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

Optimal. Leaf size=148 \[ -\frac {76 c^2 \tan (2 a+2 b x)}{105 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}+\frac {2 \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{35 b}+\frac {19 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{105 b} \]

[Out]

2/35*(-c+c*sec(2*b*x+2*a))^(3/2)*tan(2*b*x+2*a)/b+1/7*(-c+c*sec(2*b*x+2*a))^(5/2)*tan(2*b*x+2*a)/b/c-76/105*c^
2*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+19/105*c*(-c+c*sec(2*b*x+2*a))^(1/2)*tan(2*b*x+2*a)/b

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Rubi [A]  time = 0.35, antiderivative size = 148, 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, 3800, 4001, 3793, 3792} \[ -\frac {76 c^2 \tan (2 a+2 b x)}{105 b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}+\frac {2 \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{35 b}+\frac {19 c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{105 b} \]

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]

(-76*c^2*Tan[2*a + 2*b*x])/(105*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (19*c*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a
+ 2*b*x])/(105*b) + (2*(-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(35*b) + ((-c + c*Sec[2*a + 2*b*x])^
(5/2)*Tan[2*a + 2*b*x])/(7*b*c)

Rule 3792

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

Rule 3793

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

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(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*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, 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 \sec ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx &=\int \sec ^3(2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \, dx\\ &=\frac {(-c+c \sec (2 a+2 b x))^{5/2} \tan (2 a+2 b x)}{7 b c}+\frac {2 \int \sec (2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \left (\frac {5 c}{2}+c \sec (2 a+2 b x)\right ) \, dx}{7 c}\\ &=\frac {2 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{35 b}+\frac {(-c+c \sec (2 a+2 b x))^{5/2} \tan (2 a+2 b x)}{7 b c}+\frac {19}{35} \int \sec (2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \, dx\\ &=\frac {19 c \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{105 b}+\frac {2 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{35 b}+\frac {(-c+c \sec (2 a+2 b x))^{5/2} \tan (2 a+2 b x)}{7 b c}-\frac {1}{105} (76 c) \int \sec (2 a+2 b x) \sqrt {-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac {76 c^2 \tan (2 a+2 b x)}{105 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {19 c \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{105 b}+\frac {2 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{35 b}+\frac {(-c+c \sec (2 a+2 b x))^{5/2} \tan (2 a+2 b x)}{7 b c}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 73, normalized size = 0.49 \[ -\frac {\cot (a+b x) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \left (76 \cot (a+b x) \cot (2 (a+b x))-15 \sec ^2(2 (a+b x))+24 \sec (2 (a+b x))-28\right )}{105 b} \]

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]

-1/105*(Cot[a + b*x]*(-28 + 76*Cot[a + b*x]*Cot[2*(a + b*x)] + 24*Sec[2*(a + b*x)] - 15*Sec[2*(a + b*x)]^2)*(c
*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2))/b

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fricas [A]  time = 0.97, size = 111, normalized size = 0.75 \[ -\frac {2 \, \sqrt {2} {\left (105 \, c \tan \left (b x + a\right )^{6} - 140 \, c \tan \left (b x + a\right )^{4} + 133 \, c \tan \left (b x + a\right )^{2} - 38 \, c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{105 \, {\left (b \tan \left (b x + a\right )^{7} - 3 \, b \tan \left (b x + a\right )^{5} + 3 \, b \tan \left (b x + a\right )^{3} - b \tan \left (b x + a\right )\right )}} \]

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]

-2/105*sqrt(2)*(105*c*tan(b*x + a)^6 - 140*c*tan(b*x + a)^4 + 133*c*tan(b*x + a)^2 - 38*c)*sqrt(-c*tan(b*x + a
)^2/(tan(b*x + a)^2 - 1))/(b*tan(b*x + a)^7 - 3*b*tan(b*x + a)^5 + 3*b*tan(b*x + a)^3 - b*tan(b*x + a))

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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 [A]  time = 0.94, size = 95, normalized size = 0.64 \[ -\frac {2 \sqrt {2}\, \left (416 \left (\cos ^{6}\left (b x +a \right )\right )-728 \left (\cos ^{4}\left (b x +a \right )\right )+455 \left (\cos ^{2}\left (b x +a \right )\right )-105\right ) \cos \left (b x +a \right ) \left (\frac {c \left (\sin ^{2}\left (b x +a \right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}\right )^{\frac {3}{2}}}{105 b \left (2 \left (\cos ^{2}\left (b x +a \right )\right )-1\right )^{2} \sin \left (b x +a \right )^{3}} \]

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]

-2/105*2^(1/2)/b*(416*cos(b*x+a)^6-728*cos(b*x+a)^4+455*cos(b*x+a)^2-105)*cos(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/sin(b*x+a)^3

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maxima [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="maxima")

[Out]

Timed out

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mupad [B]  time = 9.26, size = 479, normalized size = 3.24 \[ \frac {\left (\frac {c\,8{}\mathrm {i}}{7\,b}-\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,8{}\mathrm {i}}{7\,b}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}^3}-\frac {\left (\frac {c\,4{}\mathrm {i}}{5\,b}-\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,92{}\mathrm {i}}{35\,b}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}^2}-\frac {\left (\frac {c\,4{}\mathrm {i}}{3\,b}+\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,52{}\mathrm {i}}{105\,b}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}-\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}\,104{}\mathrm {i}}{105\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((c*8i)/(7*b) - (c*exp(a*2i + b*x*2i)*8i)/(7*b))*((c*(exp(a*2i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i - 1i
))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i) - 1)*(exp(a*4i + b*x*4i) +
1)^3) - (((c*4i)/(5*b) - (c*exp(a*2i + b*x*2i)*92i)/(35*b))*((c*(exp(a*2i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*
4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i) - 1)*(exp(a*4i
+ b*x*4i) + 1)^2) - (((c*4i)/(3*b) + (c*exp(a*2i + b*x*2i)*52i)/(105*b))*((c*(exp(a*2i + b*x*2i)*1i - 1i)*(exp
(a*4i + b*x*4i)*1i - 1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2))/((exp(a*2i + b*x*2i) - 1
)*(exp(a*4i + b*x*4i) + 1)) - (c*exp(a*2i + b*x*2i)*((c*(exp(a*2i + b*x*2i)*1i - 1i)*(exp(a*4i + b*x*4i)*1i -
1i))/((exp(a*2i + b*x*2i) + 1)*(exp(a*4i + b*x*4i) + 1)))^(1/2)*104i)/(105*b*(exp(a*2i + b*x*2i) - 1))

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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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