3.121 \(\int \frac{\sec ^6(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=262 \[ \frac{\left (a^2+b^2\right ) \sec ^3(c+d x)}{3 b^3 d}+\frac{\left (a^2+b^2\right )^2 \sec (c+d x)}{b^5 d}-\frac{a \left (a^2+b^2\right )^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac{a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{a \left (a^2+b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b^4 d}-\frac{\left (a^2+b^2\right )^{5/2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d}-\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 b^2 d}-\frac{a \tan (c+d x) \sec ^3(c+d x)}{4 b^2 d}-\frac{3 a \tan (c+d x) \sec (c+d x)}{8 b^2 d}+\frac{\sec ^5(c+d x)}{5 b d} \]

[Out]

(-3*a*ArcTanh[Sin[c + d*x]])/(8*b^2*d) - (a*(a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(2*b^4*d) - (a*(a^2 + b^2)^2*Ar
cTanh[Sin[c + d*x]])/(b^6*d) - ((a^2 + b^2)^(5/2)*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/
(b^6*d) + ((a^2 + b^2)^2*Sec[c + d*x])/(b^5*d) + ((a^2 + b^2)*Sec[c + d*x]^3)/(3*b^3*d) + Sec[c + d*x]^5/(5*b*
d) - (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*b^2*d) - (a*(a^2 + b^2)*Sec[c + d*x]*Tan[c + d*x])/(2*b^4*d) - (a*Sec[
c + d*x]^3*Tan[c + d*x])/(4*b^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.256162, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3104, 3768, 3770, 3074, 206} \[ \frac{\left (a^2+b^2\right ) \sec ^3(c+d x)}{3 b^3 d}+\frac{\left (a^2+b^2\right )^2 \sec (c+d x)}{b^5 d}-\frac{a \left (a^2+b^2\right )^2 \tanh ^{-1}(\sin (c+d x))}{b^6 d}-\frac{a \left (a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{a \left (a^2+b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b^4 d}-\frac{\left (a^2+b^2\right )^{5/2} \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{b^6 d}-\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 b^2 d}-\frac{a \tan (c+d x) \sec ^3(c+d x)}{4 b^2 d}-\frac{3 a \tan (c+d x) \sec (c+d x)}{8 b^2 d}+\frac{\sec ^5(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(-3*a*ArcTanh[Sin[c + d*x]])/(8*b^2*d) - (a*(a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(2*b^4*d) - (a*(a^2 + b^2)^2*Ar
cTanh[Sin[c + d*x]])/(b^6*d) - ((a^2 + b^2)^(5/2)*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/
(b^6*d) + ((a^2 + b^2)^2*Sec[c + d*x])/(b^5*d) + ((a^2 + b^2)*Sec[c + d*x]^3)/(3*b^3*d) + Sec[c + d*x]^5/(5*b*
d) - (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*b^2*d) - (a*(a^2 + b^2)*Sec[c + d*x]*Tan[c + d*x])/(2*b^4*d) - (a*Sec[
c + d*x]^3*Tan[c + d*x])/(4*b^2*d)

Rule 3104

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[Cos[c + d*x]^(m + 1)/(b*d*(m + 1)), x] + (-Dist[a/b^2, Int[Cos[c + d*x]^(m + 1), x], x] + Dist[(a^2 + b
^2)/b^2, Int[Cos[c + d*x]^(m + 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[
a^2 + b^2, 0] && LtQ[m, -1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 5.24765, size = 661, normalized size = 2.52 \[ \frac{\sec (c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (\frac{2 b^3 \left (20 a^2+29 b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}-\frac{2 b^3 \left (20 a^2+29 b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{b^2 \left (20 a^2 b-60 a^3-105 a b^2+29 b^3\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{b^2 \left (20 a^2 b+60 a^3+105 a b^2+29 b^3\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 b \left (260 a^2 b^2+120 a^4+149 b^4\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-\frac{2 b \left (260 a^2 b^2+120 a^4+149 b^4\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+480 \left (a^2+b^2\right )^{5/2} \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )+30 a \left (20 a^2 b^2+8 a^4+15 b^4\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-30 a \left (20 a^2 b^2+8 a^4+15 b^4\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+520 a^2 b^3+240 a^4 b+\frac{3 b^4 (2 b-5 a)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{3 b^4 (5 a+2 b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{12 b^5 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5}-\frac{12 b^5 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5}+298 b^5\right )}{240 b^6 d (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(Sec[c + d*x]*(240*a^4*b + 520*a^2*b^3 + 298*b^5 + 480*(a^2 + b^2)^(5/2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqr
t[a^2 + b^2]] + 30*a*(8*a^4 + 20*a^2*b^2 + 15*b^4)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 30*a*(8*a^4 + 20
*a^2*b^2 + 15*b^4)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (3*b^4*(-5*a + 2*b))/(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2])^4 + (b^2*(-60*a^3 + 20*a^2*b - 105*a*b^2 + 29*b^3))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (12*b
^5*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5 + (2*b^3*(20*a^2 + 29*b^2)*Sin[(c + d*x)/2])/(Cos
[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (2*b*(120*a^4 + 260*a^2*b^2 + 149*b^4)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]) - (12*b^5*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5 + (3*b^4*(5*a + 2*b
))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - (2*b^3*(20*a^2 + 29*b^2)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^3 + (b^2*(60*a^3 + 20*a^2*b + 105*a*b^2 + 29*b^3))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (
2*b*(120*a^4 + 260*a^2*b^2 + 149*b^4)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))*(a*Cos[c + d*x]
+ b*Sin[c + d*x]))/(240*b^6*d*(a + b*Tan[c + d*x]))

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Maple [B]  time = 0.201, size = 994, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

2/d/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-15/8/d*a/b^2*ln(tan(1/2*d*x+1/2*
c)+1)+15/8/d*a/b^2*ln(tan(1/2*d*x+1/2*c)-1)+1/5/d/b/(tan(1/2*d*x+1/2*c)+1)^5-1/2/d/b/(tan(1/2*d*x+1/2*c)+1)^4-
1/5/d/b/(tan(1/2*d*x+1/2*c)-1)^5-1/2/d/b/(tan(1/2*d*x+1/2*c)-1)^4+2/d/b^6/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan
(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))*a^6+15/8/d/b/(tan(1/2*d*x+1/2*c)+1)-15/8/d/b/(tan(1/2*d*x+1/2*c)-1)-1/d/
b^5/(tan(1/2*d*x+1/2*c)-1)*a^4-1/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)*a^3+1/d*a^5/b^6*ln(tan(1/2*d*x+1/2*c)-1)+1/d/b
^5/(tan(1/2*d*x+1/2*c)+1)*a^4-1/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)*a^3-1/d*a^5/b^6*ln(tan(1/2*d*x+1/2*c)+1)-1/4/d/
b^2/(tan(1/2*d*x+1/2*c)-1)^4*a-1/3/d/b^3/(tan(1/2*d*x+1/2*c)-1)^3*a^2-1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)^3*a+1/4
/d/b^2/(tan(1/2*d*x+1/2*c)+1)^4*a+1/3/d/b^3/(tan(1/2*d*x+1/2*c)+1)^3*a^2-1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^3*a+
1/2/d/b^4/(tan(1/2*d*x+1/2*c)+1)^2*a^3-1/2/d/b^3/(tan(1/2*d*x+1/2*c)+1)^2*a^2-1/2/d/b^4/(tan(1/2*d*x+1/2*c)-1)
^2*a^3-1/2/d/b^3/(tan(1/2*d*x+1/2*c)-1)^2*a^2+11/8/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2*a+5/2/d/b^3/(tan(1/2*d*x+1/2
*c)+1)*a^2-9/8/d/b^2/(tan(1/2*d*x+1/2*c)+1)*a-5/2/d*a^3/b^4*ln(tan(1/2*d*x+1/2*c)+1)-11/8/d/b^2/(tan(1/2*d*x+1
/2*c)-1)^2*a-5/2/d/b^3/(tan(1/2*d*x+1/2*c)-1)*a^2-9/8/d/b^2/(tan(1/2*d*x+1/2*c)-1)*a+5/2/d*a^3/b^4*ln(tan(1/2*
d*x+1/2*c)-1)+13/12/d/b/(tan(1/2*d*x+1/2*c)+1)^3-9/8/d/b/(tan(1/2*d*x+1/2*c)+1)^2-13/12/d/b/(tan(1/2*d*x+1/2*c
)-1)^3-9/8/d/b/(tan(1/2*d*x+1/2*c)-1)^2+6/d/b^2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+
b^2)^(1/2))*a^2+6/d/b^4/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))*a^4

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.4155, size = 834, normalized size = 3.18 \begin{align*} \frac{120 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}} \cos \left (d x + c\right )^{5} \log \left (-\frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) - 15 \,{\left (8 \, a^{5} + 20 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \,{\left (8 \, a^{5} + 20 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 48 \, b^{5} + 240 \,{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 80 \,{\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} - 30 \,{\left (2 \, a b^{4} \cos \left (d x + c\right ) +{\left (4 \, a^{3} b^{2} + 7 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{240 \, b^{6} d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(120*(a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)*cos(d*x + c)^5*log(-(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2
- b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 + 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c
)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)) - 15*(8*a^5 + 20*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^5*log(si
n(d*x + c) + 1) + 15*(8*a^5 + 20*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 48*b^5 + 240*(a^4
*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^4 + 80*(a^2*b^3 + b^5)*cos(d*x + c)^2 - 30*(2*a*b^4*cos(d*x + c) + (4*a^3*b
^2 + 7*a*b^4)*cos(d*x + c)^3)*sin(d*x + c))/(b^6*d*cos(d*x + c)^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6/(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.38785, size = 748, normalized size = 2.85 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/120*(15*(8*a^5 + 20*a^3*b^2 + 15*a*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^6 - 15*(8*a^5 + 20*a^3*b^2 + 1
5*a*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^6 + 120*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*log(abs(2*a*tan(1/2*
d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b
^2)*b^6) + 2*(60*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 135*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*a^4*tan(1/2*d*x + 1/2*c
)^8 + 360*a^2*b^2*tan(1/2*d*x + 1/2*c)^8 + 360*b^4*tan(1/2*d*x + 1/2*c)^8 - 120*a^3*b*tan(1/2*d*x + 1/2*c)^7 -
150*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 480*a^4*tan(1/2*d*x + 1/2*c)^6 - 1200*a^2*b^2*tan(1/2*d*x + 1/2*c)^6 - 720
*b^4*tan(1/2*d*x + 1/2*c)^6 + 720*a^4*tan(1/2*d*x + 1/2*c)^4 + 1600*a^2*b^2*tan(1/2*d*x + 1/2*c)^4 + 1120*b^4*
tan(1/2*d*x + 1/2*c)^4 + 120*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 150*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 480*a^4*tan(1/2
*d*x + 1/2*c)^2 - 1040*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 560*b^4*tan(1/2*d*x + 1/2*c)^2 - 60*a^3*b*tan(1/2*d*x
+ 1/2*c) - 135*a*b^3*tan(1/2*d*x + 1/2*c) + 120*a^4 + 280*a^2*b^2 + 184*b^4)/((tan(1/2*d*x + 1/2*c)^2 - 1)^5*b
^5))/d