3.520 \(\int \sec ^2(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=22 \[ \frac{(a+b \tan (c+d x))^3}{3 b d} \]

[Out]

(a + b*Tan[c + d*x])^3/(3*b*d)

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Rubi [A]  time = 0.0365484, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3506, 32} \[ \frac{(a+b \tan (c+d x))^3}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + b*Tan[c + d*x])^2,x]

[Out]

(a + b*Tan[c + d*x])^3/(3*b*d)

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [B]  time = 0.0419874, size = 46, normalized size = 2.09 \[ \frac{a^2 \tan (c+d x)}{d}+\frac{a b \tan ^2(c+d x)}{d}+\frac{b^2 \tan ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + b*Tan[c + d*x])^2,x]

[Out]

(a^2*Tan[c + d*x])/d + (a*b*Tan[c + d*x]^2)/d + (b^2*Tan[c + d*x]^3)/(3*d)

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Maple [B]  time = 0.048, size = 48, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{ab}{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{a}^{2}\tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*tan(d*x+c))^2,x)

[Out]

1/d*(1/3*b^2*sin(d*x+c)^3/cos(d*x+c)^3+a*b/cos(d*x+c)^2+a^2*tan(d*x+c))

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Maxima [A]  time = 1.36252, size = 27, normalized size = 1.23 \begin{align*} \frac{{\left (b \tan \left (d x + c\right ) + a\right )}^{3}}{3 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*tan(d*x + c) + a)^3/(b*d)

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Fricas [B]  time = 1.71534, size = 131, normalized size = 5.95 \begin{align*} \frac{3 \, a b \cos \left (d x + c\right ) +{\left ({\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a*b*cos(d*x + c) + ((3*a^2 - b^2)*cos(d*x + c)^2 + b^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*sec(c + d*x)**2, x)

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Giac [B]  time = 1.36354, size = 55, normalized size = 2.5 \begin{align*} \frac{b^{2} \tan \left (d x + c\right )^{3} + 3 \, a b \tan \left (d x + c\right )^{2} + 3 \, a^{2} \tan \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b^2*tan(d*x + c)^3 + 3*a*b*tan(d*x + c)^2 + 3*a^2*tan(d*x + c))/d