3.319 \(\int (a+b \sec ^2(e+f x)) \tan ^2(e+f x) \, dx\)

Optimal. Leaf size=32 \[ \frac{a \tan (e+f x)}{f}-a x+\frac{b \tan ^3(e+f x)}{3 f} \]

[Out]

-(a*x) + (a*Tan[e + f*x])/f + (b*Tan[e + f*x]^3)/(3*f)

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Rubi [A]  time = 0.0520089, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4141, 1802, 203} \[ \frac{a \tan (e+f x)}{f}-a x+\frac{b \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x) + (a*Tan[e + f*x])/f + (b*Tan[e + f*x]^3)/(3*f)

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \left (a+b \sec ^2(e+f x)\right ) \tan ^2(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b \left (1+x^2\right )\right )}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (a+b x^2-\frac{a}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a \tan (e+f x)}{f}+\frac{b \tan ^3(e+f x)}{3 f}-\frac{a \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-a x+\frac{a \tan (e+f x)}{f}+\frac{b \tan ^3(e+f x)}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.0153752, size = 41, normalized size = 1.28 \[ -\frac{a \tan ^{-1}(\tan (e+f x))}{f}+\frac{a \tan (e+f x)}{f}+\frac{b \tan ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*ArcTan[Tan[e + f*x]])/f) + (a*Tan[e + f*x])/f + (b*Tan[e + f*x]^3)/(3*f)

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Maple [A]  time = 0.04, size = 41, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ( a \left ( \tan \left ( fx+e \right ) -fx-e \right ) +{\frac{b \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(a*(tan(f*x+e)-f*x-e)+1/3*b*sin(f*x+e)^3/cos(f*x+e)^3)

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Maxima [A]  time = 1.47493, size = 45, normalized size = 1.41 \begin{align*} \frac{b \tan \left (f x + e\right )^{3} - 3 \,{\left (f x + e\right )} a + 3 \, a \tan \left (f x + e\right )}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*tan(f*x + e)^3 - 3*(f*x + e)*a + 3*a*tan(f*x + e))/f

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Fricas [A]  time = 0.486491, size = 130, normalized size = 4.06 \begin{align*} -\frac{3 \, a f x \cos \left (f x + e\right )^{3} -{\left ({\left (3 \, a - b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sin \left (f x + e\right )}{3 \, f \cos \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.66331, size = 42, normalized size = 1.31 \begin{align*} a \left (\begin{cases} - x + \frac{\tan{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \tan ^{2}{\left (e \right )} & \text{otherwise} \end{cases}\right ) + b \left (\begin{cases} x \tan ^{2}{\left (e \right )} \sec ^{2}{\left (e \right )} & \text{for}\: f = 0 \\\frac{\tan ^{3}{\left (e + f x \right )}}{3 f} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((-x + tan(e + f*x)/f, Ne(f, 0)), (x*tan(e)**2, True)) + b*Piecewise((x*tan(e)**2*sec(e)**2, Eq(f,
0)), (tan(e + f*x)**3/(3*f), True))

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Giac [A]  time = 1.60415, size = 49, normalized size = 1.53 \begin{align*} \frac{b \tan \left (f x + e\right )^{3} - 3 \,{\left (f x + e\right )} a + 3 \, a \tan \left (f x + e\right )}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*tan(f*x + e)^3 - 3*(f*x + e)*a + 3*a*tan(f*x + e))/f