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

Optimal. Leaf size=128 \[ -\frac {7 \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt {2} b c^{3/2}}+\frac {\tan (2 a+2 b x)}{b c \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]

[Out]

-7/8*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/c^(3/2)*2^(1/2)-1/4*tan(2*b*x+2
*a)/b/(-c+c*sec(2*b*x+2*a))^(3/2)+tan(2*b*x+2*a)/b/c/(-c+c*sec(2*b*x+2*a))^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4397, 3799, 4001, 3795, 207} \[ -\frac {7 \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt {2} b c^{3/2}}+\frac {\tan (2 a+2 b x)}{b c \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3799

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(
a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-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 4397

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

Rubi steps

\begin {align*} \int \frac {\sec ^3(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx &=\int \frac {\sec ^3(2 a+2 b x)}{(-c+c \sec (2 a+2 b x))^{3/2}} \, dx\\ &=-\frac {\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {\int \frac {\sec (2 a+2 b x) \left (\frac {3 c}{2}+2 c \sec (2 a+2 b x)\right )}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{2 c^2}\\ &=-\frac {\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {\tan (2 a+2 b x)}{b c \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {7 \int \frac {\sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx}{4 c}\\ &=-\frac {\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {\tan (2 a+2 b x)}{b c \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {7 \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 {7 \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 {\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac {\tan (2 a+2 b x)}{b c \sqrt {-c+c \sec (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 94, normalized size = 0.73 \[ \frac {\tan (2 (a+b x)) \left (4 \sec (2 (a+b x))+7 \sin ^2(a+b x) \tan ^{-1}\left (\sqrt {\tan ^2(a+b x)-1}\right ) \sqrt {\tan ^2(a+b x)-1} \sec (2 (a+b x))-5\right )}{4 b (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-5 + 4*Sec[2*(a + b*x)] + 7*ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sec[2*(a + b*x)]*Sin[a + b*x]^2*Sqrt[-1 + Tan[
a + b*x]^2])*Tan[2*(a + b*x)])/(4*b*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2))

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fricas [A]  time = 2.00, size = 276, normalized size = 2.16 \[ \left [\frac {7 \, \sqrt {2} \sqrt {c} \log \left (\frac {c \tan \left (b x + a\right )^{3} - 2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) \tan \left (b x + a\right )^{3} + 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (9 \, \tan \left (b x + a\right )^{2} - 1\right )}}{16 \, b c^{2} \tan \left (b x + a\right )^{3}}, -\frac {7 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} - \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (9 \, \tan \left (b x + a\right )^{2} - 1\right )}}{8 \, b c^{2} \tan \left (b x + a\right )^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(7*sqrt(2)*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)*tan(b*x + a)^3 + 2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x
+ a)^2 - 1))*(9*tan(b*x + a)^2 - 1))/(b*c^2*tan(b*x + a)^3), -1/8*(7*sqrt(2)*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)))*tan(b*x + a)^3 - sqrt(2)*sqrt(-c*t
an(b*x + a)^2/(tan(b*x + a)^2 - 1))*(9*tan(b*x + a)^2 - 1))/(b*c^2*tan(b*x + a)^3)]

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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)^3/(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.15, size = 930, normalized size = 7.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*2^(1/2)/b*(7*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)/(cos(b*x+a)+1)^2)^(1/2)*cos(b*
x+a)^3+7*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/((2*c
os(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)/sin(b*x+a)^2)*cos(b*x+a)^3+7*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)/si
n(b*x+a)^2)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+7*cos(b*x+a)^2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(
1/2)*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)-7*cos(b*x+a)*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)/(cos(b*x+a)+1)^2)^(1/2)-7*cos(b
*x+a)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*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)+20*cos(b*x+a)^3-7*ln(-2*(cos(b*x+a)^2*((2*cos(b*x+a)^2-1)/(c
os(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)/(cos(b*x+a)+1)^2)^(1/2)-7*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)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-18*cos(b*x+a))*sin(b*
x+a)/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(3/2)/(2*cos(b*x+a)^2-1)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (2 \, b x + 2 \, a\right )^{3}}{\left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (2\,a+2\,b\,x\right )}^3\,{\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**3/(c*tan(b*x+a)*tan(2*b*x+2*a))**(3/2),x)

[Out]

Timed out

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