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

Optimal. Leaf size=75 \[ \frac {c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {4 c^2 \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}} \]

[Out]

-4/3*c^2*tan(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/3*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 = 0.11, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {4397, 3793, 3792} \[ \frac {c \tan (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}}{3 b}-\frac {4 c^2 \tan (2 a+2 b x)}{3 b \sqrt {c \sec (2 a+2 b x)-c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4397

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

Rubi steps

\begin {align*} \int \sec (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx &=\int \sec (2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \, dx\\ &=\frac {c \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b}-\frac {1}{3} (4 c) \int \sec (2 a+2 b x) \sqrt {-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac {4 c^2 \tan (2 a+2 b x)}{3 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {c \sqrt {-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 51, normalized size = 0.68 \[ -\frac {\cot (a+b x) (4 \cot (a+b x) \cot (2 (a+b x))-1) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Cot[a + b*x]*(-1 + 4*Cot[a + b*x]*Cot[2*(a + b*x)])*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2))/b

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fricas [A]  time = 0.88, size = 67, normalized size = 0.89 \[ -\frac {2 \, \sqrt {2} {\left (3 \, c \tan \left (b x + a\right )^{2} - 2 \, c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{3 \, {\left (b \tan \left (b x + a\right )^{3} - b \tan \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(2)*(3*c*tan(b*x + a)^2 - 2*c)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))/(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)*(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 = 0.83, size = 61, normalized size = 0.81 \[ -\frac {2 \sqrt {2}\, \left (5 \left (\cos ^{2}\left (b x +a \right )\right )-3\right ) \cos \left (b x +a \right ) \left (\frac {c \left (\sin ^{2}\left (b x +a \right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}\right )^{\frac {3}{2}}}{3 b \sin \left (b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*2^(1/2)/b*(5*cos(b*x+a)^2-3)*cos(b*x+a)*(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(-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)*(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 = 7.87, size = 158, normalized size = 2.11 \[ -\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,3{}\mathrm {i}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}\,5{}\mathrm {i}+5{}\mathrm {i}\right )\,\sqrt {\frac {c\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+1\right )}}}{3\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}-1\right )} \]

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),x)

[Out]

-(c*(exp(a*2i + b*x*2i)*3i + exp(a*4i + b*x*4i)*3i + exp(a*6i + b*x*6i)*5i + 5i)*((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))/(3*b*(exp(a*2i +
b*x*2i) - exp(a*4i + b*x*4i) + exp(a*6i + b*x*6i) - 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)*(c*tan(b*x+a)*tan(2*b*x+2*a))**(3/2),x)

[Out]

Timed out

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