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

Optimal. Leaf size=208 \[ \frac{c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{17 c^2 \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{63 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{34 c^2 \tan (2 a+2 b x)}{45 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{34 \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{105 b}+\frac{68 c \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{315 b} \]

[Out]

(34*c^2*Tan[2*a + 2*b*x])/(45*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) - (17*c^2*Sec[2*a + 2*b*x]^3*Tan[2*a + 2*b*x])/
(63*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (c^2*Sec[2*a + 2*b*x]^4*Tan[2*a + 2*b*x])/(9*b*Sqrt[-c + c*Sec[2*a + 2*
b*x]]) + (68*c*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(315*b) + (34*(-c + c*Sec[2*a + 2*b*x])^(3/2)*T
an[2*a + 2*b*x])/(105*b)

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Rubi [A]  time = 0.527679, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4397, 3814, 21, 3803, 3800, 4001, 3792} \[ \frac{c^2 \tan (2 a+2 b x) \sec ^4(2 a+2 b x)}{9 b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{17 c^2 \tan (2 a+2 b x) \sec ^3(2 a+2 b x)}{63 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{34 c^2 \tan (2 a+2 b x)}{45 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{34 \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{105 b}+\frac{68 c \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{315 b} \]

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]

(34*c^2*Tan[2*a + 2*b*x])/(45*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) - (17*c^2*Sec[2*a + 2*b*x]^3*Tan[2*a + 2*b*x])/
(63*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (c^2*Sec[2*a + 2*b*x]^4*Tan[2*a + 2*b*x])/(9*b*Sqrt[-c + c*Sec[2*a + 2*
b*x]]) + (68*c*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(315*b) + (34*(-c + c*Sec[2*a + 2*b*x])^(3/2)*T
an[2*a + 2*b*x])/(105*b)

Rule 4397

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

Rule 3814

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[b/(m + n - 1), Int[(a
+ b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3803

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

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 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]

Rubi steps

\begin{align*} \int \sec ^4(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx &=\int \sec ^4(2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \, dx\\ &=\frac{c^2 \sec ^4(2 a+2 b x) \tan (2 a+2 b x)}{9 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{1}{9} (2 c) \int \frac{\sec ^4(2 a+2 b x) \left (\frac{17 c}{2}-\frac{17}{2} c \sec (2 a+2 b x)\right )}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx\\ &=\frac{c^2 \sec ^4(2 a+2 b x) \tan (2 a+2 b x)}{9 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{1}{9} (17 c) \int \sec ^4(2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac{17 c^2 \sec ^3(2 a+2 b x) \tan (2 a+2 b x)}{63 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \sec ^4(2 a+2 b x) \tan (2 a+2 b x)}{9 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{1}{21} (34 c) \int \sec ^3(2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac{17 c^2 \sec ^3(2 a+2 b x) \tan (2 a+2 b x)}{63 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \sec ^4(2 a+2 b x) \tan (2 a+2 b x)}{9 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{34 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{105 b}+\frac{68}{105} \int \sec (2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \left (\frac{3 c}{2}+c \sec (2 a+2 b x)\right ) \, dx\\ &=-\frac{17 c^2 \sec ^3(2 a+2 b x) \tan (2 a+2 b x)}{63 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \sec ^4(2 a+2 b x) \tan (2 a+2 b x)}{9 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{68 c \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{315 b}+\frac{34 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{105 b}+\frac{1}{45} (34 c) \int \sec (2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac{34 c^2 \tan (2 a+2 b x)}{45 b \sqrt{-c+c \sec (2 a+2 b x)}}-\frac{17 c^2 \sec ^3(2 a+2 b x) \tan (2 a+2 b x)}{63 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c^2 \sec ^4(2 a+2 b x) \tan (2 a+2 b x)}{9 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{68 c \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{315 b}+\frac{34 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{105 b}\\ \end{align*}

Mathematica [A]  time = 0.367165, size = 85, normalized size = 0.41 \[ \frac{\cot (a+b x) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \left (188 \cot (a+b x) \cot (2 (a+b x))+35 \sec ^3(2 (a+b x))-50 \sec ^2(2 (a+b x))+52 \sec (2 (a+b x))-84\right )}{315 b} \]

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]*(-84 + 188*Cot[a + b*x]*Cot[2*(a + b*x)] + 52*Sec[2*(a + b*x)] - 50*Sec[2*(a + b*x)]^2 + 35*Sec[
2*(a + b*x)]^3)*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2))/(315*b)

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Maple [A]  time = 0.519, size = 105, normalized size = 0.5 \begin{align*}{\frac{2\,\sqrt{2} \left ( 2176\, \left ( \cos \left ( bx+a \right ) \right ) ^{8}-4896\, \left ( \cos \left ( bx+a \right ) \right ) ^{6}+4284\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}-1785\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+315 \right ) \cos \left ( bx+a \right ) }{315\,b \left ( 2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1 \right ) ^{3} \left ( \sin \left ( bx+a \right ) \right ) ^{3}} \left ({\frac{c \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}} \right ) ^{{\frac{3}{2}}}} \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]

2/315*2^(1/2)/b*(2176*cos(b*x+a)^8-4896*cos(b*x+a)^6+4284*cos(b*x+a)^4-1785*cos(b*x+a)^2+315)*cos(b*x+a)*(c*si
n(b*x+a)^2/(2*cos(b*x+a)^2-1))^(3/2)/(2*cos(b*x+a)^2-1)^3/sin(b*x+a)^3

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Maxima [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, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.06705, size = 348, normalized size = 1.67 \begin{align*} \frac{2 \, \sqrt{2}{\left (315 \, c \tan \left (b x + a\right )^{8} - 525 \, c \tan \left (b x + a\right )^{6} + 819 \, c \tan \left (b x + a\right )^{4} - 423 \, c \tan \left (b x + a\right )^{2} + 94 \, c\right )} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{315 \,{\left (b \tan \left (b x + a\right )^{9} - 4 \, b \tan \left (b x + a\right )^{7} + 6 \, b \tan \left (b x + a\right )^{5} - 4 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\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]

2/315*sqrt(2)*(315*c*tan(b*x + a)^8 - 525*c*tan(b*x + a)^6 + 819*c*tan(b*x + a)^4 - 423*c*tan(b*x + a)^2 + 94*
c)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))/(b*tan(b*x + a)^9 - 4*b*tan(b*x + a)^7 + 6*b*tan(b*x + a)^5 -
4*b*tan(b*x + a)^3 + b*tan(b*x + a))

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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(-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, algorithm="giac")

[Out]

Timed out