3.445 \(\int \sec ^4(c+d x) (a+b \tan ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=74 \[ \frac{a^2 \tan (c+d x)}{d}+\frac{b (2 a+b) \tan ^5(c+d x)}{5 d}+\frac{a (a+2 b) \tan ^3(c+d x)}{3 d}+\frac{b^2 \tan ^7(c+d x)}{7 d} \]

[Out]

(a^2*Tan[c + d*x])/d + (a*(a + 2*b)*Tan[c + d*x]^3)/(3*d) + (b*(2*a + b)*Tan[c + d*x]^5)/(5*d) + (b^2*Tan[c +
d*x]^7)/(7*d)

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Rubi [A]  time = 0.0693627, antiderivative size = 74, 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, 373} \[ \frac{a^2 \tan (c+d x)}{d}+\frac{b (2 a+b) \tan ^5(c+d x)}{5 d}+\frac{a (a+2 b) \tan ^3(c+d x)}{3 d}+\frac{b^2 \tan ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

(a^2*Tan[c + d*x])/d + (a*(a + 2*b)*Tan[c + d*x]^3)/(3*d) + (b*(2*a + b)*Tan[c + d*x]^5)/(5*d) + (b^2*Tan[c +
d*x]^7)/(7*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 373

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

Rubi steps

\begin{align*} \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \left (a+b x^2\right )^2 \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2+a (a+2 b) x^2+b (2 a+b) x^4+b^2 x^6\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{a^2 \tan (c+d x)}{d}+\frac{a (a+2 b) \tan ^3(c+d x)}{3 d}+\frac{b (2 a+b) \tan ^5(c+d x)}{5 d}+\frac{b^2 \tan ^7(c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.514383, size = 83, normalized size = 1.12 \[ \frac{\tan (c+d x) \left (\left (35 a^2-14 a b+3 b^2\right ) \sec ^2(c+d x)+70 a^2+6 b (7 a-4 b) \sec ^4(c+d x)-28 a b+15 b^2 \sec ^6(c+d x)+6 b^2\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((70*a^2 - 28*a*b + 6*b^2 + (35*a^2 - 14*a*b + 3*b^2)*Sec[c + d*x]^2 + 6*(7*a - 4*b)*b*Sec[c + d*x]^4 + 15*b^2
*Sec[c + d*x]^6)*Tan[c + d*x])/(105*d)

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Maple [A]  time = 0.059, size = 111, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \right ) +2\,ab \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+2/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) -{a}^{2} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x)

[Out]

1/d*(b^2*(1/7*sin(d*x+c)^5/cos(d*x+c)^7+2/35*sin(d*x+c)^5/cos(d*x+c)^5)+2*a*b*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2
/15*sin(d*x+c)^3/cos(d*x+c)^3)-a^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.13133, size = 89, normalized size = 1.2 \begin{align*} \frac{15 \, b^{2} \tan \left (d x + c\right )^{7} + 21 \,{\left (2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{5} + 35 \,{\left (a^{2} + 2 \, a b\right )} \tan \left (d x + c\right )^{3} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/105*(15*b^2*tan(d*x + c)^7 + 21*(2*a*b + b^2)*tan(d*x + c)^5 + 35*(a^2 + 2*a*b)*tan(d*x + c)^3 + 105*a^2*tan
(d*x + c))/d

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Fricas [A]  time = 1.51317, size = 231, normalized size = 3.12 \begin{align*} \frac{{\left (2 \,{\left (35 \, a^{2} - 14 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{6} +{\left (35 \, a^{2} - 14 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 6 \,{\left (7 \, a b - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/105*(2*(35*a^2 - 14*a*b + 3*b^2)*cos(d*x + c)^6 + (35*a^2 - 14*a*b + 3*b^2)*cos(d*x + c)^4 + 6*(7*a*b - 4*b^
2)*cos(d*x + c)^2 + 15*b^2)*sin(d*x + c)/(d*cos(d*x + c)^7)

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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 ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+b*tan(d*x+c)**2)**2,x)

[Out]

Integral((a + b*tan(c + d*x)**2)**2*sec(c + d*x)**4, x)

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Giac [A]  time = 2.26377, size = 108, normalized size = 1.46 \begin{align*} \frac{15 \, b^{2} \tan \left (d x + c\right )^{7} + 42 \, a b \tan \left (d x + c\right )^{5} + 21 \, b^{2} \tan \left (d x + c\right )^{5} + 35 \, a^{2} \tan \left (d x + c\right )^{3} + 70 \, a b \tan \left (d x + c\right )^{3} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/105*(15*b^2*tan(d*x + c)^7 + 42*a*b*tan(d*x + c)^5 + 21*b^2*tan(d*x + c)^5 + 35*a^2*tan(d*x + c)^3 + 70*a*b*
tan(d*x + c)^3 + 105*a^2*tan(d*x + c))/d