3.626 \(\int \frac{\sec ^4(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx\)

Optimal. Leaf size=180 \[ \frac{7 \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{12 b c^2}-\frac{11 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}+\frac{13 \tan (2 a+2 b x)}{6 b c \sqrt{c \sec (2 a+2 b x)-c}} \]

[Out]

(-11*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]])])/(4*Sqrt[2]*b*c^(3/2)) - (Sec
[2*a + 2*b*x]^2*Tan[2*a + 2*b*x])/(4*b*(-c + c*Sec[2*a + 2*b*x])^(3/2)) + (13*Tan[2*a + 2*b*x])/(6*b*c*Sqrt[-c
+ c*Sec[2*a + 2*b*x]]) + (7*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(12*b*c^2)

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Rubi [A]  time = 0.513003, antiderivative size = 180, 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, 3816, 4010, 4001, 3795, 207} \[ \frac{7 \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{12 b c^2}-\frac{11 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{\tan (2 a+2 b x) \sec ^2(2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}+\frac{13 \tan (2 a+2 b x)}{6 b c \sqrt{c \sec (2 a+2 b x)-c}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[2*(a + b*x)]^4/(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]

[Out]

(-11*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]])])/(4*Sqrt[2]*b*c^(3/2)) - (Sec
[2*a + 2*b*x]^2*Tan[2*a + 2*b*x])/(4*b*(-c + c*Sec[2*a + 2*b*x])^(3/2)) + (13*Tan[2*a + 2*b*x])/(6*b*c*Sqrt[-c
+ c*Sec[2*a + 2*b*x]]) + (7*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(12*b*c^2)

Rule 4397

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

Rule 3816

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)), In
t[(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] /; Fr
eeQ[{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 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))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx &=\int \frac{\sec ^4(2 a+2 b x)}{(-c+c \sec (2 a+2 b x))^{3/2}} \, dx\\ &=-\frac{\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac{\int \frac{\sec ^2(2 a+2 b x) \left (2 c+\frac{7}{2} c \sec (2 a+2 b x)\right )}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx}{2 c^2}\\ &=-\frac{\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac{7 \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{12 b c^2}+\frac{\int \frac{\sec (2 a+2 b x) \left (\frac{7 c^2}{4}+\frac{13}{2} c^2 \sec (2 a+2 b x)\right )}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx}{3 c^3}\\ &=-\frac{\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac{13 \tan (2 a+2 b x)}{6 b c \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{7 \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{12 b c^2}+\frac{11 \int \frac{\sec (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx}{4 c}\\ &=-\frac{\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac{13 \tan (2 a+2 b x)}{6 b c \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{7 \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{12 b c^2}-\frac{11 \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 )}{4 b c}\\ &=-\frac{11 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{\sec ^2(2 a+2 b x) \tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac{13 \tan (2 a+2 b x)}{6 b c \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{7 \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{12 b c^2}\\ \end{align*}

Mathematica [A]  time = 1.36145, size = 100, normalized size = 0.56 \[ -\frac{\cot (a+b x) \sqrt{c \tan (a+b x) \tan (2 (a+b x))} \left (\csc ^2(a+b x) ((19 \cos (4 (a+b x))+11) \sec (2 (a+b x))-24)-66 \tan ^{-1}\left (\sqrt{\tan ^2(a+b x)-1}\right ) \sqrt{\tan ^2(a+b x)-1}\right )}{48 b c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[2*(a + b*x)]^4/(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]

[Out]

-(Cot[a + b*x]*(Csc[a + b*x]^2*(-24 + (11 + 19*Cos[4*(a + b*x)])*Sec[2*(a + b*x)]) - 66*ArcTan[Sqrt[-1 + Tan[a
+ b*x]^2]]*Sqrt[-1 + Tan[a + b*x]^2])*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(48*b*c^2)

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Maple [B]  time = 0.454, size = 1211, normalized size = 6.7 \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))^(3/2),x)

[Out]

1/96*2^(1/2)/b*4^(1/2)/(-3+2*2^(1/2))^3/(3+2*2^(1/2))^3*(-1+cos(b*x+a))^2*(132*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)^5+152*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos(b*x+a)^5+132*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)^5-132*cos(b
*x+a)^4*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)-132*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)^4-132*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)^3-200*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos(b*x+a)^3-132*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)^3+132*cos(b*x+a)^2*l
n(-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)+132*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+33*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
)+54*cos(b*x+a)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+33*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)-33*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)-33*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)))/(2*cos(b*x+a)^2-1)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(3/2)/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1
))^(3/2)/sin(b*x+a)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (2 \, b x + 2 \, a\right )^{4}}{\left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(2*b*x + 2*a)^4/(c*tan(2*b*x + 2*a)*tan(b*x + a))^(3/2), x)

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Fricas [A]  time = 2.90706, size = 886, normalized size = 4.92 \begin{align*} \left [\frac{33 \, \sqrt{2}{\left (\tan \left (b x + a\right )^{5} - \tan \left (b x + a\right )^{3}\right )} \sqrt{c} \log \left (\frac{c \tan \left (b x + a\right )^{3} - 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 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) + 2 \, \sqrt{2}{\left (27 \, \tan \left (b x + a\right )^{4} - 46 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{48 \,{\left (b c^{2} \tan \left (b x + a\right )^{5} - b c^{2} \tan \left (b x + a\right )^{3}\right )}}, -\frac{33 \, \sqrt{2}{\left (\tan \left (b x + a\right )^{5} - \tan \left (b x + a\right )^{3}\right )} \sqrt{-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{-c}}{c \tan \left (b x + a\right )}\right ) - \sqrt{2}{\left (27 \, \tan \left (b x + a\right )^{4} - 46 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{24 \,{\left (b c^{2} \tan \left (b x + a\right )^{5} - b c^{2} \tan \left (b x + a\right )^{3}\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))^(3/2),x, algorithm="fricas")

[Out]

[1/48*(33*sqrt(2)*(tan(b*x + a)^5 - tan(b*x + a)^3)*sqrt(c)*log((c*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*c*tan(b*x + a))/tan(b*x + a)^3) + 2*sqrt(2)*(27*tan(b*x
+ a)^4 - 46*tan(b*x + a)^2 + 3)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*c^2*tan(b*x + a)^5 - b*c^2*ta
n(b*x + a)^3), -1/24*(33*sqrt(2)*(tan(b*x + a)^5 - tan(b*x + a)^3)*sqrt(-c)*arctan(sqrt(-c*tan(b*x + a)^2/(tan
(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(-c)/(c*tan(b*x + a))) - sqrt(2)*(27*tan(b*x + a)^4 - 46*tan(b*x +
a)^2 + 3)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*c^2*tan(b*x + a)^5 - b*c^2*tan(b*x + a)^3)]

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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))**(3/2),x)

[Out]

Timed out

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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))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError