3.1.14 \(\int (b \tan ^4(e+f x))^{3/2} \, dx\) [14]

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

[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 = 0.03, 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 = {3739, 3554, 8} \begin {gather*} -\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} \end {gather*}

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 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 \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 = 0.52, size = 66, normalized size = 0.60 \begin {gather*} \frac {\cot (e+f x) \left (3-5 \cot ^2(e+f x)+15 \cot ^4(e+f x)-15 \text {ArcTan}(\tan (e+f x)) \cot ^5(e+f x)\right ) \left (b \tan ^4(e+f x)\right )^{3/2}}{15 f} \end {gather*}

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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Maple [A]
time = 0.03, size = 64, normalized size = 0.58

method	result	size
derivativedivides	\(-\frac {\left (b \left (\tan ^{4}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (-3 \left (\tan ^{5}\left (f x +e \right )\right )+5 \left (\tan ^{3}\left (f x +e \right )\right )+15 \arctan \left (\tan \left (f x +e \right )\right )-15 \tan \left (f x +e \right )\right )}{15 f \tan \left (f x +e \right )^{6}}\)	\(64\)
default	\(-\frac {\left (b \left (\tan ^{4}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (-3 \left (\tan ^{5}\left (f x +e \right )\right )+5 \left (\tan ^{3}\left (f x +e \right )\right )+15 \arctan \left (\tan \left (f x +e \right )\right )-15 \tan \left (f x +e \right )\right )}{15 f \tan \left (f x +e \right )^{6}}\)	\(64\)
risch	\(\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, x}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {2 i b \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, \left (45 \,{\mathrm e}^{8 i \left (f x +e \right )}+90 \,{\mathrm e}^{6 i \left (f x +e \right )}+140 \,{\mathrm e}^{4 i \left (f x +e \right )}+70 \,{\mathrm e}^{2 i \left (f x +e \right )}+23\right )}{15 \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3} f}\)	\(170\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.50, size = 57, normalized size = 0.52 \begin {gather*} \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} \end {gather*}

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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Fricas [A]
time = 2.14, size = 67, normalized size = 0.61 \begin {gather*} \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}} \end {gather*}

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

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1079 vs. \(2 (106) = 212\).
time = 2.41, size = 1079, normalized size = 9.81 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*pi - 60*f*x*tan(f*x)^5*tan(e)^5 - 15*pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f*x) -
2*tan(e))*tan(f*x)^5*tan(e)^5 - 15*pi*tan(f*x)^5*tan(e)^5 + 30*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x) + tan(e)
))*tan(f*x)^5*tan(e)^5 + 30*arctan((tan(f*x) + tan(e))/(tan(f*x)*tan(e) - 1))*tan(f*x)^5*tan(e)^5 + 300*f*x*ta
n(f*x)^4*tan(e)^4 + 75*pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f*x) - 2*tan(e))*tan(f*x)^4*ta
n(e)^4 + 75*pi*tan(f*x)^4*tan(e)^4 - 150*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x) + tan(e)))*tan(f*x)^4*tan(e)^4
- 150*arctan((tan(f*x) + tan(e))/(tan(f*x)*tan(e) - 1))*tan(f*x)^4*tan(e)^4 - 60*tan(f*x)^5*tan(e)^4 - 60*tan
(f*x)^4*tan(e)^5 - 600*f*x*tan(f*x)^3*tan(e)^3 - 150*pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(
f*x) - 2*tan(e))*tan(f*x)^3*tan(e)^3 + 20*tan(f*x)^5*tan(e)^2 - 150*pi*tan(f*x)^3*tan(e)^3 + 300*arctan((tan(f
*x)*tan(e) - 1)/(tan(f*x) + tan(e)))*tan(f*x)^3*tan(e)^3 + 300*arctan((tan(f*x) + tan(e))/(tan(f*x)*tan(e) - 1
))*tan(f*x)^3*tan(e)^3 + 300*tan(f*x)^4*tan(e)^3 + 300*tan(f*x)^3*tan(e)^4 + 20*tan(f*x)^2*tan(e)^5 + 600*f*x*
tan(f*x)^2*tan(e)^2 + 150*pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f*x) - 2*tan(e))*tan(f*x)^2
*tan(e)^2 - 12*tan(f*x)^5 - 100*tan(f*x)^4*tan(e) + 150*pi*tan(f*x)^2*tan(e)^2 - 300*arctan((tan(f*x)*tan(e) -
1)/(tan(f*x) + tan(e)))*tan(f*x)^2*tan(e)^2 - 300*arctan((tan(f*x) + tan(e))/(tan(f*x)*tan(e) - 1))*tan(f*x)^
2*tan(e)^2 - 600*tan(f*x)^3*tan(e)^2 - 600*tan(f*x)^2*tan(e)^3 - 100*tan(f*x)*tan(e)^4 - 12*tan(e)^5 - 300*f*x
*tan(f*x)*tan(e) - 75*pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f*x) - 2*tan(e))*tan(f*x)*tan(e
) + 20*tan(f*x)^3 - 75*pi*tan(f*x)*tan(e) + 150*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x) + tan(e)))*tan(f*x)*tan
(e) + 150*arctan((tan(f*x) + tan(e))/(tan(f*x)*tan(e) - 1))*tan(f*x)*tan(e) + 300*tan(f*x)^2*tan(e) + 300*tan(
f*x)*tan(e)^2 + 20*tan(e)^3 + 60*f*x + 15*pi*sgn(2*tan(f*x)^2*tan(e) + 2*tan(f*x)*tan(e)^2 - 2*tan(f*x) - 2*ta
n(e)) - 30*arctan((tan(f*x)*tan(e) - 1)/(tan(f*x) + tan(e))) - 30*arctan((tan(f*x) + tan(e))/(tan(f*x)*tan(e)
- 1)) - 60*tan(f*x) - 60*tan(e))*b^(3/2)/(f*tan(f*x)^5*tan(e)^5 - 5*f*tan(f*x)^4*tan(e)^4 + 10*f*tan(f*x)^3*ta
n(e)^3 - 10*f*tan(f*x)^2*tan(e)^2 + 5*f*tan(f*x)*tan(e) - f)

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

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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