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]

-(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)

________________________________________________________________________________________

Rubi [A]  time = 0.599777, antiderivative size = 175, normalized size of antiderivative = 1., 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 4397

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

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 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 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 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 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])

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 = 0.726858, 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)
]])

________________________________________________________________________________________

Maple [B]  time = 0.577, size = 980, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)

________________________________________________________________________________________

Fricas [A]  time = 2.43407, size = 994, normalized size = 5.68 \begin{align*} \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 ] \end{align*}

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))]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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]

Exception raised: TypeError