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

Optimal. Leaf size=49 \[ \frac{a^2 \tan (c+d x)}{d}+\frac{2 a b \tan ^3(c+d x)}{3 d}+\frac{b^2 \tan ^5(c+d x)}{5 d} \]

[Out]

(a^2*Tan[c + d*x])/d + (2*a*b*Tan[c + d*x]^3)/(3*d) + (b^2*Tan[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.0535059, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3675, 194} \[ \frac{a^2 \tan (c+d x)}{d}+\frac{2 a b \tan ^3(c+d x)}{3 d}+\frac{b^2 \tan ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Tan[c + d*x])/d + (2*a*b*Tan[c + d*x]^3)/(3*d) + (b^2*Tan[c + d*x]^5)/(5*d)

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b x^2\right )^2 \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{a^2 \tan (c+d x)}{d}+\frac{2 a b \tan ^3(c+d x)}{3 d}+\frac{b^2 \tan ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.140486, size = 49, normalized size = 1. \[ \frac{a^2 \tan (c+d x)}{d}+\frac{2 a b \tan ^3(c+d x)}{3 d}+\frac{b^2 \tan ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Tan[c + d*x])/d + (2*a*b*Tan[c + d*x]^3)/(3*d) + (b^2*Tan[c + d*x]^5)/(5*d)

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Maple [A]  time = 0.053, size = 57, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{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)^2,x)

[Out]

1/d*(1/5*b^2*sin(d*x+c)^5/cos(d*x+c)^5+2/3*a*b*sin(d*x+c)^3/cos(d*x+c)^3+a^2*tan(d*x+c))

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Maxima [A]  time = 1.02884, size = 57, normalized size = 1.16 \begin{align*} \frac{3 \, b^{2} \tan \left (d x + c\right )^{5} + 10 \, a b \tan \left (d x + c\right )^{3} + 15 \, a^{2} \tan \left (d x + c\right )}{15 \, 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)^2,x, algorithm="maxima")

[Out]

1/15*(3*b^2*tan(d*x + c)^5 + 10*a*b*tan(d*x + c)^3 + 15*a^2*tan(d*x + c))/d

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Fricas [A]  time = 1.44614, size = 167, normalized size = 3.41 \begin{align*} \frac{{\left ({\left (15 \, a^{2} - 10 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (5 \, a b - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \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)^2,x, algorithm="fricas")

[Out]

1/15*((15*a^2 - 10*a*b + 3*b^2)*cos(d*x + c)^4 + 2*(5*a*b - 3*b^2)*cos(d*x + c)^2 + 3*b^2)*sin(d*x + c)/(d*cos
(d*x + c)^5)

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

[Out]

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

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Giac [A]  time = 1.77604, size = 57, normalized size = 1.16 \begin{align*} \frac{3 \, b^{2} \tan \left (d x + c\right )^{5} + 10 \, a b \tan \left (d x + c\right )^{3} + 15 \, a^{2} \tan \left (d x + c\right )}{15 \, 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)^2,x, algorithm="giac")

[Out]

1/15*(3*b^2*tan(d*x + c)^5 + 10*a*b*tan(d*x + c)^3 + 15*a^2*tan(d*x + c))/d