3.3.99 \(\int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [299]

Optimal. Leaf size=255 \[ -\frac {\log (\cos (c+d x))}{a^2 d}+\frac {\left (a^2-b^2\right )^3 \left (7 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^8 d}-\frac {2 a \left (3 a^4-8 a^2 b^2+6 b^4\right ) \sec (c+d x)}{b^7 d}+\frac {\left (5 a^4-12 a^2 b^2+6 b^4\right ) \sec ^2(c+d x)}{2 b^6 d}-\frac {4 a \left (a^2-2 b^2\right ) \sec ^3(c+d x)}{3 b^5 d}+\frac {\left (3 a^2-4 b^2\right ) \sec ^4(c+d x)}{4 b^4 d}-\frac {2 a \sec ^5(c+d x)}{5 b^3 d}+\frac {\sec ^6(c+d x)}{6 b^2 d}+\frac {\left (a^2-b^2\right )^4}{a b^8 d (a+b \sec (c+d x))} \]

[Out]

-ln(cos(d*x+c))/a^2/d+(a^2-b^2)^3*(7*a^2+b^2)*ln(a+b*sec(d*x+c))/a^2/b^8/d-2*a*(3*a^4-8*a^2*b^2+6*b^4)*sec(d*x
+c)/b^7/d+1/2*(5*a^4-12*a^2*b^2+6*b^4)*sec(d*x+c)^2/b^6/d-4/3*a*(a^2-2*b^2)*sec(d*x+c)^3/b^5/d+1/4*(3*a^2-4*b^
2)*sec(d*x+c)^4/b^4/d-2/5*a*sec(d*x+c)^5/b^3/d+1/6*sec(d*x+c)^6/b^2/d+(a^2-b^2)^4/a/b^8/d/(a+b*sec(d*x+c))

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Rubi [A]
time = 0.15, antiderivative size = 255, 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 = {3970, 908} \begin {gather*} \frac {\left (a^2-b^2\right )^4}{a b^8 d (a+b \sec (c+d x))}+\frac {\left (a^2-b^2\right )^3 \left (7 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^8 d}-\frac {4 a \left (a^2-2 b^2\right ) \sec ^3(c+d x)}{3 b^5 d}+\frac {\left (3 a^2-4 b^2\right ) \sec ^4(c+d x)}{4 b^4 d}-\frac {\log (\cos (c+d x))}{a^2 d}-\frac {2 a \left (3 a^4-8 a^2 b^2+6 b^4\right ) \sec (c+d x)}{b^7 d}+\frac {\left (5 a^4-12 a^2 b^2+6 b^4\right ) \sec ^2(c+d x)}{2 b^6 d}-\frac {2 a \sec ^5(c+d x)}{5 b^3 d}+\frac {\sec ^6(c+d x)}{6 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^9/(a + b*Sec[c + d*x])^2,x]

[Out]

-(Log[Cos[c + d*x]]/(a^2*d)) + ((a^2 - b^2)^3*(7*a^2 + b^2)*Log[a + b*Sec[c + d*x]])/(a^2*b^8*d) - (2*a*(3*a^4
- 8*a^2*b^2 + 6*b^4)*Sec[c + d*x])/(b^7*d) + ((5*a^4 - 12*a^2*b^2 + 6*b^4)*Sec[c + d*x]^2)/(2*b^6*d) - (4*a*(
a^2 - 2*b^2)*Sec[c + d*x]^3)/(3*b^5*d) + ((3*a^2 - 4*b^2)*Sec[c + d*x]^4)/(4*b^4*d) - (2*a*Sec[c + d*x]^5)/(5*
b^3*d) + Sec[c + d*x]^6/(6*b^2*d) + (a^2 - b^2)^4/(a*b^8*d*(a + b*Sec[c + d*x]))

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 3970

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[(b^2 - x^2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^9(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (b^2-x^2\right )^4}{x (a+x)^2} \, dx,x,b \sec (c+d x)\right )}{b^8 d}\\ &=\frac {\text {Subst}\left (\int \left (-2 a \left (3 a^4-8 a^2 b^2+6 b^4\right )+\frac {b^8}{a^2 x}+\left (5 a^4-12 a^2 b^2+6 b^4\right ) x-4 a \left (a^2-2 b^2\right ) x^2+\left (3 a^2-4 b^2\right ) x^3-2 a x^4+x^5-\frac {\left (a^2-b^2\right )^4}{a (a+x)^2}+\frac {\left (a^2-b^2\right )^3 \left (7 a^2+b^2\right )}{a^2 (a+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{b^8 d}\\ &=-\frac {\log (\cos (c+d x))}{a^2 d}+\frac {\left (a^2-b^2\right )^3 \left (7 a^2+b^2\right ) \log (a+b \sec (c+d x))}{a^2 b^8 d}-\frac {2 a \left (3 a^4-8 a^2 b^2+6 b^4\right ) \sec (c+d x)}{b^7 d}+\frac {\left (5 a^4-12 a^2 b^2+6 b^4\right ) \sec ^2(c+d x)}{2 b^6 d}-\frac {4 a \left (a^2-2 b^2\right ) \sec ^3(c+d x)}{3 b^5 d}+\frac {\left (3 a^2-4 b^2\right ) \sec ^4(c+d x)}{4 b^4 d}-\frac {2 a \sec ^5(c+d x)}{5 b^3 d}+\frac {\sec ^6(c+d x)}{6 b^2 d}+\frac {\left (a^2-b^2\right )^4}{a b^8 d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(528\) vs. \(2(255)=510\).
time = 6.30, size = 528, normalized size = 2.07 \begin {gather*} -\frac {(-a+b)^4 (a+b)^4 (b+a \cos (c+d x)) \sec ^2(c+d x)}{a^2 b^7 d (a+b \sec (c+d x))^2}+\frac {\left (-7 a^6+20 a^4 b^2-18 a^2 b^4+4 b^6\right ) (b+a \cos (c+d x))^2 \log (\cos (c+d x)) \sec ^2(c+d x)}{b^8 d (a+b \sec (c+d x))^2}+\frac {\left (7 a^8-20 a^6 b^2+18 a^4 b^4-4 a^2 b^6-b^8\right ) (b+a \cos (c+d x))^2 \log (b+a \cos (c+d x)) \sec ^2(c+d x)}{a^2 b^8 d (a+b \sec (c+d x))^2}-\frac {2 a \left (3 a^4-8 a^2 b^2+6 b^4\right ) (b+a \cos (c+d x))^2 \sec ^3(c+d x)}{b^7 d (a+b \sec (c+d x))^2}+\frac {\left (5 a^4-12 a^2 b^2+6 b^4\right ) (b+a \cos (c+d x))^2 \sec ^4(c+d x)}{2 b^6 d (a+b \sec (c+d x))^2}+\frac {4 a \left (-a^2+2 b^2\right ) (b+a \cos (c+d x))^2 \sec ^5(c+d x)}{3 b^5 d (a+b \sec (c+d x))^2}+\frac {\left (3 a^2-4 b^2\right ) (b+a \cos (c+d x))^2 \sec ^6(c+d x)}{4 b^4 d (a+b \sec (c+d x))^2}-\frac {2 a (b+a \cos (c+d x))^2 \sec ^7(c+d x)}{5 b^3 d (a+b \sec (c+d x))^2}+\frac {(b+a \cos (c+d x))^2 \sec ^8(c+d x)}{6 b^2 d (a+b \sec (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^9/(a + b*Sec[c + d*x])^2,x]

[Out]

-(((-a + b)^4*(a + b)^4*(b + a*Cos[c + d*x])*Sec[c + d*x]^2)/(a^2*b^7*d*(a + b*Sec[c + d*x])^2)) + ((-7*a^6 +
20*a^4*b^2 - 18*a^2*b^4 + 4*b^6)*(b + a*Cos[c + d*x])^2*Log[Cos[c + d*x]]*Sec[c + d*x]^2)/(b^8*d*(a + b*Sec[c
+ d*x])^2) + ((7*a^8 - 20*a^6*b^2 + 18*a^4*b^4 - 4*a^2*b^6 - b^8)*(b + a*Cos[c + d*x])^2*Log[b + a*Cos[c + d*x
]]*Sec[c + d*x]^2)/(a^2*b^8*d*(a + b*Sec[c + d*x])^2) - (2*a*(3*a^4 - 8*a^2*b^2 + 6*b^4)*(b + a*Cos[c + d*x])^
2*Sec[c + d*x]^3)/(b^7*d*(a + b*Sec[c + d*x])^2) + ((5*a^4 - 12*a^2*b^2 + 6*b^4)*(b + a*Cos[c + d*x])^2*Sec[c
+ d*x]^4)/(2*b^6*d*(a + b*Sec[c + d*x])^2) + (4*a*(-a^2 + 2*b^2)*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^5)/(3*b^5
*d*(a + b*Sec[c + d*x])^2) + ((3*a^2 - 4*b^2)*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^6)/(4*b^4*d*(a + b*Sec[c + d
*x])^2) - (2*a*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^7)/(5*b^3*d*(a + b*Sec[c + d*x])^2) + ((b + a*Cos[c + d*x])
^2*Sec[c + d*x]^8)/(6*b^2*d*(a + b*Sec[c + d*x])^2)

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Maple [A]
time = 0.20, size = 287, normalized size = 1.13

method	result	size
derivativedivides	\(\frac {-\frac {a^{8}-4 a^{6} b^{2}+6 a^{4} b^{4}-4 b^{6} a^{2}+b^{8}}{a^{2} b^{7} \left (b +a \cos \left (d x +c \right )\right )}+\frac {\left (7 a^{8}-20 a^{6} b^{2}+18 a^{4} b^{4}-4 b^{6} a^{2}-b^{8}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{8} a^{2}}-\frac {-3 a^{2}+4 b^{2}}{4 b^{4} \cos \left (d x +c \right )^{4}}-\frac {-5 a^{4}+12 b^{2} a^{2}-6 b^{4}}{2 b^{6} \cos \left (d x +c \right )^{2}}+\frac {\left (-7 a^{6}+20 a^{4} b^{2}-18 a^{2} b^{4}+4 b^{6}\right ) \ln \left (\cos \left (d x +c \right )\right )}{b^{8}}+\frac {1}{6 b^{2} \cos \left (d x +c \right )^{6}}-\frac {2 a}{5 b^{3} \cos \left (d x +c \right )^{5}}-\frac {4 a \left (a^{2}-2 b^{2}\right )}{3 b^{5} \cos \left (d x +c \right )^{3}}-\frac {2 a \left (3 a^{4}-8 b^{2} a^{2}+6 b^{4}\right )}{b^{7} \cos \left (d x +c \right )}}{d}\)	\(287\)
default	\(\frac {-\frac {a^{8}-4 a^{6} b^{2}+6 a^{4} b^{4}-4 b^{6} a^{2}+b^{8}}{a^{2} b^{7} \left (b +a \cos \left (d x +c \right )\right )}+\frac {\left (7 a^{8}-20 a^{6} b^{2}+18 a^{4} b^{4}-4 b^{6} a^{2}-b^{8}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{b^{8} a^{2}}-\frac {-3 a^{2}+4 b^{2}}{4 b^{4} \cos \left (d x +c \right )^{4}}-\frac {-5 a^{4}+12 b^{2} a^{2}-6 b^{4}}{2 b^{6} \cos \left (d x +c \right )^{2}}+\frac {\left (-7 a^{6}+20 a^{4} b^{2}-18 a^{2} b^{4}+4 b^{6}\right ) \ln \left (\cos \left (d x +c \right )\right )}{b^{8}}+\frac {1}{6 b^{2} \cos \left (d x +c \right )^{6}}-\frac {2 a}{5 b^{3} \cos \left (d x +c \right )^{5}}-\frac {4 a \left (a^{2}-2 b^{2}\right )}{3 b^{5} \cos \left (d x +c \right )^{3}}-\frac {2 a \left (3 a^{4}-8 b^{2} a^{2}+6 b^{4}\right )}{b^{7} \cos \left (d x +c \right )}}{d}\)	\(287\)
risch	\(\text {Expression too large to display}\)	\(1191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^9/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-(a^8-4*a^6*b^2+6*a^4*b^4-4*a^2*b^6+b^8)/a^2/b^7/(b+a*cos(d*x+c))+(7*a^8-20*a^6*b^2+18*a^4*b^4-4*a^2*b^6-
b^8)/b^8/a^2*ln(b+a*cos(d*x+c))-1/4*(-3*a^2+4*b^2)/b^4/cos(d*x+c)^4-1/2*(-5*a^4+12*a^2*b^2-6*b^4)/b^6/cos(d*x+
c)^2+(-7*a^6+20*a^4*b^2-18*a^2*b^4+4*b^6)/b^8*ln(cos(d*x+c))+1/6/b^2/cos(d*x+c)^6-2/5/b^3*a/cos(d*x+c)^5-4/3*a
*(a^2-2*b^2)/b^5/cos(d*x+c)^3-2*a*(3*a^4-8*a^2*b^2+6*b^4)/b^7/cos(d*x+c))

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Maxima [A]
time = 0.29, size = 321, normalized size = 1.26 \begin {gather*} -\frac {\frac {14 \, a^{3} b^{5} \cos \left (d x + c\right ) - 10 \, a^{2} b^{6} + 60 \, {\left (7 \, a^{8} - 20 \, a^{6} b^{2} + 18 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \cos \left (d x + c\right )^{6} + 30 \, {\left (7 \, a^{7} b - 20 \, a^{5} b^{3} + 18 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{5} - 10 \, {\left (7 \, a^{6} b^{2} - 20 \, a^{4} b^{4} + 18 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (7 \, a^{5} b^{3} - 20 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (7 \, a^{4} b^{4} - 20 \, a^{2} b^{6}\right )} \cos \left (d x + c\right )^{2}}{a^{3} b^{7} \cos \left (d x + c\right )^{7} + a^{2} b^{8} \cos \left (d x + c\right )^{6}} + \frac {60 \, {\left (7 \, a^{6} - 20 \, a^{4} b^{2} + 18 \, a^{2} b^{4} - 4 \, b^{6}\right )} \log \left (\cos \left (d x + c\right )\right )}{b^{8}} - \frac {60 \, {\left (7 \, a^{8} - 20 \, a^{6} b^{2} + 18 \, a^{4} b^{4} - 4 \, a^{2} b^{6} - b^{8}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{2} b^{8}}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^9/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/60*((14*a^3*b^5*cos(d*x + c) - 10*a^2*b^6 + 60*(7*a^8 - 20*a^6*b^2 + 18*a^4*b^4 - 4*a^2*b^6 + b^8)*cos(d*x
+ c)^6 + 30*(7*a^7*b - 20*a^5*b^3 + 18*a^3*b^5)*cos(d*x + c)^5 - 10*(7*a^6*b^2 - 20*a^4*b^4 + 18*a^2*b^6)*cos(
d*x + c)^4 + 5*(7*a^5*b^3 - 20*a^3*b^5)*cos(d*x + c)^3 - 3*(7*a^4*b^4 - 20*a^2*b^6)*cos(d*x + c)^2)/(a^3*b^7*c
os(d*x + c)^7 + a^2*b^8*cos(d*x + c)^6) + 60*(7*a^6 - 20*a^4*b^2 + 18*a^2*b^4 - 4*b^6)*log(cos(d*x + c))/b^8 -
60*(7*a^8 - 20*a^6*b^2 + 18*a^4*b^4 - 4*a^2*b^6 - b^8)*log(a*cos(d*x + c) + b)/(a^2*b^8))/d

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Fricas [A]
time = 2.98, size = 423, normalized size = 1.66 \begin {gather*} -\frac {14 \, a^{3} b^{6} \cos \left (d x + c\right ) - 10 \, a^{2} b^{7} + 60 \, {\left (7 \, a^{8} b - 20 \, a^{6} b^{3} + 18 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} \cos \left (d x + c\right )^{6} + 30 \, {\left (7 \, a^{7} b^{2} - 20 \, a^{5} b^{4} + 18 \, a^{3} b^{6}\right )} \cos \left (d x + c\right )^{5} - 10 \, {\left (7 \, a^{6} b^{3} - 20 \, a^{4} b^{5} + 18 \, a^{2} b^{7}\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (7 \, a^{5} b^{4} - 20 \, a^{3} b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (7 \, a^{4} b^{5} - 20 \, a^{2} b^{7}\right )} \cos \left (d x + c\right )^{2} - 60 \, {\left ({\left (7 \, a^{9} - 20 \, a^{7} b^{2} + 18 \, a^{5} b^{4} - 4 \, a^{3} b^{6} - a b^{8}\right )} \cos \left (d x + c\right )^{7} + {\left (7 \, a^{8} b - 20 \, a^{6} b^{3} + 18 \, a^{4} b^{5} - 4 \, a^{2} b^{7} - b^{9}\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + 60 \, {\left ({\left (7 \, a^{9} - 20 \, a^{7} b^{2} + 18 \, a^{5} b^{4} - 4 \, a^{3} b^{6}\right )} \cos \left (d x + c\right )^{7} + {\left (7 \, a^{8} b - 20 \, a^{6} b^{3} + 18 \, a^{4} b^{5} - 4 \, a^{2} b^{7}\right )} \cos \left (d x + c\right )^{6}\right )} \log \left (-\cos \left (d x + c\right )\right )}{60 \, {\left (a^{3} b^{8} d \cos \left (d x + c\right )^{7} + a^{2} b^{9} d \cos \left (d x + c\right )^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^9/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/60*(14*a^3*b^6*cos(d*x + c) - 10*a^2*b^7 + 60*(7*a^8*b - 20*a^6*b^3 + 18*a^4*b^5 - 4*a^2*b^7 + b^9)*cos(d*x
+ c)^6 + 30*(7*a^7*b^2 - 20*a^5*b^4 + 18*a^3*b^6)*cos(d*x + c)^5 - 10*(7*a^6*b^3 - 20*a^4*b^5 + 18*a^2*b^7)*c
os(d*x + c)^4 + 5*(7*a^5*b^4 - 20*a^3*b^6)*cos(d*x + c)^3 - 3*(7*a^4*b^5 - 20*a^2*b^7)*cos(d*x + c)^2 - 60*((7
*a^9 - 20*a^7*b^2 + 18*a^5*b^4 - 4*a^3*b^6 - a*b^8)*cos(d*x + c)^7 + (7*a^8*b - 20*a^6*b^3 + 18*a^4*b^5 - 4*a^
2*b^7 - b^9)*cos(d*x + c)^6)*log(a*cos(d*x + c) + b) + 60*((7*a^9 - 20*a^7*b^2 + 18*a^5*b^4 - 4*a^3*b^6)*cos(d
*x + c)^7 + (7*a^8*b - 20*a^6*b^3 + 18*a^4*b^5 - 4*a^2*b^7)*cos(d*x + c)^6)*log(-cos(d*x + c)))/(a^3*b^8*d*cos
(d*x + c)^7 + a^2*b^9*d*cos(d*x + c)^6)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{9}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**9/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**9/(a + b*sec(c + d*x))**2, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1696 vs. \(2 (245) = 490\).
time = 6.04, size = 1696, normalized size = 6.65 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^9/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/60*(60*(7*a^9 - 7*a^8*b - 20*a^7*b^2 + 20*a^6*b^3 + 18*a^5*b^4 - 18*a^4*b^5 - 4*a^3*b^6 + 4*a^2*b^7 - a*b^8
+ b^9)*log(abs(a + b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)))/(a^
3*b^8 - a^2*b^9) + 60*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^2 - 60*(7*a^6 - 20*a^4*b^2 + 18*a
^2*b^4 - 4*b^6)*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1))/b^8 - 60*(7*a^9 + 9*a^8*b - 18*a^7*b^2 -
26*a^6*b^3 + 12*a^5*b^4 + 24*a^4*b^5 + 2*a^3*b^6 - 6*a^2*b^7 - 3*a*b^8 - b^9 + 7*a^9*(cos(d*x + c) - 1)/(cos(d
*x + c) + 1) - 7*a^8*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 20*a^7*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1
) + 20*a^6*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 18*a^5*b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 18*a
^4*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 4*a^3*b^6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 4*a^2*b^7*(co
s(d*x + c) - 1)/(cos(d*x + c) + 1) - a*b^8*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b^9*(cos(d*x + c) - 1)/(cos
(d*x + c) + 1))/((a + b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*a
^2*b^8) + (1029*a^6 - 720*a^5*b - 2940*a^4*b^2 + 1760*a^3*b^3 + 2646*a^2*b^4 - 1168*a*b^5 - 588*b^6 + 6174*a^6
*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 3600*a^5*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 18240*a^4*b^2*(cos
(d*x + c) - 1)/(cos(d*x + c) + 1) + 9120*a^3*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 16956*a^2*b^4*(cos(d*
x + c) - 1)/(cos(d*x + c) + 1) - 6288*a*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 3888*b^6*(cos(d*x + c) - 1
)/(cos(d*x + c) + 1) + 15435*a^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 7200*a^5*b*(cos(d*x + c) - 1)^2/(
cos(d*x + c) + 1)^2 - 46500*a^4*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 18240*a^3*b^3*(cos(d*x + c) -
1)^2/(cos(d*x + c) + 1)^2 + 44730*a^2*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 12960*a*b^5*(cos(d*x + c
) - 1)^2/(cos(d*x + c) + 1)^2 - 10740*b^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 20580*a^6*(cos(d*x + c)
- 1)^3/(cos(d*x + c) + 1)^3 - 7200*a^5*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 62400*a^4*b^2*(cos(d*x +
c) - 1)^3/(cos(d*x + c) + 1)^3 + 17600*a^3*b^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 60840*a^2*b^4*(cos(
d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 11680*a*b^5*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 15520*b^6*(cos(
d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 15435*a^6*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 3600*a^5*b*(cos(d
*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 46500*a^4*b^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 8160*a^3*b^3*(
cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 44730*a^2*b^4*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 4560*a*b^
5*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 10740*b^6*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 6174*a^6*(
cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 - 720*a^5*b*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 - 18240*a^4*b^2
*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 1440*a^3*b^3*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 16956*a^
2*b^4*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 - 720*a*b^5*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 - 3888*b
^6*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 1029*a^6*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 - 2940*a^4*b
^2*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 + 2646*a^2*b^4*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 - 588*b^
6*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6)/(b^8*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^6))/d

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Mupad [B]
time = 4.77, size = 760, normalized size = 2.98 \begin {gather*} \frac {\frac {2\,\left (-105\,a^7-105\,a^6\,b+265\,a^5\,b^2+265\,a^4\,b^3-191\,a^3\,b^4-191\,a^2\,b^5+15\,a\,b^6+15\,b^7\right )}{15\,a\,b^7}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (-42\,a^7-7\,a^6\,b+113\,a^5\,b^2+13\,a^4\,b^3-95\,a^3\,b^4-5\,a^2\,b^5+19\,a\,b^6+6\,b^7\right )}{a\,b^7}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (-210\,a^7-105\,a^6\,b+523\,a^5\,b^2+244\,a^4\,b^3-362\,a^3\,b^4-145\,a^2\,b^5+7\,a\,b^6+30\,b^7\right )}{3\,a\,b^7}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (-315\,a^7-105\,a^6\,b+809\,a^5\,b^2+223\,a^4\,b^3-613\,a^3\,b^4-99\,a^2\,b^5+91\,a\,b^6+45\,b^7\right )}{3\,a\,b^7}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-525\,a^7-350\,a^6\,b+1290\,a^5\,b^2+860\,a^4\,b^3-862\,a^3\,b^4-598\,a^2\,b^5+10\,a\,b^6+75\,b^7\right )}{5\,a\,b^7}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-630\,a^7-525\,a^6\,b+1555\,a^5\,b^2+1325\,a^4\,b^3-1067\,a^3\,b^4-955\,a^2\,b^5+45\,a\,b^6+90\,b^7\right )}{15\,a\,b^7}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (-7\,a^7+20\,a^5\,b^2-18\,a^3\,b^4+4\,a\,b^6+b^7\right )}{a\,b^7}}{d\,\left (\left (b-a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+\left (7\,a-5\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\left (9\,b-21\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (35\,a-5\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (-35\,a-5\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (21\,a+9\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-7\,a-5\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,\left (7\,a^6-20\,a^4\,b^2+18\,a^2\,b^4-4\,b^6\right )}{b^8\,d}+\frac {\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,{\left (a^2-b^2\right )}^3\,\left (7\,a^2+b^2\right )}{a^2\,b^8\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^9/(a + b/cos(c + d*x))^2,x)

[Out]

((2*(15*a*b^6 - 105*a^6*b - 105*a^7 + 15*b^7 - 191*a^2*b^5 - 191*a^3*b^4 + 265*a^4*b^3 + 265*a^5*b^2))/(15*a*b
^7) - (2*tan(c/2 + (d*x)/2)^10*(19*a*b^6 - 7*a^6*b - 42*a^7 + 6*b^7 - 5*a^2*b^5 - 95*a^3*b^4 + 13*a^4*b^3 + 11
3*a^5*b^2))/(a*b^7) - (4*tan(c/2 + (d*x)/2)^6*(7*a*b^6 - 105*a^6*b - 210*a^7 + 30*b^7 - 145*a^2*b^5 - 362*a^3*
b^4 + 244*a^4*b^3 + 523*a^5*b^2))/(3*a*b^7) + (2*tan(c/2 + (d*x)/2)^8*(91*a*b^6 - 105*a^6*b - 315*a^7 + 45*b^7
- 99*a^2*b^5 - 613*a^3*b^4 + 223*a^4*b^3 + 809*a^5*b^2))/(3*a*b^7) + (2*tan(c/2 + (d*x)/2)^4*(10*a*b^6 - 350*
a^6*b - 525*a^7 + 75*b^7 - 598*a^2*b^5 - 862*a^3*b^4 + 860*a^4*b^3 + 1290*a^5*b^2))/(5*a*b^7) - (2*tan(c/2 + (
d*x)/2)^2*(45*a*b^6 - 525*a^6*b - 630*a^7 + 90*b^7 - 955*a^2*b^5 - 1067*a^3*b^4 + 1325*a^4*b^3 + 1555*a^5*b^2)
)/(15*a*b^7) + (2*tan(c/2 + (d*x)/2)^12*(4*a*b^6 - 7*a^7 + b^7 - 18*a^3*b^4 + 20*a^5*b^2))/(a*b^7))/(d*(a + b
- tan(c/2 + (d*x)/2)^14*(a - b) - tan(c/2 + (d*x)/2)^2*(7*a + 5*b) + tan(c/2 + (d*x)/2)^12*(7*a - 5*b) + tan(c
/2 + (d*x)/2)^4*(21*a + 9*b) - tan(c/2 + (d*x)/2)^10*(21*a - 9*b) - tan(c/2 + (d*x)/2)^6*(35*a + 5*b) + tan(c/
2 + (d*x)/2)^8*(35*a - 5*b))) + log(tan(c/2 + (d*x)/2)^2 + 1)/(a^2*d) - (log(tan(c/2 + (d*x)/2)^2 - 1)*(7*a^6
- 4*b^6 + 18*a^2*b^4 - 20*a^4*b^2))/(b^8*d) + (log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(a
^2 - b^2)^3*(7*a^2 + b^2))/(a^2*b^8*d)

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