3.6.31 \(\int \sec ^8(c+d x) (a+b \tan (c+d x))^3 \, dx\) [531]

Optimal. Leaf size=194 \[ \frac {3 a^2 b \sec ^8(c+d x)}{8 d}+\frac {a^3 \tan (c+d x)}{d}+\frac {a \left (a^2+b^2\right ) \tan ^3(c+d x)}{d}+\frac {b^3 \tan ^4(c+d x)}{4 d}+\frac {3 a \left (a^2+3 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b^3 \tan ^6(c+d x)}{2 d}+\frac {a \left (a^2+9 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac {3 b^3 \tan ^8(c+d x)}{8 d}+\frac {a b^2 \tan ^9(c+d x)}{3 d}+\frac {b^3 \tan ^{10}(c+d x)}{10 d} \]

[Out]

3/8*a^2*b*sec(d*x+c)^8/d+a^3*tan(d*x+c)/d+a*(a^2+b^2)*tan(d*x+c)^3/d+1/4*b^3*tan(d*x+c)^4/d+3/5*a*(a^2+3*b^2)*
tan(d*x+c)^5/d+1/2*b^3*tan(d*x+c)^6/d+1/7*a*(a^2+9*b^2)*tan(d*x+c)^7/d+3/8*b^3*tan(d*x+c)^8/d+1/3*a*b^2*tan(d*
x+c)^9/d+1/10*b^3*tan(d*x+c)^10/d

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Rubi [A]
time = 0.12, antiderivative size = 194, 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 = {3587, 710, 1824} \begin {gather*} \frac {a^3 \tan (c+d x)}{d}+\frac {a \left (a^2+9 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac {3 a \left (a^2+3 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {a \left (a^2+b^2\right ) \tan ^3(c+d x)}{d}+\frac {3 a^2 b \sec ^8(c+d x)}{8 d}+\frac {a b^2 \tan ^9(c+d x)}{3 d}+\frac {b^3 \tan ^{10}(c+d x)}{10 d}+\frac {3 b^3 \tan ^8(c+d x)}{8 d}+\frac {b^3 \tan ^6(c+d x)}{2 d}+\frac {b^3 \tan ^4(c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + b*Tan[c + d*x])^3,x]

[Out]

(3*a^2*b*Sec[c + d*x]^8)/(8*d) + (a^3*Tan[c + d*x])/d + (a*(a^2 + b^2)*Tan[c + d*x]^3)/d + (b^3*Tan[c + d*x]^4
)/(4*d) + (3*a*(a^2 + 3*b^2)*Tan[c + d*x]^5)/(5*d) + (b^3*Tan[c + d*x]^6)/(2*d) + (a*(a^2 + 9*b^2)*Tan[c + d*x
]^7)/(7*d) + (3*b^3*Tan[c + d*x]^8)/(8*d) + (a*b^2*Tan[c + d*x]^9)/(3*d) + (b^3*Tan[c + d*x]^10)/(10*d)

Rule 710

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 1824

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 3587

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))^3 \, dx &=\frac {\text {Subst}\left (\int (a+x)^3 \left (1+\frac {x^2}{b^2}\right )^3 \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {3 a^2 b \sec ^8(c+d x)}{8 d}+\frac {\text {Subst}\left (\int \left (1+\frac {x^2}{b^2}\right )^3 \left (-3 a^2 x+(a+x)^3\right ) \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {3 a^2 b \sec ^8(c+d x)}{8 d}+\frac {\text {Subst}\left (\int \left (a^3+\frac {3 a \left (a^2+b^2\right ) x^2}{b^2}+x^3+\frac {3 a \left (a^2+3 b^2\right ) x^4}{b^4}+\frac {3 x^5}{b^2}+\frac {a \left (a^2+9 b^2\right ) x^6}{b^6}+\frac {3 x^7}{b^4}+\frac {3 a x^8}{b^6}+\frac {x^9}{b^6}\right ) \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac {3 a^2 b \sec ^8(c+d x)}{8 d}+\frac {a^3 \tan (c+d x)}{d}+\frac {a \left (a^2+b^2\right ) \tan ^3(c+d x)}{d}+\frac {b^3 \tan ^4(c+d x)}{4 d}+\frac {3 a \left (a^2+3 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b^3 \tan ^6(c+d x)}{2 d}+\frac {a \left (a^2+9 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac {3 b^3 \tan ^8(c+d x)}{8 d}+\frac {a b^2 \tan ^9(c+d x)}{3 d}+\frac {b^3 \tan ^{10}(c+d x)}{10 d}\\ \end {align*}

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Mathematica [A]
time = 2.14, size = 177, normalized size = 0.91 \begin {gather*} \frac {\frac {1}{4} \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^4-\frac {6}{5} a \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^5+\frac {1}{2} \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) (a+b \tan (c+d x))^6-\frac {4}{7} a \left (5 a^2+3 b^2\right ) (a+b \tan (c+d x))^7+\frac {3}{8} \left (5 a^2+b^2\right ) (a+b \tan (c+d x))^8-\frac {2}{3} a (a+b \tan (c+d x))^9+\frac {1}{10} (a+b \tan (c+d x))^{10}}{b^7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + b*Tan[c + d*x])^3,x]

[Out]

(((a^2 + b^2)^3*(a + b*Tan[c + d*x])^4)/4 - (6*a*(a^2 + b^2)^2*(a + b*Tan[c + d*x])^5)/5 + ((a^2 + b^2)*(5*a^2
+ b^2)*(a + b*Tan[c + d*x])^6)/2 - (4*a*(5*a^2 + 3*b^2)*(a + b*Tan[c + d*x])^7)/7 + (3*(5*a^2 + b^2)*(a + b*T
an[c + d*x])^8)/8 - (2*a*(a + b*Tan[c + d*x])^9)/3 + (a + b*Tan[c + d*x])^10/10)/(b^7*d)

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Maple [A]
time = 0.26, size = 219, normalized size = 1.13

method	result	size
derivativedivides	\(\frac {b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{10 \cos \left (d x +c \right )^{10}}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{40 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{20 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{40 \cos \left (d x +c \right )^{4}}\right )+3 b^{2} a \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )+\frac {3 a^{2} b}{8 \cos \left (d x +c \right )^{8}}-a^{3} \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )}{d}\)	\(219\)
default	\(\frac {b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{10 \cos \left (d x +c \right )^{10}}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{40 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{20 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{40 \cos \left (d x +c \right )^{4}}\right )+3 b^{2} a \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )+\frac {3 a^{2} b}{8 \cos \left (d x +c \right )^{8}}-a^{3} \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )}{d}\)	\(219\)
risch	\(-\frac {32 \left (10 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 i a^{3}-315 a^{2} b \,{\mathrm e}^{12 i \left (d x +c \right )}+105 b^{3} {\mathrm e}^{12 i \left (d x +c \right )}-525 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+45 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-630 a^{2} b \,{\mathrm e}^{10 i \left (d x +c \right )}-126 b^{3} {\mathrm e}^{10 i \left (d x +c \right )}-105 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-378 i a^{3} {\mathrm e}^{10 i \left (d x +c \right )}-315 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+105 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-30 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+120 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+i a \,b^{2}+315 i a \,b^{2} {\mathrm e}^{12 i \left (d x +c \right )}+126 i a \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}-105 i a^{3} {\mathrm e}^{12 i \left (d x +c \right )}-135 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-360 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{10}}\)	\(306\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(b^3*(1/10*sin(d*x+c)^4/cos(d*x+c)^10+3/40*sin(d*x+c)^4/cos(d*x+c)^8+1/20*sin(d*x+c)^4/cos(d*x+c)^6+1/40*s
in(d*x+c)^4/cos(d*x+c)^4)+3*b^2*a*(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)+3/8*a^2*b/cos(d*x+c)^8-a^3*(-16/35-1/7*sec(d*x+c)^6-6/35*s
ec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]
time = 0.27, size = 176, normalized size = 0.91 \begin {gather*} \frac {84 \, b^{3} \tan \left (d x + c\right )^{10} + 280 \, a b^{2} \tan \left (d x + c\right )^{9} + 315 \, {\left (a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{8} + 120 \, {\left (a^{3} + 9 \, a b^{2}\right )} \tan \left (d x + c\right )^{7} + 420 \, {\left (3 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{6} + 504 \, {\left (a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{5} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 210 \, {\left (9 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{4} + 840 \, a^{3} \tan \left (d x + c\right ) + 840 \, {\left (a^{3} + a b^{2}\right )} \tan \left (d x + c\right )^{3}}{840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/840*(84*b^3*tan(d*x + c)^10 + 280*a*b^2*tan(d*x + c)^9 + 315*(a^2*b + b^3)*tan(d*x + c)^8 + 120*(a^3 + 9*a*b
^2)*tan(d*x + c)^7 + 420*(3*a^2*b + b^3)*tan(d*x + c)^6 + 504*(a^3 + 3*a*b^2)*tan(d*x + c)^5 + 1260*a^2*b*tan(
d*x + c)^2 + 210*(9*a^2*b + b^3)*tan(d*x + c)^4 + 840*a^3*tan(d*x + c) + 840*(a^3 + a*b^2)*tan(d*x + c)^3)/d

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Fricas [A]
time = 0.44, size = 150, normalized size = 0.77 \begin {gather*} \frac {84 \, b^{3} + 105 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (16 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{9} + 8 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{7} + 6 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{5} + 35 \, a b^{2} \cos \left (d x + c\right ) + 5 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{840 \, d \cos \left (d x + c\right )^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/840*(84*b^3 + 105*(3*a^2*b - b^3)*cos(d*x + c)^2 + 8*(16*(3*a^3 - a*b^2)*cos(d*x + c)^9 + 8*(3*a^3 - a*b^2)*
cos(d*x + c)^7 + 6*(3*a^3 - a*b^2)*cos(d*x + c)^5 + 35*a*b^2*cos(d*x + c) + 5*(3*a^3 - a*b^2)*cos(d*x + c)^3)*
sin(d*x + c))/(d*cos(d*x + c)^10)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sec ^{8}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8*(a+b*tan(d*x+c))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*sec(c + d*x)**8, x)

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Giac [A]
time = 0.92, size = 220, normalized size = 1.13 \begin {gather*} \frac {84 \, b^{3} \tan \left (d x + c\right )^{10} + 280 \, a b^{2} \tan \left (d x + c\right )^{9} + 315 \, a^{2} b \tan \left (d x + c\right )^{8} + 315 \, b^{3} \tan \left (d x + c\right )^{8} + 120 \, a^{3} \tan \left (d x + c\right )^{7} + 1080 \, a b^{2} \tan \left (d x + c\right )^{7} + 1260 \, a^{2} b \tan \left (d x + c\right )^{6} + 420 \, b^{3} \tan \left (d x + c\right )^{6} + 504 \, a^{3} \tan \left (d x + c\right )^{5} + 1512 \, a b^{2} \tan \left (d x + c\right )^{5} + 1890 \, a^{2} b \tan \left (d x + c\right )^{4} + 210 \, b^{3} \tan \left (d x + c\right )^{4} + 840 \, a^{3} \tan \left (d x + c\right )^{3} + 840 \, a b^{2} \tan \left (d x + c\right )^{3} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 840 \, a^{3} \tan \left (d x + c\right )}{840 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/840*(84*b^3*tan(d*x + c)^10 + 280*a*b^2*tan(d*x + c)^9 + 315*a^2*b*tan(d*x + c)^8 + 315*b^3*tan(d*x + c)^8 +
120*a^3*tan(d*x + c)^7 + 1080*a*b^2*tan(d*x + c)^7 + 1260*a^2*b*tan(d*x + c)^6 + 420*b^3*tan(d*x + c)^6 + 504
*a^3*tan(d*x + c)^5 + 1512*a*b^2*tan(d*x + c)^5 + 1890*a^2*b*tan(d*x + c)^4 + 210*b^3*tan(d*x + c)^4 + 840*a^3
*tan(d*x + c)^3 + 840*a*b^2*tan(d*x + c)^3 + 1260*a^2*b*tan(d*x + c)^2 + 840*a^3*tan(d*x + c))/d

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Mupad [B]
time = 3.68, size = 175, normalized size = 0.90 \begin {gather*} \frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {3\,a^3}{5}+\frac {9\,a\,b^2}{5}\right )+{\mathrm {tan}\left (c+d\,x\right )}^7\,\left (\frac {a^3}{7}+\frac {9\,a\,b^2}{7}\right )+{\mathrm {tan}\left (c+d\,x\right )}^6\,\left (\frac {3\,a^2\,b}{2}+\frac {b^3}{2}\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {9\,a^2\,b}{4}+\frac {b^3}{4}\right )+a^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^{10}}{10}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^9}{3}+a\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a^2+b^2\right )+\frac {3\,b\,{\mathrm {tan}\left (c+d\,x\right )}^8\,\left (a^2+b^2\right )}{8}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^3/cos(c + d*x)^8,x)

[Out]

(tan(c + d*x)^5*((9*a*b^2)/5 + (3*a^3)/5) + tan(c + d*x)^7*((9*a*b^2)/7 + a^3/7) + tan(c + d*x)^6*((3*a^2*b)/2
+ b^3/2) + tan(c + d*x)^4*((9*a^2*b)/4 + b^3/4) + a^3*tan(c + d*x) + (b^3*tan(c + d*x)^10)/10 + (3*a^2*b*tan(
c + d*x)^2)/2 + (a*b^2*tan(c + d*x)^9)/3 + a*tan(c + d*x)^3*(a^2 + b^2) + (3*b*tan(c + d*x)^8*(a^2 + b^2))/8)/
d

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