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

Optimal. Leaf size=80 \[ \frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}+\frac {c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4397, 3775, 21, 3774, 207} \[ \frac {c^2 \tan (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}+\frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3775

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[a/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 2)*(a*(n - 1) + b*(3*n - 4)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 4397

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

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 86, normalized size = 1.08 \[ \frac {c \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \left (2 \cot (a+b x)+\sqrt {2} \sqrt {\cos (2 (a+b x))} \csc (a+b x) \tanh ^{-1}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(2*Cot[a + b*x] + Sqrt[2]*ArcTanh[(Sqrt[2]*Cos[a + b*x])/Sqrt[Cos[2*(a + b*x)]]]*Sqrt[Cos[2*(a + b*x)]]*Csc
[a + b*x])*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(2*b)

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fricas [A]  time = 1.03, size = 296, normalized size = 3.70 \[ \left [\frac {c^{\frac {3}{2}} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} + 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right ) + 4 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} c}{4 \, b \tan \left (b x + a\right )}, -\frac {\sqrt {-c} c \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right ) - 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} c}{2 \, b \tan \left (b x + a\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(c^(3/2)*log(-(c*tan(b*x + a)^5 - 14*c*tan(b*x + a)^3 + 4*sqrt(2)*(tan(b*x + a)^4 - 4*tan(b*x + a)^2 + 3)
*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*sqrt(c) + 17*c*tan(b*x + a))/(tan(b*x + a)^5 + 2*tan(b*x + a)^3
+ tan(b*x + a)))*tan(b*x + a) + 4*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*c)/(b*tan(b*x + a)), -1
/2*(sqrt(-c)*c*arctan(2*sqrt(2)*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)^3 - 3*c*tan(b*x + a)))*tan(b*x + a) - 2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*c)/(
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((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 = 0.95, size = 253, normalized size = 3.16 \[ \frac {\sqrt {2}\, \left (2 \left (\cos ^{2}\left (b x +a \right )\right )-1\right ) \left (\sqrt {2}\, \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \cos \left (b x +a \right )+\sqrt {2}\, \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \arctanh \left (\frac {\cos \left (b x +a \right ) \sqrt {4}\, \left (-1+\cos \left (b x +a \right )\right ) \sqrt {2}}{2 \sin \left (b x +a \right )^{2} \sqrt {\frac {2 \left (\cos ^{2}\left (b x +a \right )\right )-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right )-2 \cos \left (b x +a \right )\right ) \left (\frac {c \left (\sin ^{2}\left (b x +a \right )\right )}{2 \left (\cos ^{2}\left (b x +a \right )\right )-1}\right )^{\frac {3}{2}}}{b \sin \left (b x +a \right )^{3} \left (2+\sqrt {2}\right ) \left (\sqrt {2}-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2^(1/2)/b*(2*cos(b*x+a)^2-1)*(2^(1/2)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*arctanh(1/2*cos(b*x+a)*4^(1/
2)*(-1+cos(b*x+a))/sin(b*x+a)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*2^(1/2))*cos(b*x+a)+2^(1/2)*((2*co
s(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*arctanh(1/2*cos(b*x+a)*4^(1/2)*(-1+cos(b*x+a))/sin(b*x+a)^2/((2*cos(b*x+
a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*2^(1/2))-2*cos(b*x+a))*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(3/2)/sin(b*x+a)^3/
(2+2^(1/2))/(2^(1/2)-2)

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maxima [B]  time = 0.89, size = 1317, normalized size = 16.46 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*(c*log(sqrt(cos(4*b*x + 4*a)^2
+ sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + s
qrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4
*b*x + 4*a) - 1))^2 + 2*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*sin(1/2*arcta
n2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + 1) - c*log(sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos
(4*b*x + 4*a) + 1)*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + sqrt(cos(4*b*x + 4*a)^2 + sin
(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 - 2*(cos
(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*
b*x + 4*a) - 1)) + 1) + c*log(((cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + sin(1/2*arctan2(
sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2)*cos(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))^2 + (cos(1/2
*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1
))^2)*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))^2)*sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2
*cos(4*b*x + 4*a) + 1) + 2*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*(cos(1/2*a
rctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + cos(1/
2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))) + 1)
- c*log(((cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + sin(1/2*arctan2(sin(4*b*x + 4*a), -co
s(4*b*x + 4*a) - 1))^2)*cos(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))^2 + (cos(1/2*arctan2(sin(4*b*x +
4*a), -cos(4*b*x + 4*a) - 1))^2 + sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2)*sin(1/2*arctan2
(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))^2)*sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1
) - 2*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*(cos(1/2*arctan2(sin(4*b*x + 4*
a), cos(4*b*x + 4*a)))*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + cos(1/2*arctan2(sin(4*b*x +
4*a), -cos(4*b*x + 4*a) - 1))*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))) + 1))*sqrt(c) + 8*(c*cos(
1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a)))*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + c
*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))
) + c*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*sqrt(c))/((cos(4*b*x + 4*a)^2 + sin(4*b*x + 4
*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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