3.1 \(\int (b \tan ^2(e+f x))^{5/2} \, dx\)
Optimal. Leaf size=98 \[ -\frac {b^2 \tan (e+f x) \sqrt {b \tan ^2(e+f x)}}{2 f}+\frac {b^2 \tan ^3(e+f x) \sqrt {b \tan ^2(e+f x)}}{4 f}-\frac {b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \log (\cos (e+f x))}{f} \]
[Out]
-b^2*cot(f*x+e)*ln(cos(f*x+e))*(b*tan(f*x+e)^2)^(1/2)/f-1/2*b^2*(b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f+1/4*b^2*(b
*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3/f
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Rubi [A] time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used =
{3658, 3473, 3475} \[ \frac {b^2 \tan ^3(e+f x) \sqrt {b \tan ^2(e+f x)}}{4 f}-\frac {b^2 \tan (e+f x) \sqrt {b \tan ^2(e+f x)}}{2 f}-\frac {b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)} \log (\cos (e+f x))}{f} \]
Antiderivative was successfully verified.
[In]
Int[(b*Tan[e + f*x]^2)^(5/2),x]
[Out]
-((b^2*Cot[e + f*x]*Log[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]^2])/f) - (b^2*Tan[e + f*x]*Sqrt[b*Tan[e + f*x]^2])/(
2*f) + (b^2*Tan[e + f*x]^3*Sqrt[b*Tan[e + f*x]^2])/(4*f)
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 3475
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]
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 ^2(e+f x)\right )^{5/2} \, dx &=\left (b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)}\right ) \int \tan ^5(e+f x) \, dx\\ &=\frac {b^2 \tan ^3(e+f x) \sqrt {b \tan ^2(e+f x)}}{4 f}-\left (b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)}\right ) \int \tan ^3(e+f x) \, dx\\ &=-\frac {b^2 \tan (e+f x) \sqrt {b \tan ^2(e+f x)}}{2 f}+\frac {b^2 \tan ^3(e+f x) \sqrt {b \tan ^2(e+f x)}}{4 f}+\left (b^2 \cot (e+f x) \sqrt {b \tan ^2(e+f x)}\right ) \int \tan (e+f x) \, dx\\ &=-\frac {b^2 \cot (e+f x) \log (\cos (e+f x)) \sqrt {b \tan ^2(e+f x)}}{f}-\frac {b^2 \tan (e+f x) \sqrt {b \tan ^2(e+f x)}}{2 f}+\frac {b^2 \tan ^3(e+f x) \sqrt {b \tan ^2(e+f x)}}{4 f}\\ \end {align*}
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Mathematica [A] time = 0.36, size = 56, normalized size = 0.57 \[ -\frac {\cot (e+f x) \left (b \tan ^2(e+f x)\right )^{5/2} \left (2 \cot ^2(e+f x)+4 \cot ^4(e+f x) \log (\cos (e+f x))-1\right )}{4 f} \]
Antiderivative was successfully verified.
[In]
Integrate[(b*Tan[e + f*x]^2)^(5/2),x]
[Out]
-1/4*(Cot[e + f*x]*(-1 + 2*Cot[e + f*x]^2 + 4*Cot[e + f*x]^4*Log[Cos[e + f*x]])*(b*Tan[e + f*x]^2)^(5/2))/f
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fricas [A] time = 0.74, size = 74, normalized size = 0.76 \[ \frac {{\left (b^{2} \tan \left (f x + e\right )^{4} - 2 \, b^{2} \tan \left (f x + e\right )^{2} - 2 \, b^{2} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2}}}{4 \, f \tan \left (f x + e\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")
[Out]
1/4*(b^2*tan(f*x + e)^4 - 2*b^2*tan(f*x + e)^2 - 2*b^2*log(1/(tan(f*x + e)^2 + 1)) - 3*b^2)*sqrt(b*tan(f*x + e
)^2)/(f*tan(f*x + e))
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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)^2)^(5/2),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)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)sqrt(b)*(-3*b^2*si
gn(tan(f*x+exp(1)))-2*b^2*sign(tan(f*x+exp(1)))*tan(f*x)^2+b^2*sign(tan(f*x+exp(1)))*tan(f*x)^4-2*b^2*sign(tan
(f*x+exp(1)))*tan(exp(1))^2+b^2*sign(tan(f*x+exp(1)))*tan(exp(1))^4-2*b^2*sign(tan(f*x+exp(1)))*ln((4*tan(f*x)
^2*tan(exp(1))^2-8*tan(f*x)^3*tan(exp(1))+4*tan(f*x)^4*tan(exp(1))^2+4*tan(f*x)^2-8*tan(f*x)*tan(exp(1))+4)/(t
an(exp(1))^2+1))-4*b^2*sign(tan(f*x+exp(1)))*tan(f*x)^2*tan(exp(1))^2-2*b^2*sign(tan(f*x+exp(1)))*tan(f*x)^2*t
an(exp(1))^4+8*b^2*sign(tan(f*x+exp(1)))*tan(f*x)^3*tan(exp(1))^3+8*b^2*sign(tan(f*x+exp(1)))*tan(f*x)^3*tan(e
xp(1))-2*b^2*sign(tan(f*x+exp(1)))*tan(f*x)^4*tan(exp(1))^2-3*b^2*sign(tan(f*x+exp(1)))*tan(f*x)^4*tan(exp(1))
^4+8*b^2*sign(tan(f*x+exp(1)))*tan(f*x)*tan(exp(1))^3+8*b^2*sign(tan(f*x+exp(1)))*tan(f*x)*tan(exp(1))-12*b^2*
sign(tan(f*x+exp(1)))*tan(f*x)^2*tan(exp(1))^2*ln((4*tan(f*x)^2*tan(exp(1))^2-8*tan(f*x)^3*tan(exp(1))+4*tan(f
*x)^4*tan(exp(1))^2+4*tan(f*x)^2-8*tan(f*x)*tan(exp(1))+4)/(tan(exp(1))^2+1))+8*b^2*sign(tan(f*x+exp(1)))*tan(
f*x)^3*tan(exp(1))^3*ln((4*tan(f*x)^2*tan(exp(1))^2-8*tan(f*x)^3*tan(exp(1))+4*tan(f*x)^4*tan(exp(1))^2+4*tan(
f*x)^2-8*tan(f*x)*tan(exp(1))+4)/(tan(exp(1))^2+1))-2*b^2*sign(tan(f*x+exp(1)))*tan(f*x)^4*tan(exp(1))^4*ln((4
*tan(f*x)^2*tan(exp(1))^2-8*tan(f*x)^3*tan(exp(1))+4*tan(f*x)^4*tan(exp(1))^2+4*tan(f*x)^2-8*tan(f*x)*tan(exp(
1))+4)/(tan(exp(1))^2+1))+8*b^2*sign(tan(f*x+exp(1)))*tan(f*x)*tan(exp(1))*ln((4*tan(f*x)^2*tan(exp(1))^2-8*ta
n(f*x)^3*tan(exp(1))+4*tan(f*x)^4*tan(exp(1))^2+4*tan(f*x)^2-8*tan(f*x)*tan(exp(1))+4)/(tan(exp(1))^2+1)))/(4*
f*tan(f*x)^4*tan(exp(1))^4-16*f*tan(f*x)^3*tan(exp(1))^3+24*f*tan(f*x)^2*tan(exp(1))^2-16*f*tan(f*x)*tan(exp(1
))+4*f)
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maple [A] time = 0.29, size = 58, normalized size = 0.59 \[ \frac {\left (b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {5}{2}} \left (\tan ^{4}\left (f x +e \right )-2 \left (\tan ^{2}\left (f x +e \right )\right )+2 \ln \left (1+\tan ^{2}\left (f x +e \right )\right )\right )}{4 f \tan \left (f x +e \right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*tan(f*x+e)^2)^(5/2),x)
[Out]
1/4/f*(b*tan(f*x+e)^2)^(5/2)*(tan(f*x+e)^4-2*tan(f*x+e)^2+2*ln(1+tan(f*x+e)^2))/tan(f*x+e)^5
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maxima [A] time = 0.61, size = 47, normalized size = 0.48 \[ \frac {b^{\frac {5}{2}} \tan \left (f x + e\right )^{4} - 2 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{2} + 2 \, b^{\frac {5}{2}} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{4 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")
[Out]
1/4*(b^(5/2)*tan(f*x + e)^4 - 2*b^(5/2)*tan(f*x + e)^2 + 2*b^(5/2)*log(tan(f*x + e)^2 + 1))/f
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*tan(e + f*x)^2)^(5/2),x)
[Out]
int((b*tan(e + f*x)^2)^(5/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*tan(f*x+e)**2)**(5/2),x)
[Out]
Integral((b*tan(e + f*x)**2)**(5/2), x)
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