3.1.15 \(\int \sqrt {b \tan ^4(e+f x)} \, dx\) [15]

Optimal. Leaf size=50 \[ \frac {\cot (e+f x) \sqrt {b \tan ^4(e+f x)}}{f}-x \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \begin {gather*} \frac {\cot (e+f x) \sqrt {b \tan ^4(e+f x)}}{f}-x \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[e + f*x]^4],x]

[Out]

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

Rule 8

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

Rule 3554

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 3739

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

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Mathematica [A]
time = 0.07, size = 41, normalized size = 0.82 \begin {gather*} -\frac {\cot (e+f x) (-1+\text {ArcTan}(\tan (e+f x)) \cot (e+f x)) \sqrt {b \tan ^4(e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Tan[e + f*x]^4],x]

[Out]

-((Cot[e + f*x]*(-1 + ArcTan[Tan[e + f*x]]*Cot[e + f*x])*Sqrt[b*Tan[e + f*x]^4])/f)

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Maple [A]
time = 0.03, size = 42, normalized size = 0.84

method	result	size
derivativedivides	\(-\frac {\sqrt {b \left (\tan ^{4}\left (f x +e \right )\right )}\, \left (-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f \tan \left (f x +e \right )^{2}}\)	\(42\)
default	\(-\frac {\sqrt {b \left (\tan ^{4}\left (f x +e \right )\right )}\, \left (-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f \tan \left (f x +e \right )^{2}}\)	\(42\)
risch	\(\frac {\sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} x}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {2 i \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} f}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 28, normalized size = 0.56 \begin {gather*} -\frac {{\left (f x + e\right )} \sqrt {b} - \sqrt {b} \tan \left (f x + e\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(1/2),x, algorithm="maxima")

[Out]

-((f*x + e)*sqrt(b) - sqrt(b)*tan(f*x + e))/f

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Fricas [A]
time = 1.96, size = 40, normalized size = 0.80 \begin {gather*} -\frac {\sqrt {b \tan \left (f x + e\right )^{4}} {\left (f x - \tan \left (f x + e\right )\right )}}{f \tan \left (f x + e\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(1/2),x, algorithm="fricas")

[Out]

-sqrt(b*tan(f*x + e)^4)*(f*x - tan(f*x + e))/(f*tan(f*x + e)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {b \tan ^{4}{\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)**4)**(1/2),x)

[Out]

Integral(sqrt(b*tan(e + f*x)**4), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (50) = 100\).
time = 0.57, size = 250, normalized size = 5.00 \begin {gather*} \frac {{\left (\pi - 4 \, f x \tan \left (f x\right ) \tan \left (e\right ) - \pi \mathrm {sgn}\left (2 \, \tan \left (f x\right )^{2} \tan \left (e\right ) + 2 \, \tan \left (f x\right ) \tan \left (e\right )^{2} - 2 \, \tan \left (f x\right ) - 2 \, \tan \left (e\right )\right ) \tan \left (f x\right ) \tan \left (e\right ) - \pi \tan \left (f x\right ) \tan \left (e\right ) + 2 \, \arctan \left (\frac {\tan \left (f x\right ) \tan \left (e\right ) - 1}{\tan \left (f x\right ) + \tan \left (e\right )}\right ) \tan \left (f x\right ) \tan \left (e\right ) + 2 \, \arctan \left (\frac {\tan \left (f x\right ) + \tan \left (e\right )}{\tan \left (f x\right ) \tan \left (e\right ) - 1}\right ) \tan \left (f x\right ) \tan \left (e\right ) + 4 \, f x + \pi \mathrm {sgn}\left (2 \, \tan \left (f x\right )^{2} \tan \left (e\right ) + 2 \, \tan \left (f x\right ) \tan \left (e\right )^{2} - 2 \, \tan \left (f x\right ) - 2 \, \tan \left (e\right )\right ) - 2 \, \arctan \left (\frac {\tan \left (f x\right ) \tan \left (e\right ) - 1}{\tan \left (f x\right ) + \tan \left (e\right )}\right ) - 2 \, \arctan \left (\frac {\tan \left (f x\right ) + \tan \left (e\right )}{\tan \left (f x\right ) \tan \left (e\right ) - 1}\right ) - 4 \, \tan \left (f x\right ) - 4 \, \tan \left (e\right )\right )} \sqrt {b}}{4 \, {\left (f \tan \left (f x\right ) \tan \left (e\right ) - f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(1/2),x, algorithm="giac")

[Out]

1/4*(pi - 4*f*x*tan(f*x)*tan(e) - pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f*x) - 2*tan(e))*ta
n(f*x)*tan(e) - pi*tan(f*x)*tan(e) + 2*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x) + tan(e)))*tan(f*x)*tan(e) + 2*a
rctan((tan(f*x) + tan(e))/(tan(f*x)*tan(e) - 1))*tan(f*x)*tan(e) + 4*f*x + pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(
f*x)*tan(e)^2 - 2*tan(f*x) - 2*tan(e)) - 2*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x) + tan(e))) - 2*arctan((tan(f
*x) + tan(e))/(tan(f*x)*tan(e) - 1)) - 4*tan(f*x) - 4*tan(e))*sqrt(b)/(f*tan(f*x)*tan(e) - f)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x)^4)^(1/2),x)

[Out]

int((b*tan(e + f*x)^4)^(1/2), x)

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