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3.14 \(\int (b \tan ^4(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=110 \[ -\frac {b \tan (e+f x) \sqrt {b \tan ^4(e+f x)}}{3 f}+\frac {b \tan ^3(e+f x) \sqrt {b \tan ^4(e+f x)}}{5 f}-b x \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}+\frac {b \cot (e+f x) \sqrt {b \tan ^4(e+f x)}}{f} \]

[Out]

b*cot(f*x+e)*(b*tan(f*x+e)^4)^(1/2)/f-b*x*cot(f*x+e)^2*(b*tan(f*x+e)^4)^(1/2)-1/3*b*(b*tan(f*x+e)^4)^(1/2)*tan
(f*x+e)/f+1/5*b*(b*tan(f*x+e)^4)^(1/2)*tan(f*x+e)^3/f

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Rubi [A]  time = 0.04, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3658, 3473, 8} \[ \frac {b \tan ^3(e+f x) \sqrt {b \tan ^4(e+f x)}}{5 f}-\frac {b \tan (e+f x) \sqrt {b \tan ^4(e+f x)}}{3 f}-b x \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}+\frac {b \cot (e+f x) \sqrt {b \tan ^4(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^4)^(3/2),x]

[Out]

(b*Cot[e + f*x]*Sqrt[b*Tan[e + f*x]^4])/f - b*x*Cot[e + f*x]^2*Sqrt[b*Tan[e + f*x]^4] - (b*Tan[e + f*x]*Sqrt[b
*Tan[e + f*x]^4])/(3*f) + (b*Tan[e + f*x]^3*Sqrt[b*Tan[e + f*x]^4])/(5*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

\begin {align*} \int \left (b \tan ^4(e+f x)\right )^{3/2} \, dx &=\left (b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}\right ) \int \tan ^6(e+f x) \, dx\\ &=\frac {b \tan ^3(e+f x) \sqrt {b \tan ^4(e+f x)}}{5 f}-\left (b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}\right ) \int \tan ^4(e+f x) \, dx\\ &=-\frac {b \tan (e+f x) \sqrt {b \tan ^4(e+f x)}}{3 f}+\frac {b \tan ^3(e+f x) \sqrt {b \tan ^4(e+f x)}}{5 f}+\left (b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}\right ) \int \tan ^2(e+f x) \, dx\\ &=\frac {b \cot (e+f x) \sqrt {b \tan ^4(e+f x)}}{f}-\frac {b \tan (e+f x) \sqrt {b \tan ^4(e+f x)}}{3 f}+\frac {b \tan ^3(e+f x) \sqrt {b \tan ^4(e+f x)}}{5 f}-\left (b \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}\right ) \int 1 \, dx\\ &=\frac {b \cot (e+f x) \sqrt {b \tan ^4(e+f x)}}{f}-b x \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}-\frac {b \tan (e+f x) \sqrt {b \tan ^4(e+f x)}}{3 f}+\frac {b \tan ^3(e+f x) \sqrt {b \tan ^4(e+f x)}}{5 f}\\ \end {align*}

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Mathematica [A]  time = 0.75, size = 66, normalized size = 0.60 \[ \frac {\cot (e+f x) \left (b \tan ^4(e+f x)\right )^{3/2} \left (15 \cot ^4(e+f x)-5 \cot ^2(e+f x)-15 \tan ^{-1}(\tan (e+f x)) \cot ^5(e+f x)+3\right )}{15 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^4)^(3/2),x]

[Out]

(Cot[e + f*x]*(3 - 5*Cot[e + f*x]^2 + 15*Cot[e + f*x]^4 - 15*ArcTan[Tan[e + f*x]]*Cot[e + f*x]^5)*(b*Tan[e + f
*x]^4)^(3/2))/(15*f)

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fricas [A]  time = 0.76, size = 62, normalized size = 0.56 \[ \frac {{\left (3 \, b \tan \left (f x + e\right )^{5} - 5 \, b \tan \left (f x + e\right )^{3} - 15 \, b f x + 15 \, b \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{4}}}{15 \, f \tan \left (f x + e\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(3/2),x, algorithm="fricas")

[Out]

1/15*(3*b*tan(f*x + e)^5 - 5*b*tan(f*x + e)^3 - 15*b*f*x + 15*b*tan(f*x + e))*sqrt(b*tan(f*x + e)^4)/(f*tan(f*
x + e)^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)sqrt(b)*b*(-60*f*x*tan(exp(1))^5*tan(f*x)^5+300*f*x*tan(exp(1)
)^4*tan(f*x)^4-600*f*x*tan(exp(1))^3*tan(f*x)^3+600*f*x*tan(exp(1))^2*tan(f*x)^2-300*f*x*tan(exp(1))*tan(f*x)+
60*f*x-15*pi*sign(2*tan(exp(1))^2*tan(f*x)+2*tan(exp(1))*tan(f*x)^2-2*tan(exp(1))-2*tan(f*x))*tan(exp(1))^5*ta
n(f*x)^5+75*pi*sign(2*tan(exp(1))^2*tan(f*x)+2*tan(exp(1))*tan(f*x)^2-2*tan(exp(1))-2*tan(f*x))*tan(exp(1))^4*
tan(f*x)^4-150*pi*sign(2*tan(exp(1))^2*tan(f*x)+2*tan(exp(1))*tan(f*x)^2-2*tan(exp(1))-2*tan(f*x))*tan(exp(1))
^3*tan(f*x)^3+150*pi*sign(2*tan(exp(1))^2*tan(f*x)+2*tan(exp(1))*tan(f*x)^2-2*tan(exp(1))-2*tan(f*x))*tan(exp(
1))^2*tan(f*x)^2-75*pi*sign(2*tan(exp(1))^2*tan(f*x)+2*tan(exp(1))*tan(f*x)^2-2*tan(exp(1))-2*tan(f*x))*tan(ex
p(1))*tan(f*x)+15*pi*sign(2*tan(exp(1))^2*tan(f*x)+2*tan(exp(1))*tan(f*x)^2-2*tan(exp(1))-2*tan(f*x))-15*pi*ta
n(exp(1))^5*tan(f*x)^5+75*pi*tan(exp(1))^4*tan(f*x)^4-150*pi*tan(exp(1))^3*tan(f*x)^3+150*pi*tan(exp(1))^2*tan
(f*x)^2-75*pi*tan(exp(1))*tan(f*x)+15*pi+30*atan((tan(exp(1))*tan(f*x)-1)/(tan(exp(1))+tan(f*x)))*tan(exp(1))^
5*tan(f*x)^5-150*atan((tan(exp(1))*tan(f*x)-1)/(tan(exp(1))+tan(f*x)))*tan(exp(1))^4*tan(f*x)^4+300*atan((tan(
exp(1))*tan(f*x)-1)/(tan(exp(1))+tan(f*x)))*tan(exp(1))^3*tan(f*x)^3-300*atan((tan(exp(1))*tan(f*x)-1)/(tan(ex
p(1))+tan(f*x)))*tan(exp(1))^2*tan(f*x)^2+150*atan((tan(exp(1))*tan(f*x)-1)/(tan(exp(1))+tan(f*x)))*tan(exp(1)
)*tan(f*x)-30*atan((tan(exp(1))*tan(f*x)-1)/(tan(exp(1))+tan(f*x)))+30*atan((tan(exp(1))+tan(f*x))/(tan(exp(1)
)*tan(f*x)-1))*tan(exp(1))^5*tan(f*x)^5-150*atan((tan(exp(1))+tan(f*x))/(tan(exp(1))*tan(f*x)-1))*tan(exp(1))^
4*tan(f*x)^4+300*atan((tan(exp(1))+tan(f*x))/(tan(exp(1))*tan(f*x)-1))*tan(exp(1))^3*tan(f*x)^3-300*atan((tan(
exp(1))+tan(f*x))/(tan(exp(1))*tan(f*x)-1))*tan(exp(1))^2*tan(f*x)^2+150*atan((tan(exp(1))+tan(f*x))/(tan(exp(
1))*tan(f*x)-1))*tan(exp(1))*tan(f*x)-30*atan((tan(exp(1))+tan(f*x))/(tan(exp(1))*tan(f*x)-1))-60*tan(exp(1))^
5*tan(f*x)^4+20*tan(exp(1))^5*tan(f*x)^2-12*tan(exp(1))^5-60*tan(exp(1))^4*tan(f*x)^5+300*tan(exp(1))^4*tan(f*
x)^3-100*tan(exp(1))^4*tan(f*x)+300*tan(exp(1))^3*tan(f*x)^4-600*tan(exp(1))^3*tan(f*x)^2+20*tan(exp(1))^3+20*
tan(exp(1))^2*tan(f*x)^5-600*tan(exp(1))^2*tan(f*x)^3+300*tan(exp(1))^2*tan(f*x)-100*tan(exp(1))*tan(f*x)^4+30
0*tan(exp(1))*tan(f*x)^2-60*tan(exp(1))-12*tan(f*x)^5+20*tan(f*x)^3-60*tan(f*x))/(60*f*tan(exp(1))^5*tan(f*x)^
5-300*f*tan(exp(1))^4*tan(f*x)^4+600*f*tan(exp(1))^3*tan(f*x)^3-600*f*tan(exp(1))^2*tan(f*x)^2+300*f*tan(exp(1
))*tan(f*x)-60*f)

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maple [A]  time = 0.21, size = 64, normalized size = 0.58 \[ -\frac {\left (b \left (\tan ^{4}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (-3 \left (\tan ^{5}\left (f x +e \right )\right )+5 \left (\tan ^{3}\left (f x +e \right )\right )+15 \arctan \left (\tan \left (f x +e \right )\right )-15 \tan \left (f x +e \right )\right )}{15 f \tan \left (f x +e \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^4)^(3/2),x)

[Out]

-1/15/f*(b*tan(f*x+e)^4)^(3/2)*(-3*tan(f*x+e)^5+5*tan(f*x+e)^3+15*arctan(tan(f*x+e))-15*tan(f*x+e))/tan(f*x+e)
^6

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maxima [A]  time = 0.85, size = 53, normalized size = 0.48 \[ \frac {3 \, b^{\frac {3}{2}} \tan \left (f x + e\right )^{5} - 5 \, b^{\frac {3}{2}} \tan \left (f x + e\right )^{3} - 15 \, {\left (f x + e\right )} b^{\frac {3}{2}} + 15 \, b^{\frac {3}{2}} \tan \left (f x + e\right )}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(3/2),x, algorithm="maxima")

[Out]

1/15*(3*b^(3/2)*tan(f*x + e)^5 - 5*b^(3/2)*tan(f*x + e)^3 - 15*(f*x + e)*b^(3/2) + 15*b^(3/2)*tan(f*x + e))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x)^4)^(3/2),x)

[Out]

int((b*tan(e + f*x)^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)**4)**(3/2),x)

[Out]

Integral((b*tan(e + f*x)**4)**(3/2), x)

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