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]

34/105*(-c+c*sec(2*b*x+2*a))^(3/2)*tan(2*b*x+2*a)/b+34/45*c^2*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)-17/
63*c^2*sec(2*b*x+2*a)^3*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/9*c^2*sec(2*b*x+2*a)^4*tan(2*b*x+2*a)/b
/(-c+c*sec(2*b*x+2*a))^(1/2)+68/315*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 = 0.53, antiderivative size = 208, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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 ^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*}

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Mathematica [A]  time = 0.35, 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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fricas [A]  time = 1.05, size = 132, normalized size = 0.63 \[ \frac {2 \, \sqrt {2} {\left (315 \, c \tan \left (b x + a\right )^{8} - 525 \, c \tan \left (b x + a\right )^{6} + 819 \, c \tan \left (b x + a\right )^{4} - 423 \, c \tan \left (b x + a\right )^{2} + 94 \, c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{315 \, {\left (b \tan \left (b x + a\right )^{9} - 4 \, b \tan \left (b x + a\right )^{7} + 6 \, b \tan \left (b x + a\right )^{5} - 4 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}} \]

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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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)^4*(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 = 1.12, size = 105, normalized size = 0.50 \[ \frac {2 \sqrt {2}\, \left (2176 \left (\cos ^{8}\left (b x +a \right )\right )-4896 \left (\cos ^{6}\left (b x +a \right )\right )+4284 \left (\cos ^{4}\left (b x +a \right )\right )-1785 \left (\cos ^{2}\left (b x +a \right )\right )+315\right ) \cos \left (b x +a \right ) \left (\frac {c \left (\sin ^{2}\left (b x +a \right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}\right )^{\frac {3}{2}}}{315 b \left (2 \left (\cos ^{2}\left (b x +a \right )\right )-1\right )^{3} \sin \left (b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(2*b*x+2*a)^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 = 0.00, 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)^4*(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 = 10.18, size = 594, normalized size = 2.86 \[ \frac {\left (\frac {c\,16{}\mathrm {i}}{9\,b}+\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,16{}\mathrm {i}}{9\,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 )}^4}-\frac {\left (\frac {c\,40{}\mathrm {i}}{7\,b}+\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,88{}\mathrm {i}}{63\,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\,24{}\mathrm {i}}{5\,b}-\frac {c\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,176{}\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 )}^2}+\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 )}}\,272{}\mathrm {i}}{315\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\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 )}}\,136{}\mathrm {i}}{315\,b\,\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 )} \]

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)^4,x)

[Out]

(((c*16i)/(9*b) + (c*exp(a*2i + b*x*2i)*16i)/(9*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)^4) - (((c*40i)/(7*b) + (c*exp(a*2i + b*x*2i)*88i)/(63*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*24i)/(5*b) - (c*exp(a*2i + b*x*2i)*176i)/(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)^2) + (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)*272i)/(315*b*(exp(a*2i + b*x*2i) - 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)*136i)/(315*b*(exp(a*2i + b*x*2i) - 1)*(exp(a*4i + b*x*4i) + 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)**4*(c*tan(b*x+a)*tan(2*b*x+2*a))**(3/2),x)

[Out]

Timed out

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