3.515 \(\int \frac{\sec ^5(c+d x)}{(a+b \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=259 \[ -\frac{a \left (-7 a^4 b^2+8 a^2 b^4+2 a^6-8 b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{7/2} (a+b)^{7/2}}-\frac{a^2 \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac{a^3 \left (3 a^2-8 b^2\right ) \tan (c+d x)}{6 b^3 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac{a^2 \left (-28 a^2 b^2+9 a^4+34 b^4\right ) \tan (c+d x)}{6 b^3 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\tanh ^{-1}(\sin (c+d x))}{b^4 d} \]

[Out]

ArcTanh[Sin[c + d*x]]/(b^4*d) - (a*(2*a^6 - 7*a^4*b^2 + 8*a^2*b^4 - 8*b^6)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/
2])/Sqrt[a + b]])/((a - b)^(7/2)*b^4*(a + b)^(7/2)*d) - (a^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(
a + b*Sec[c + d*x])^3) + (a^3*(3*a^2 - 8*b^2)*Tan[c + d*x])/(6*b^3*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) - (
a^2*(9*a^4 - 28*a^2*b^2 + 34*b^4)*Tan[c + d*x])/(6*b^3*(a^2 - b^2)^3*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.751795, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3845, 4090, 4080, 3998, 3770, 3831, 2659, 208} \[ -\frac{a \left (-7 a^4 b^2+8 a^2 b^4+2 a^6-8 b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{7/2} (a+b)^{7/2}}-\frac{a^2 \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac{a^3 \left (3 a^2-8 b^2\right ) \tan (c+d x)}{6 b^3 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac{a^2 \left (-28 a^2 b^2+9 a^4+34 b^4\right ) \tan (c+d x)}{6 b^3 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\tanh ^{-1}(\sin (c+d x))}{b^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a + b*Sec[c + d*x])^4,x]

[Out]

ArcTanh[Sin[c + d*x]]/(b^4*d) - (a*(2*a^6 - 7*a^4*b^2 + 8*a^2*b^4 - 8*b^6)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/
2])/Sqrt[a + b]])/((a - b)^(7/2)*b^4*(a + b)^(7/2)*d) - (a^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(
a + b*Sec[c + d*x])^3) + (a^3*(3*a^2 - 8*b^2)*Tan[c + d*x])/(6*b^3*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^2) - (
a^2*(9*a^4 - 28*a^2*b^2 + 34*b^4)*Tan[c + d*x])/(6*b^3*(a^2 - b^2)^3*d*(a + b*Sec[c + d*x]))

Rule 3845

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a^2*
d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[d
^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b*(
m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && N
eQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

Rule 4090

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e
+ f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*C
sc[e + f*x])^(m + 1)*Simp[b*(m + 1)*(-(a*(b*B - a*C)) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) +
C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 4080

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f
*x])^(m + 1))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e +
f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

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

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

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

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x)}{(a+b \sec (c+d x))^4} \, dx &=-\frac{a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac{\int \frac{\sec ^2(c+d x) \left (2 a^2-3 a b \sec (c+d x)-3 \left (a^2-b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac{a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a^3 \left (3 a^2-8 b^2\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{\int \frac{\sec (c+d x) \left (2 a^2 b \left (3 a^2-8 b^2\right )+a \left (3 a^4-10 a^2 b^2+12 b^4\right ) \sec (c+d x)-6 b \left (a^2-b^2\right )^2 \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a^3 \left (3 a^2-8 b^2\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\int \frac{\sec (c+d x) \left (3 a b^2 \left (a^4-2 a^2 b^2+6 b^4\right )+6 b \left (a^2-b^2\right )^3 \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 b^4 \left (a^2-b^2\right )^3}\\ &=-\frac{a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a^3 \left (3 a^2-8 b^2\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\int \sec (c+d x) \, dx}{b^4}-\frac{\left (a \left (2 a^6-7 a^4 b^2+8 a^2 b^4-8 b^6\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^3}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a^3 \left (3 a^2-8 b^2\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (a \left (2 a^6-7 a^4 b^2+8 a^2 b^4-8 b^6\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b^5 \left (a^2-b^2\right )^3}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a^3 \left (3 a^2-8 b^2\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (a \left (2 a^6-7 a^4 b^2+8 a^2 b^4-8 b^6\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^5 \left (a^2-b^2\right )^3 d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{a \left (2 a^6-7 a^4 b^2+8 a^2 b^4-8 b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{7/2} b^4 (a+b)^{7/2} d}-\frac{a^2 \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{a^3 \left (3 a^2-8 b^2\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{a^2 \left (9 a^4-28 a^2 b^2+34 b^4\right ) \tan (c+d x)}{6 b^3 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 4.18369, size = 250, normalized size = 0.97 \[ \frac{-\frac{a^2 b \sin (c+d x) \left (a^2 \left (-17 a^2 b^2+6 a^4+26 b^4\right ) \cos ^2(c+d x)+15 a b \left (-3 a^2 b^2+a^4+4 b^4\right ) \cos (c+d x)-32 a^2 b^4+11 a^4 b^2+36 b^6\right )}{(a-b)^3 (a+b)^3 (a \cos (c+d x)+b)^3}+\frac{6 a \left (-7 a^4 b^2+8 a^2 b^4+2 a^6-8 b^6\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}-6 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{6 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a + b*Sec[c + d*x])^4,x]

[Out]

((6*a*(2*a^6 - 7*a^4*b^2 + 8*a^2*b^4 - 8*b^6)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2
)^(7/2) - 6*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - (a^2*b*(11
*a^4*b^2 - 32*a^2*b^4 + 36*b^6 + 15*a*b*(a^4 - 3*a^2*b^2 + 4*b^4)*Cos[c + d*x] + a^2*(6*a^4 - 17*a^2*b^2 + 26*
b^4)*Cos[c + d*x]^2)*Sin[c + d*x])/((a - b)^3*(a + b)^3*(b + a*Cos[c + d*x])^3))/(6*b^4*d)

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Maple [B]  time = 0.069, size = 1429, normalized size = 5.5 \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)^5/(a+b*sec(d*x+c))^4,x)

[Out]

2/d*a^6/b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+
1/2*c)^5-1/d*a^5/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan
(1/2*d*x+1/2*c)^5-6/d*a^4/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b
^3)*tan(1/2*d*x+1/2*c)^5+4/d*a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3/(a-b)/(a^3+3*a^2*b+3*a*
b^2+b^3)*tan(1/2*d*x+1/2*c)^5+12/d*a^2*b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3/(a-b)/(a^3+3*a^
2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5-4/d*a^6/b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3/(a^2+2
*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+44/3/d*a^4/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b
)^3/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-24/d*a^2*b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)
^2*b-a-b)^3/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+2/d*a^6/b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d
*x+1/2*c)^2*b-a-b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)+1/d*a^5/b^2/(tan(1/2*d*x+1/2*c)^2*a-ta
n(1/2*d*x+1/2*c)^2*b-a-b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)-6/d*a^4/b/(tan(1/2*d*x+1/2*c)^2
*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)-4/d*a^3/(tan(1/2*d*x+1/2*c
)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)+12/d*a^2*b/(tan(1/2*d*x
+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)-2/d*a^7/b^4/(a^6-
3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+7/d*a^5/b^2
/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-8/d*a
^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+8/d
*a*b^2/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))
+1/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)-1/d/b^4*ln(tan(1/2*d*x+1/2*c)-1)

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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)^5/(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 20.7488, size = 3977, normalized size = 15.36 \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)^5/(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

[1/12*(3*(2*a^7*b^3 - 7*a^5*b^5 + 8*a^3*b^7 - 8*a*b^9 + (2*a^10 - 7*a^8*b^2 + 8*a^6*b^4 - 8*a^4*b^6)*cos(d*x +
c)^3 + 3*(2*a^9*b - 7*a^7*b^3 + 8*a^5*b^5 - 8*a^3*b^7)*cos(d*x + c)^2 + 3*(2*a^8*b^2 - 7*a^6*b^4 + 8*a^4*b^6
- 8*a^2*b^8)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2
- b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + 6*
(a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11 + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*cos
(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4
+ 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c))*log(sin(d*x + c) + 1) - 6*(a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 -
4*a^2*b^9 + b^11 + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*cos(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3
+ 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*
cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(11*a^8*b^3 - 43*a^6*b^5 + 68*a^4*b^7 - 36*a^2*b^9 + (6*a^10*b - 23*a
^8*b^3 + 43*a^6*b^5 - 26*a^4*b^7)*cos(d*x + c)^2 + 15*(a^9*b^2 - 4*a^7*b^4 + 7*a^5*b^6 - 4*a^3*b^8)*cos(d*x +
c))*sin(d*x + c))/((a^11*b^4 - 4*a^9*b^6 + 6*a^7*b^8 - 4*a^5*b^10 + a^3*b^12)*d*cos(d*x + c)^3 + 3*(a^10*b^5 -
4*a^8*b^7 + 6*a^6*b^9 - 4*a^4*b^11 + a^2*b^13)*d*cos(d*x + c)^2 + 3*(a^9*b^6 - 4*a^7*b^8 + 6*a^5*b^10 - 4*a^3
*b^12 + a*b^14)*d*cos(d*x + c) + (a^8*b^7 - 4*a^6*b^9 + 6*a^4*b^11 - 4*a^2*b^13 + b^15)*d), -1/6*(3*(2*a^7*b^3
- 7*a^5*b^5 + 8*a^3*b^7 - 8*a*b^9 + (2*a^10 - 7*a^8*b^2 + 8*a^6*b^4 - 8*a^4*b^6)*cos(d*x + c)^3 + 3*(2*a^9*b
- 7*a^7*b^3 + 8*a^5*b^5 - 8*a^3*b^7)*cos(d*x + c)^2 + 3*(2*a^8*b^2 - 7*a^6*b^4 + 8*a^4*b^6 - 8*a^2*b^8)*cos(d*
x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - 3*(a^8*b^
3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11 + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*cos(d*x +
c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^
5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x + c))*log(sin(d*x + c) + 1) + 3*(a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b
^9 + b^11 + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*cos(d*x + c)^3 + 3*(a^10*b - 4*a^8*b^3 + 6*a^
6*b^5 - 4*a^4*b^7 + a^2*b^9)*cos(d*x + c)^2 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*cos(d*x
+ c))*log(-sin(d*x + c) + 1) + (11*a^8*b^3 - 43*a^6*b^5 + 68*a^4*b^7 - 36*a^2*b^9 + (6*a^10*b - 23*a^8*b^3 +
43*a^6*b^5 - 26*a^4*b^7)*cos(d*x + c)^2 + 15*(a^9*b^2 - 4*a^7*b^4 + 7*a^5*b^6 - 4*a^3*b^8)*cos(d*x + c))*sin(d
*x + c))/((a^11*b^4 - 4*a^9*b^6 + 6*a^7*b^8 - 4*a^5*b^10 + a^3*b^12)*d*cos(d*x + c)^3 + 3*(a^10*b^5 - 4*a^8*b^
7 + 6*a^6*b^9 - 4*a^4*b^11 + a^2*b^13)*d*cos(d*x + c)^2 + 3*(a^9*b^6 - 4*a^7*b^8 + 6*a^5*b^10 - 4*a^3*b^12 + a
*b^14)*d*cos(d*x + c) + (a^8*b^7 - 4*a^6*b^9 + 6*a^4*b^11 - 4*a^2*b^13 + b^15)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+b*sec(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)**5/(a + b*sec(c + d*x))**4, x)

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Giac [B]  time = 1.53588, size = 755, normalized size = 2.92 \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)^5/(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(3*(2*a^7 - 7*a^5*b^2 + 8*a^3*b^4 - 8*a*b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*
tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^6*b^4 - 3*a^4*b^6 + 3*a^2*b^8 - b^10)*sq
rt(-a^2 + b^2)) + (6*a^8*tan(1/2*d*x + 1/2*c)^5 - 15*a^7*b*tan(1/2*d*x + 1/2*c)^5 - 6*a^6*b^2*tan(1/2*d*x + 1/
2*c)^5 + 45*a^5*b^3*tan(1/2*d*x + 1/2*c)^5 - 6*a^4*b^4*tan(1/2*d*x + 1/2*c)^5 - 60*a^3*b^5*tan(1/2*d*x + 1/2*c
)^5 + 36*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 - 12*a^8*tan(1/2*d*x + 1/2*c)^3 + 56*a^6*b^2*tan(1/2*d*x + 1/2*c)^3 -
116*a^4*b^4*tan(1/2*d*x + 1/2*c)^3 + 72*a^2*b^6*tan(1/2*d*x + 1/2*c)^3 + 6*a^8*tan(1/2*d*x + 1/2*c) + 15*a^7*b
*tan(1/2*d*x + 1/2*c) - 6*a^6*b^2*tan(1/2*d*x + 1/2*c) - 45*a^5*b^3*tan(1/2*d*x + 1/2*c) - 6*a^4*b^4*tan(1/2*d
*x + 1/2*c) + 60*a^3*b^5*tan(1/2*d*x + 1/2*c) + 36*a^2*b^6*tan(1/2*d*x + 1/2*c))/((a^6*b^3 - 3*a^4*b^5 + 3*a^2
*b^7 - b^9)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^3) + 3*log(abs(tan(1/2*d*x + 1/2*c)
+ 1))/b^4 - 3*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4)/d