3.38 \(\int \sqrt{b \tan ^4(c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.020658, antiderivative size = 50, normalized size of antiderivative = 1., 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 = {3658, 3473, 8} \[ \frac{\cot (c+d x) \sqrt{b \tan ^4(c+d x)}}{d}-x \cot ^2(c+d x) \sqrt{b \tan ^4(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[c + d*x]^4],x]

[Out]

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

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
]])

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 8

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

Rubi steps

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

Mathematica [A]  time = 0.0884631, size = 41, normalized size = 0.82 \[ -\frac{\cot (c+d x) \sqrt{b \tan ^4(c+d x)} \left (\tan ^{-1}(\tan (c+d x)) \cot (c+d x)-1\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Tan[c + d*x]^4],x]

[Out]

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

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Maple [A]  time = 0.022, size = 42, normalized size = 0.8 \begin{align*} -{\frac{-\tan \left ( dx+c \right ) +\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ( \tan \left ( dx+c \right ) \right ) ^{2}}\sqrt{b \left ( \tan \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(d*x+c)^4)^(1/2),x)

[Out]

-1/d*(b*tan(d*x+c)^4)^(1/2)*(-tan(d*x+c)+arctan(tan(d*x+c)))/tan(d*x+c)^2

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Maxima [A]  time = 1.49983, size = 35, normalized size = 0.7 \begin{align*} -\frac{{\left (d x + c\right )} \sqrt{b} - \sqrt{b} \tan \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((d*x + c)*sqrt(b) - sqrt(b)*tan(d*x + c))/d

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Fricas [A]  time = 1.34459, size = 88, normalized size = 1.76 \begin{align*} -\frac{\sqrt{b \tan \left (d x + c\right )^{4}}{\left (d x - \tan \left (d x + c\right )\right )}}{d \tan \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(b*tan(d*x + c)^4)*(d*x - tan(d*x + c))/(d*tan(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan ^{4}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((tan(d*x+c)**4*b)**(1/2),x)

[Out]

Integral(sqrt(b*tan(c + d*x)**4), x)

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Giac [B]  time = 1.42438, size = 309, normalized size = 6.18 \begin{align*} \frac{{\left (\pi - 4 \, d x \tan \left (d x\right ) \tan \left (c\right ) - \pi \mathrm{sgn}\left (2 \, \tan \left (d x\right )^{2} \tan \left (c\right ) + 2 \, \tan \left (d x\right ) \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right ) - 2 \, \tan \left (c\right )\right ) \tan \left (d x\right ) \tan \left (c\right ) - \pi \tan \left (d x\right ) \tan \left (c\right ) + 2 \, \arctan \left (\frac{\tan \left (d x\right ) \tan \left (c\right ) - 1}{\tan \left (d x\right ) + \tan \left (c\right )}\right ) \tan \left (d x\right ) \tan \left (c\right ) + 2 \, \arctan \left (\frac{\tan \left (d x\right ) + \tan \left (c\right )}{\tan \left (d x\right ) \tan \left (c\right ) - 1}\right ) \tan \left (d x\right ) \tan \left (c\right ) + 4 \, d x + \pi \mathrm{sgn}\left (2 \, \tan \left (d x\right )^{2} \tan \left (c\right ) + 2 \, \tan \left (d x\right ) \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right ) - 2 \, \tan \left (c\right )\right ) - 2 \, \arctan \left (\frac{\tan \left (d x\right ) \tan \left (c\right ) - 1}{\tan \left (d x\right ) + \tan \left (c\right )}\right ) - 2 \, \arctan \left (\frac{\tan \left (d x\right ) + \tan \left (c\right )}{\tan \left (d x\right ) \tan \left (c\right ) - 1}\right ) - 4 \, \tan \left (d x\right ) - 4 \, \tan \left (c\right )\right )} \sqrt{b}}{4 \,{\left (d \tan \left (d x\right ) \tan \left (c\right ) - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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