)}{15 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[2*(a + b*x)]^4/Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]

[Out]

-(ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]])]/(Sqrt[2]*b*Sqrt[c])) + (14*Tan[2
*a + 2*b*x])/(15*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (Sec[2*a + 2*b*x]^2*Tan[2*a + 2*b*x])/(5*b*Sqrt[-c + c*Sec
[2*a + 2*b*x]]) + (Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(15*b*c)

Rule 207

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

Rule 3795

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 3822

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*d^2
*Cot[e + f*x]*(d*Csc[e + f*x])^(n - 2))/(f*(2*n - 3)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[d^2/(b*(2*n - 3)), I
nt[((d*Csc[e + f*x])^(n - 2)*(2*b*(n - 2) - a*Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b,
d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 4001

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 4010

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)), I
nt[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] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps

\begin {align*} \int \frac {\sec ^4(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx &=\int \frac {\sec ^4(2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx\\ &=\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\int \frac {\sec ^2(2 a+2 b x) (4 c+c \sec (2 a+2 b x))}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{5 c}\\ &=\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b c}+\frac {2 \int \frac {\sec (2 a+2 b x) \left (\frac {c^2}{2}+7 c^2 \sec (2 a+2 b x)\right )}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{15 c^2}\\ &=\frac {14 \tan (2 a+2 b x)}{15 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b c}+\int \frac {\sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx\\ &=\frac {14 \tan (2 a+2 b x)}{15 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b c}-\frac {\operatorname {Subst}\left (\int \frac {1}{-2 c+x^2} \, dx,x,-\frac {c \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{b}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {14 \tan (2 a+2 b x)}{15 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{5 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.67, size = 112, normalized size = 0.64 \[ \frac {\sin (a+b x) \cos (a+b x) \sec ^3(2 (a+b x)) \left (4 \cos (2 (a+b x))+26 \cos (4 (a+b x))+30 \cos ^2(2 (a+b x)) \tan ^{-1}\left (\sqrt {\tan ^2(a+b x)-1}\right ) \sqrt {\tan ^2(a+b x)-1}+38\right )}{30 b \sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[2*(a + b*x)]^4/Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]

[Out]

(Cos[a + b*x]*Sec[2*(a + b*x)]^3*Sin[a + b*x]*(38 + 4*Cos[2*(a + b*x)] + 26*Cos[4*(a + b*x)] + 30*ArcTan[Sqrt[
-1 + Tan[a + b*x]^2]]*Cos[2*(a + b*x)]^2*Sqrt[-1 + Tan[a + b*x]^2]))/(30*b*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)
]])

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 380, normalized size = 2.17 \[ \left [\frac {4 \, \sqrt {2} {\left (15 \, \tan \left (b x + a\right )^{4} - 20 \, \tan \left (b x + a\right )^{2} + 17\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} + \frac {15 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{5} - 2 \, c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \log \left (\frac {\tan \left (b x + a\right )^{3} - \frac {2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{\sqrt {c}} - 2 \, \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right )}{\sqrt {c}}}{60 \, {\left (b c \tan \left (b x + a\right )^{5} - 2 \, b c \tan \left (b x + a\right )^{3} + b c \tan \left (b x + a\right )\right )}}, -\frac {15 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{5} - 2 \, c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {1}{c}}}{\tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} {\left (15 \, \tan \left (b x + a\right )^{4} - 20 \, \tan \left (b x + a\right )^{2} + 17\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{30 \, {\left (b c \tan \left (b x + a\right )^{5} - 2 \, b c \tan \left (b x + a\right )^{3} + b c \tan \left (b x + a\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*b*x+2*a)^4/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="fricas")

[Out]

[1/60*(4*sqrt(2)*(15*tan(b*x + a)^4 - 20*tan(b*x + a)^2 + 17)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)) + 1
5*sqrt(2)*(c*tan(b*x + a)^5 - 2*c*tan(b*x + a)^3 + c*tan(b*x + a))*log((tan(b*x + a)^3 - 2*sqrt(-c*tan(b*x + a
)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)/sqrt(c) - 2*tan(b*x + a))/tan(b*x + a)^3)/sqrt(c))/(b*c*tan(b*x
+ a)^5 - 2*b*c*tan(b*x + a)^3 + b*c*tan(b*x + a)), -1/30*(15*sqrt(2)*(c*tan(b*x + a)^5 - 2*c*tan(b*x + a)^3 +
c*tan(b*x + a))*sqrt(-1/c)*arctan(sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(-1/c
)/tan(b*x + a)) - 2*sqrt(2)*(15*tan(b*x + a)^4 - 20*tan(b*x + a)^2 + 17)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^
2 - 1)))/(b*c*tan(b*x + a)^5 - 2*b*c*tan(b*x + a)^3 + b*c*tan(b*x + a))]

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*b*x+2*a)^4/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 1.36, size = 984, normalized size = 5.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(2*b*x+2*a)^4/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x)

[Out]

1/120*2^(1/2)/b*(-1+cos(b*x+a))*(208*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos(b*x+a)^6+120*arctanh(1/2*
4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)/sin(b*x+a)^2)*cos(b*x+a)^6
+120*ln(-2*(cos(b*x+a)^2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^
2-1)/(cos(b*x+a)+1)^2)^(1/2)+1)/sin(b*x+a)^2)*cos(b*x+a)^6+208*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos
(b*x+a)^5-200*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos(b*x+a)^4-180*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2
-3*cos(b*x+a)+1)/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)/sin(b*x+a)^2)*cos(b*x+a)^4-180*ln(-2*(cos(b*x+a)^
2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^
(1/2)+1)/sin(b*x+a)^2)*cos(b*x+a)^4-200*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos(b*x+a)^3+60*cos(b*x+a)
^2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+90*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/((2*cos(
b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)/sin(b*x+a)^2)*cos(b*x+a)^2+90*ln(-2*(cos(b*x+a)^2*((2*cos(b*x+a)^2-1)/(cos
(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+1)/sin(b*x+a)^2)*cos
(b*x+a)^2+60*cos(b*x+a)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-15*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*c
os(b*x+a)+1)/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)/sin(b*x+a)^2)-15*ln(-2*(cos(b*x+a)^2*((2*cos(b*x+a)^2
-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+1)/sin(b*x+a
)^2))/(2*cos(b*x+a)^2-1)^3/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)/(c*(1-cos(b*x+a)^2)/(2*cos(b*x+a)^2-1))
^(1/2)/sin(b*x+a)*4^(1/2)/(-3+2*2^(1/2))^3/(3+2*2^(1/2))^3

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (2 \, b x + 2 \, a\right )^{4}}{\sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*b*x+2*a)^4/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(2*b*x + 2*a)^4/sqrt(c*tan(2*b*x + 2*a)*tan(b*x + a)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (2\,a+2\,b\,x\right )}^4\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(2*a + 2*b*x)^4*(c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2)),x)

[Out]

int(1/(cos(2*a + 2*b*x)^4*(c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2)), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*b*x+2*a)**4/(c*tan(b*x+a)*tan(2*b*x+2*a))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________