3.517 \(\int \sec ^8(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=119 \[ \frac {\left (a^2+3 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac {3 \left (a^2+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {\left (3 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {a b \sec ^8(c+d x)}{4 d}+\frac {b^2 \tan ^9(c+d x)}{9 d} \]

[Out]

1/4*a*b*sec(d*x+c)^8/d+a^2*tan(d*x+c)/d+1/3*(3*a^2+b^2)*tan(d*x+c)^3/d+3/5*(a^2+b^2)*tan(d*x+c)^5/d+1/7*(a^2+3
*b^2)*tan(d*x+c)^7/d+1/9*b^2*tan(d*x+c)^9/d

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Rubi [A]  time = 0.11, antiderivative size = 119, normalized size of antiderivative = 1.00, 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 = {3506, 696, 1810} \[ \frac {\left (a^2+3 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac {3 \left (a^2+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {\left (3 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {a b \sec ^8(c+d x)}{4 d}+\frac {b^2 \tan ^9(c+d x)}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + b*Tan[c + d*x])^2,x]

[Out]

(a*b*Sec[c + d*x]^8)/(4*d) + (a^2*Tan[c + d*x])/d + ((3*a^2 + b^2)*Tan[c + d*x]^3)/(3*d) + (3*(a^2 + b^2)*Tan[
c + d*x]^5)/(5*d) + ((a^2 + 3*b^2)*Tan[c + d*x]^7)/(7*d) + (b^2*Tan[c + d*x]^9)/(9*d)

Rule 696

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*m*d^(m - 1)*(a + c*x^2)^(p + 1))
/(2*c*(p + 1)), x] + Int[((d + e*x)^m - e*m*d^(m - 1)*x)*(a + c*x^2)^p, x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*
d^2 + a*e^2, 0] && IGtQ[p, 1] && IGtQ[m, 0] && LeQ[m, p]

Rule 1810

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

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]

Rubi steps

\begin {align*} \int \sec ^8(c+d x) (a+b \tan (c+d x))^2 \, dx &=\frac {\operatorname {Subst}\left (\int (a+x)^2 \left (1+\frac {x^2}{b^2}\right )^3 \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {a b \sec ^8(c+d x)}{4 d}+\frac {\operatorname {Subst}\left (\int \left (1+\frac {x^2}{b^2}\right )^3 \left (-2 a x+(a+x)^2\right ) \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {a b \sec ^8(c+d x)}{4 d}+\frac {\operatorname {Subst}\left (\int \left (a^2+\frac {\left (3 a^2+b^2\right ) x^2}{b^2}+\frac {3 \left (a^2+b^2\right ) x^4}{b^4}+\frac {\left (a^2+3 b^2\right ) x^6}{b^6}+\frac {x^8}{b^6}\right ) \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {a b \sec ^8(c+d x)}{4 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {\left (3 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {3 \left (a^2+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {\left (a^2+3 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac {b^2 \tan ^9(c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 0.58, size = 133, normalized size = 1.12 \[ \frac {\tan (c+d x) \left (180 \left (a^2+3 b^2\right ) \tan ^6(c+d x)+756 \left (a^2+b^2\right ) \tan ^4(c+d x)+420 \left (3 a^2+b^2\right ) \tan ^2(c+d x)+1260 a^2+315 a b \tan ^7(c+d x)+1260 a b \tan ^5(c+d x)+1890 a b \tan ^3(c+d x)+1260 a b \tan (c+d x)+140 b^2 \tan ^8(c+d x)\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + b*Tan[c + d*x])^2,x]

[Out]

(Tan[c + d*x]*(1260*a^2 + 1260*a*b*Tan[c + d*x] + 420*(3*a^2 + b^2)*Tan[c + d*x]^2 + 1890*a*b*Tan[c + d*x]^3 +
756*(a^2 + b^2)*Tan[c + d*x]^4 + 1260*a*b*Tan[c + d*x]^5 + 180*(a^2 + 3*b^2)*Tan[c + d*x]^6 + 315*a*b*Tan[c +
d*x]^7 + 140*b^2*Tan[c + d*x]^8))/(1260*d)

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fricas [A]  time = 0.68, size = 122, normalized size = 1.03 \[ \frac {315 \, a b \cos \left (d x + c\right ) + 4 \, {\left (16 \, {\left (9 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{8} + 8 \, {\left (9 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 6 \, {\left (9 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (9 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 35 \, b^{2}\right )} \sin \left (d x + c\right )}{1260 \, d \cos \left (d x + c\right )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/1260*(315*a*b*cos(d*x + c) + 4*(16*(9*a^2 - b^2)*cos(d*x + c)^8 + 8*(9*a^2 - b^2)*cos(d*x + c)^6 + 6*(9*a^2
- b^2)*cos(d*x + c)^4 + 5*(9*a^2 - b^2)*cos(d*x + c)^2 + 35*b^2)*sin(d*x + c))/(d*cos(d*x + c)^9)

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giac [A]  time = 0.82, size = 156, normalized size = 1.31 \[ \frac {140 \, b^{2} \tan \left (d x + c\right )^{9} + 315 \, a b \tan \left (d x + c\right )^{8} + 180 \, a^{2} \tan \left (d x + c\right )^{7} + 540 \, b^{2} \tan \left (d x + c\right )^{7} + 1260 \, a b \tan \left (d x + c\right )^{6} + 756 \, a^{2} \tan \left (d x + c\right )^{5} + 756 \, b^{2} \tan \left (d x + c\right )^{5} + 1890 \, a b \tan \left (d x + c\right )^{4} + 1260 \, a^{2} \tan \left (d x + c\right )^{3} + 420 \, b^{2} \tan \left (d x + c\right )^{3} + 1260 \, a b \tan \left (d x + c\right )^{2} + 1260 \, a^{2} \tan \left (d x + c\right )}{1260 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/1260*(140*b^2*tan(d*x + c)^9 + 315*a*b*tan(d*x + c)^8 + 180*a^2*tan(d*x + c)^7 + 540*b^2*tan(d*x + c)^7 + 12
60*a*b*tan(d*x + c)^6 + 756*a^2*tan(d*x + c)^5 + 756*b^2*tan(d*x + c)^5 + 1890*a*b*tan(d*x + c)^4 + 1260*a^2*t
an(d*x + c)^3 + 420*b^2*tan(d*x + c)^3 + 1260*a*b*tan(d*x + c)^2 + 1260*a^2*tan(d*x + c))/d

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maple [A]  time = 0.41, size = 138, normalized size = 1.16 \[ \frac {-a^{2} \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )+\frac {a b}{4 \cos \left (d x +c \right )^{8}}+b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8*(a+b*tan(d*x+c))^2,x)

[Out]

1/d*(-a^2*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c)+1/4*a*b/cos(d*x+c)^8+b^2*(1
/9*sin(d*x+c)^3/cos(d*x+c)^9+2/21*sin(d*x+c)^3/cos(d*x+c)^7+8/105*sin(d*x+c)^3/cos(d*x+c)^5+16/315*sin(d*x+c)^
3/cos(d*x+c)^3))

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maxima [A]  time = 0.33, size = 133, normalized size = 1.12 \[ \frac {140 \, b^{2} \tan \left (d x + c\right )^{9} + 315 \, a b \tan \left (d x + c\right )^{8} + 1260 \, a b \tan \left (d x + c\right )^{6} + 180 \, {\left (a^{2} + 3 \, b^{2}\right )} \tan \left (d x + c\right )^{7} + 1890 \, a b \tan \left (d x + c\right )^{4} + 756 \, {\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{5} + 1260 \, a b \tan \left (d x + c\right )^{2} + 420 \, {\left (3 \, a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{3} + 1260 \, a^{2} \tan \left (d x + c\right )}{1260 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/1260*(140*b^2*tan(d*x + c)^9 + 315*a*b*tan(d*x + c)^8 + 1260*a*b*tan(d*x + c)^6 + 180*(a^2 + 3*b^2)*tan(d*x
+ c)^7 + 1890*a*b*tan(d*x + c)^4 + 756*(a^2 + b^2)*tan(d*x + c)^5 + 1260*a*b*tan(d*x + c)^2 + 420*(3*a^2 + b^2
)*tan(d*x + c)^3 + 1260*a^2*tan(d*x + c))/d

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mupad [B]  time = 3.68, size = 132, normalized size = 1.11 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a^2+\frac {b^2}{3}\right )+a^2\,\mathrm {tan}\left (c+d\,x\right )+{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {3\,a^2}{5}+\frac {3\,b^2}{5}\right )+{\mathrm {tan}\left (c+d\,x\right )}^7\,\left (\frac {a^2}{7}+\frac {3\,b^2}{7}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^9}{9}+a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {3\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^4}{2}+a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^6+\frac {a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^8}{4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^2/cos(c + d*x)^8,x)

[Out]

(tan(c + d*x)^3*(a^2 + b^2/3) + a^2*tan(c + d*x) + tan(c + d*x)^5*((3*a^2)/5 + (3*b^2)/5) + tan(c + d*x)^7*(a^
2/7 + (3*b^2)/7) + (b^2*tan(c + d*x)^9)/9 + a*b*tan(c + d*x)^2 + (3*a*b*tan(c + d*x)^4)/2 + a*b*tan(c + d*x)^6
+ (a*b*tan(c + d*x)^8)/4)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sec ^{8}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*sec(c + d*x)**8, x)

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