3.186 \(\int \tan ^3(e+f x) (a+b \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=53 \[ \frac{(a-b) \tan ^2(e+f x)}{2 f}+\frac{(a-b) \log (\cos (e+f x))}{f}+\frac{b \tan ^4(e+f x)}{4 f} \]

[Out]

((a - b)*Log[Cos[e + f*x]])/f + ((a - b)*Tan[e + f*x]^2)/(2*f) + (b*Tan[e + f*x]^4)/(4*f)

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Rubi [A]  time = 0.0387884, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3631, 3473, 3475} \[ \frac{(a-b) \tan ^2(e+f x)}{2 f}+\frac{(a-b) \log (\cos (e+f x))}{f}+\frac{b \tan ^4(e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^3*(a + b*Tan[e + f*x]^2),x]

[Out]

((a - b)*Log[Cos[e + f*x]])/f + ((a - b)*Tan[e + f*x]^2)/(2*f) + (b*Tan[e + f*x]^4)/(4*f)

Rule 3631

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp
[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[A - C, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[
{a, b, e, f, A, C, m}, x] && NeQ[A*b^2 + a^2*C, 0] &&  !LeQ[m, -1]

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]

Rubi steps

\begin{align*} \int \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx &=\frac{b \tan ^4(e+f x)}{4 f}+(a-b) \int \tan ^3(e+f x) \, dx\\ &=\frac{(a-b) \tan ^2(e+f x)}{2 f}+\frac{b \tan ^4(e+f x)}{4 f}+(-a+b) \int \tan (e+f x) \, dx\\ &=\frac{(a-b) \log (\cos (e+f x))}{f}+\frac{(a-b) \tan ^2(e+f x)}{2 f}+\frac{b \tan ^4(e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.163594, size = 65, normalized size = 1.23 \[ \frac{a \left (\tan ^2(e+f x)+2 \log (\cos (e+f x))\right )}{2 f}-\frac{b \left (-\tan ^4(e+f x)+2 \tan ^2(e+f x)+4 \log (\cos (e+f x))\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^3*(a + b*Tan[e + f*x]^2),x]

[Out]

(a*(2*Log[Cos[e + f*x]] + Tan[e + f*x]^2))/(2*f) - (b*(4*Log[Cos[e + f*x]] + 2*Tan[e + f*x]^2 - Tan[e + f*x]^4
))/(4*f)

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Maple [A]  time = 0.003, size = 78, normalized size = 1.5 \begin{align*}{\frac{b \left ( \tan \left ( fx+e \right ) \right ) ^{4}}{4\,f}}+{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{2}a}{2\,f}}-{\frac{b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{2\,f}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) a}{2\,f}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) b}{2\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.14193, size = 95, normalized size = 1.79 \begin{align*} \frac{2 \,{\left (a - b\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - \frac{2 \,{\left (a - 2 \, b\right )} \sin \left (f x + e\right )^{2} - 2 \, a + 3 \, b}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(a - b)*log(sin(f*x + e)^2 - 1) - (2*(a - 2*b)*sin(f*x + e)^2 - 2*a + 3*b)/(sin(f*x + e)^4 - 2*sin(f*x
+ e)^2 + 1))/f

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Fricas [A]  time = 1.10426, size = 126, normalized size = 2.38 \begin{align*} \frac{b \tan \left (f x + e\right )^{4} + 2 \,{\left (a - b\right )} \tan \left (f x + e\right )^{2} + 2 \,{\left (a - b\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.479477, size = 88, normalized size = 1.66 \begin{align*} \begin{cases} - \frac{a \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac{a \tan ^{2}{\left (e + f x \right )}}{2 f} + \frac{b \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac{b \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac{b \tan ^{2}{\left (e + f x \right )}}{2 f} & \text{for}\: f \neq 0 \\x \left (a + b \tan ^{2}{\left (e \right )}\right ) \tan ^{3}{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*log(tan(e + f*x)**2 + 1)/(2*f) + a*tan(e + f*x)**2/(2*f) + b*log(tan(e + f*x)**2 + 1)/(2*f) + b*
tan(e + f*x)**4/(4*f) - b*tan(e + f*x)**2/(2*f), Ne(f, 0)), (x*(a + b*tan(e)**2)*tan(e)**3, True))

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Giac [B]  time = 2.99459, size = 1446, normalized size = 27.28 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 -
2*tan(f*x)*tan(e) + 1))*tan(f*x)^4*tan(e)^4 - 2*b*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan
(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)^4*tan(e)^4 + 2*a*tan(f*x)^4*tan(e)^4
- 3*b*tan(f*x)^4*tan(e)^4 - 8*a*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*
tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)^3*tan(e)^3 + 8*b*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan
(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)^3*tan(e)^3 +
2*a*tan(f*x)^4*tan(e)^2 - 2*b*tan(f*x)^4*tan(e)^2 - 4*a*tan(f*x)^3*tan(e)^3 + 8*b*tan(f*x)^3*tan(e)^3 + 2*a*t
an(f*x)^2*tan(e)^4 - 2*b*tan(f*x)^2*tan(e)^4 + 12*a*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*t
an(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)^2*tan(e)^2 - 12*b*log(4*(tan(e)^2
+ 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*t
an(f*x)^2*tan(e)^2 + b*tan(f*x)^4 - 4*a*tan(f*x)^3*tan(e) + 8*b*tan(f*x)^3*tan(e) + 4*a*tan(f*x)^2*tan(e)^2 -
4*b*tan(f*x)^2*tan(e)^2 - 4*a*tan(f*x)*tan(e)^3 + 8*b*tan(f*x)*tan(e)^3 + b*tan(e)^4 - 8*a*log(4*(tan(e)^2 + 1
)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(
f*x)*tan(e) + 8*b*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(
f*x)^2 - 2*tan(f*x)*tan(e) + 1))*tan(f*x)*tan(e) + 2*a*tan(f*x)^2 - 2*b*tan(f*x)^2 - 4*a*tan(f*x)*tan(e) + 8*b
*tan(f*x)*tan(e) + 2*a*tan(e)^2 - 2*b*tan(e)^2 + 2*a*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan(e)^2 - 2*tan(f*x)^3*
tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1)) - 2*b*log(4*(tan(e)^2 + 1)/(tan(f*x)^4*tan
(e)^2 - 2*tan(f*x)^3*tan(e) + tan(f*x)^2*tan(e)^2 + tan(f*x)^2 - 2*tan(f*x)*tan(e) + 1)) + 2*a - 3*b)/(f*tan(f
*x)^4*tan(e)^4 - 4*f*tan(f*x)^3*tan(e)^3 + 6*f*tan(f*x)^2*tan(e)^2 - 4*f*tan(f*x)*tan(e) + f)