3.431 \(\int \sec ^6(c+d x) (a+b \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=68 \[ \frac {(a+2 b) \tan ^5(c+d x)}{5 d}+\frac {(2 a+b) \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {b \tan ^7(c+d x)}{7 d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3675, 373} \[ \frac {(a+2 b) \tan ^5(c+d x)}{5 d}+\frac {(2 a+b) \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {b \tan ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + b*Tan[c + d*x]^2),x]

[Out]

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

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]

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])

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 75, normalized size = 1.10 \[ \frac {\tan (c+d x) \left (21 a \tan ^4(c+d x)+70 a \tan ^2(c+d x)+105 a+15 b \sec ^6(c+d x)-3 b \sec ^4(c+d x)-4 b \sec ^2(c+d x)-8 b\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + b*Tan[c + d*x]^2),x]

[Out]

(Tan[c + d*x]*(105*a - 8*b - 4*b*Sec[c + d*x]^2 - 3*b*Sec[c + d*x]^4 + 15*b*Sec[c + d*x]^6 + 70*a*Tan[c + d*x]
^2 + 21*a*Tan[c + d*x]^4))/(105*d)

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fricas [A]  time = 0.47, size = 74, normalized size = 1.09 \[ \frac {{\left (8 \, {\left (7 \, a - b\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (7 \, a - b\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (7 \, a - b\right )} \cos \left (d x + c\right )^{2} + 15 \, b\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

1/105*(8*(7*a - b)*cos(d*x + c)^6 + 4*(7*a - b)*cos(d*x + c)^4 + 3*(7*a - b)*cos(d*x + c)^2 + 15*b)*sin(d*x +
c)/(d*cos(d*x + c)^7)

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giac [A]  time = 1.89, size = 70, normalized size = 1.03 \[ \frac {15 \, b \tan \left (d x + c\right )^{7} + 21 \, a \tan \left (d x + c\right )^{5} + 42 \, b \tan \left (d x + c\right )^{5} + 70 \, a \tan \left (d x + c\right )^{3} + 35 \, b \tan \left (d x + c\right )^{3} + 105 \, a \tan \left (d x + c\right )}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.62, size = 94, normalized size = 1.38 \[ \frac {b \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )-a \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+b*tan(d*x+c)^2),x)

[Out]

1/d*(b*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)-a*(-8/15
-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

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maxima [A]  time = 0.34, size = 56, normalized size = 0.82 \[ \frac {15 \, b \tan \left (d x + c\right )^{7} + 21 \, {\left (a + 2 \, b\right )} \tan \left (d x + c\right )^{5} + 35 \, {\left (2 \, a + b\right )} \tan \left (d x + c\right )^{3} + 105 \, a \tan \left (d x + c\right )}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 12.01, size = 56, normalized size = 0.82 \[ \frac {\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7}+\left (\frac {a}{5}+\frac {2\,b}{5}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^5+\left (\frac {2\,a}{3}+\frac {b}{3}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+a\,\mathrm {tan}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x)^2)/cos(c + d*x)^6,x)

[Out]

(tan(c + d*x)^3*((2*a)/3 + b/3) + tan(c + d*x)^5*(a/5 + (2*b)/5) + a*tan(c + d*x) + (b*tan(c + d*x)^7)/7)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (c + d x \right )}\right ) \sec ^{6}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6*(a+b*tan(d*x+c)**2),x)

[Out]

Integral((a + b*tan(c + d*x)**2)*sec(c + d*x)**6, x)

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