3.225 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^6}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=363 \[ -\frac{2 d \left (107 c^3 d^2-472 c^2 d^3+18 c^4 d+2 c^5+456 c d^4-136 d^5\right ) \tan (e+f x)}{15 a^3 f}+\frac{d^3 \left (-90 c^2 d+40 c^3+78 c d^2-23 d^3\right ) \tanh ^{-1}(\sin (e+f x))}{2 a^3 f}+\frac{(c-d) \left (2 c^2+18 c d+115 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^3}{15 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac{d \left (18 c^2 d+2 c^3+111 c d^2-136 d^3\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{15 a^3 f}-\frac{d^2 \left (216 c^2 d^2+36 c^3 d+4 c^4-626 c d^3+345 d^4\right ) \tan (e+f x) \sec (e+f x)}{30 a^3 f}+\frac{(c-d) \tan (e+f x) (c+d \sec (e+f x))^5}{5 f (a \sec (e+f x)+a)^3}+\frac{(c-d) (2 c+13 d) \tan (e+f x) (c+d \sec (e+f x))^4}{15 a f (a \sec (e+f x)+a)^2} \]

[Out]

(d^3*(40*c^3 - 90*c^2*d + 78*c*d^2 - 23*d^3)*ArcTanh[Sin[e + f*x]])/(2*a^3*f) - (2*d*(2*c^5 + 18*c^4*d + 107*c
^3*d^2 - 472*c^2*d^3 + 456*c*d^4 - 136*d^5)*Tan[e + f*x])/(15*a^3*f) - (d^2*(4*c^4 + 36*c^3*d + 216*c^2*d^2 -
626*c*d^3 + 345*d^4)*Sec[e + f*x]*Tan[e + f*x])/(30*a^3*f) - (d*(2*c^3 + 18*c^2*d + 111*c*d^2 - 136*d^3)*(c +
d*Sec[e + f*x])^2*Tan[e + f*x])/(15*a^3*f) + ((c - d)*(2*c^2 + 18*c*d + 115*d^2)*(c + d*Sec[e + f*x])^3*Tan[e
+ f*x])/(15*f*(a^3 + a^3*Sec[e + f*x])) + ((c - d)*(2*c + 13*d)*(c + d*Sec[e + f*x])^4*Tan[e + f*x])/(15*a*f*(
a + a*Sec[e + f*x])^2) + ((c - d)*(c + d*Sec[e + f*x])^5*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3)

________________________________________________________________________________________

Rubi [A]  time = 0.538618, antiderivative size = 405, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {3987, 98, 150, 153, 147, 63, 217, 203} \[ \frac{d^3 \left (-90 c^2 d+40 c^3+78 c d^2-23 d^3\right ) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{(c-d) \left (2 c^2+18 c d+115 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^3}{15 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac{d \left (18 c^2 d+2 c^3+111 c d^2-136 d^3\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{15 a^3 f}-\frac{d \tan (e+f x) \left (d \left (216 c^2 d^2+36 c^3 d+4 c^4-626 c d^3+345 d^4\right ) \sec (e+f x)+4 \left (107 c^3 d^2-472 c^2 d^3+18 c^4 d+2 c^5+456 c d^4-136 d^5\right )\right )}{30 a^3 f}+\frac{(c-d) \tan (e+f x) (c+d \sec (e+f x))^5}{5 f (a \sec (e+f x)+a)^3}+\frac{(c-d) (2 c+13 d) \tan (e+f x) (c+d \sec (e+f x))^4}{15 a f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x])^6)/(a + a*Sec[e + f*x])^3,x]

[Out]

(d^3*(40*c^3 - 90*c^2*d + 78*c*d^2 - 23*d^3)*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e
+ f*x])/(a^2*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (d*(2*c^3 + 18*c^2*d + 111*c*d^2 - 136*d^
3)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(15*a^3*f) + ((c - d)*(2*c^2 + 18*c*d + 115*d^2)*(c + d*Sec[e + f*x])^
3*Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*x])) + ((c - d)*(2*c + 13*d)*(c + d*Sec[e + f*x])^4*Tan[e + f*x])/(
15*a*f*(a + a*Sec[e + f*x])^2) + ((c - d)*(c + d*Sec[e + f*x])^5*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) -
(d*(4*(2*c^5 + 18*c^4*d + 107*c^3*d^2 - 472*c^2*d^3 + 456*c*d^4 - 136*d^5) + d*(4*c^4 + 36*c^3*d + 216*c^2*d^2
- 626*c*d^3 + 345*d^4)*Sec[e + f*x])*Tan[e + f*x])/(30*a^3*f)

Rule 3987

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[(a^2*g*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x
]]), Subst[Int[((g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^n)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))^6}{(a+a \sec (e+f x))^3} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^6}{\sqrt{a-a x} (a+a x)^{7/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) (c+d \sec (e+f x))^5 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x)^4 \left (-a^2 \left (2 c^2+8 c d-5 d^2\right )+a^2 (3 c-8 d) d x\right )}{\sqrt{a-a x} (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) (2 c+13 d) (c+d \sec (e+f x))^4 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^5 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x)^3 \left (-a^4 \left (2 c^3+10 c^2 d+55 c d^2-52 d^3\right )+3 a^4 d \left (2 c^2+14 c d-21 d^2\right ) x\right )}{\sqrt{a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 a^4 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) \left (2 c^2+18 c d+115 d^2\right ) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+13 d) (c+d \sec (e+f x))^4 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^5 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x)^2 \left (-3 a^6 d^2 \left (2 c^2+118 c d-115 d^2\right )+3 a^6 d \left (2 c^3+18 c^2 d+111 c d^2-136 d^3\right ) x\right )}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{15 a^7 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d \left (2 c^3+18 c^2 d+111 c d^2-136 d^3\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 a^3 f}+\frac{(c-d) \left (2 c^2+18 c d+115 d^2\right ) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+13 d) (c+d \sec (e+f x))^4 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^5 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x) \left (3 a^8 d^2 \left (2 c^3+318 c^2 d-567 c d^2+272 d^3\right )-3 a^8 d \left (4 c^4+36 c^3 d+216 c^2 d^2-626 c d^3+345 d^4\right ) x\right )}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{45 a^9 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d \left (2 c^3+18 c^2 d+111 c d^2-136 d^3\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 a^3 f}+\frac{(c-d) \left (2 c^2+18 c d+115 d^2\right ) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+13 d) (c+d \sec (e+f x))^4 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^5 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{d \left (4 \left (2 c^5+18 c^4 d+107 c^3 d^2-472 c^2 d^3+456 c d^4-136 d^5\right )+d \left (4 c^4+36 c^3 d+216 c^2 d^2-626 c d^3+345 d^4\right ) \sec (e+f x)\right ) \tan (e+f x)}{30 a^3 f}-\frac{\left (d^3 \left (40 c^3-90 c^2 d+78 c d^2-23 d^3\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{2 a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d \left (2 c^3+18 c^2 d+111 c d^2-136 d^3\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 a^3 f}+\frac{(c-d) \left (2 c^2+18 c d+115 d^2\right ) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+13 d) (c+d \sec (e+f x))^4 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^5 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{d \left (4 \left (2 c^5+18 c^4 d+107 c^3 d^2-472 c^2 d^3+456 c d^4-136 d^5\right )+d \left (4 c^4+36 c^3 d+216 c^2 d^2-626 c d^3+345 d^4\right ) \sec (e+f x)\right ) \tan (e+f x)}{30 a^3 f}+\frac{\left (d^3 \left (40 c^3-90 c^2 d+78 c d^2-23 d^3\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 a-x^2}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d \left (2 c^3+18 c^2 d+111 c d^2-136 d^3\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 a^3 f}+\frac{(c-d) \left (2 c^2+18 c d+115 d^2\right ) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+13 d) (c+d \sec (e+f x))^4 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^5 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{d \left (4 \left (2 c^5+18 c^4 d+107 c^3 d^2-472 c^2 d^3+456 c d^4-136 d^5\right )+d \left (4 c^4+36 c^3 d+216 c^2 d^2-626 c d^3+345 d^4\right ) \sec (e+f x)\right ) \tan (e+f x)}{30 a^3 f}+\frac{\left (d^3 \left (40 c^3-90 c^2 d+78 c d^2-23 d^3\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{d^3 \left (40 c^3-90 c^2 d+78 c d^2-23 d^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right ) \tan (e+f x)}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{d \left (2 c^3+18 c^2 d+111 c d^2-136 d^3\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 a^3 f}+\frac{(c-d) \left (2 c^2+18 c d+115 d^2\right ) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+13 d) (c+d \sec (e+f x))^4 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^5 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{d \left (4 \left (2 c^5+18 c^4 d+107 c^3 d^2-472 c^2 d^3+456 c d^4-136 d^5\right )+d \left (4 c^4+36 c^3 d+216 c^2 d^2-626 c d^3+345 d^4\right ) \sec (e+f x)\right ) \tan (e+f x)}{30 a^3 f}\\ \end{align*}

Mathematica [B]  time = 6.70421, size = 1338, normalized size = 3.69 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^6)/(a + a*Sec[e + f*x])^3,x]

[Out]

(4*(-40*c^3*d^3 + 90*c^2*d^4 - 78*c*d^5 + 23*d^6)*Cos[e/2 + (f*x)/2]^6*Cos[e + f*x]^3*Log[Cos[e/2 + (f*x)/2] -
Sin[e/2 + (f*x)/2]]*(c + d*Sec[e + f*x])^6)/(f*(d + c*Cos[e + f*x])^6*(a + a*Sec[e + f*x])^3) - (4*(-40*c^3*d
^3 + 90*c^2*d^4 - 78*c*d^5 + 23*d^6)*Cos[e/2 + (f*x)/2]^6*Cos[e + f*x]^3*Log[Cos[e/2 + (f*x)/2] + Sin[e/2 + (f
*x)/2]]*(c + d*Sec[e + f*x])^6)/(f*(d + c*Cos[e + f*x])^6*(a + a*Sec[e + f*x])^3) + (2*Cos[e/2 + (f*x)/2]^2*Co
s[e + f*x]^3*Sec[e/2]*(c + d*Sec[e + f*x])^6*(c^6*Sin[e/2] - 6*c^5*d*Sin[e/2] + 15*c^4*d^2*Sin[e/2] - 20*c^3*d
^3*Sin[e/2] + 15*c^2*d^4*Sin[e/2] - 6*c*d^5*Sin[e/2] + d^6*Sin[e/2]))/(5*f*(d + c*Cos[e + f*x])^6*(a + a*Sec[e
+ f*x])^3) + (8*Cos[e/2 + (f*x)/2]^4*Cos[e + f*x]^3*Sec[e/2]*(c + d*Sec[e + f*x])^6*(-4*c^6*Sin[e/2] + 9*c^5*
d*Sin[e/2] + 15*c^4*d^2*Sin[e/2] - 70*c^3*d^3*Sin[e/2] + 90*c^2*d^4*Sin[e/2] - 51*c*d^5*Sin[e/2] + 11*d^6*Sin[
e/2]))/(15*f*(d + c*Cos[e + f*x])^6*(a + a*Sec[e + f*x])^3) + (2*Cos[e/2 + (f*x)/2]*Cos[e + f*x]^3*Sec[e/2]*(c
+ d*Sec[e + f*x])^6*(c^6*Sin[(f*x)/2] - 6*c^5*d*Sin[(f*x)/2] + 15*c^4*d^2*Sin[(f*x)/2] - 20*c^3*d^3*Sin[(f*x)
/2] + 15*c^2*d^4*Sin[(f*x)/2] - 6*c*d^5*Sin[(f*x)/2] + d^6*Sin[(f*x)/2]))/(5*f*(d + c*Cos[e + f*x])^6*(a + a*S
ec[e + f*x])^3) + (8*Cos[e/2 + (f*x)/2]^3*Cos[e + f*x]^3*Sec[e/2]*(c + d*Sec[e + f*x])^6*(-4*c^6*Sin[(f*x)/2]
+ 9*c^5*d*Sin[(f*x)/2] + 15*c^4*d^2*Sin[(f*x)/2] - 70*c^3*d^3*Sin[(f*x)/2] + 90*c^2*d^4*Sin[(f*x)/2] - 51*c*d^
5*Sin[(f*x)/2] + 11*d^6*Sin[(f*x)/2]))/(15*f*(d + c*Cos[e + f*x])^6*(a + a*Sec[e + f*x])^3) + (8*Cos[e/2 + (f*
x)/2]^5*Cos[e + f*x]^3*Sec[e/2]*(c + d*Sec[e + f*x])^6*(7*c^6*Sin[(f*x)/2] + 18*c^5*d*Sin[(f*x)/2] + 30*c^4*d^
2*Sin[(f*x)/2] - 440*c^3*d^3*Sin[(f*x)/2] + 855*c^2*d^4*Sin[(f*x)/2] - 642*c*d^5*Sin[(f*x)/2] + 172*d^6*Sin[(f
*x)/2]))/(15*f*(d + c*Cos[e + f*x])^6*(a + a*Sec[e + f*x])^3) + (8*d^6*Cos[e/2 + (f*x)/2]^6*Sec[e]*(c + d*Sec[
e + f*x])^6*Sin[f*x])/(3*f*(d + c*Cos[e + f*x])^6*(a + a*Sec[e + f*x])^3) - (4*Cos[e/2 + (f*x)/2]^6*Cos[e + f*
x]^2*Sec[e]*(c + d*Sec[e + f*x])^6*(-18*c*d^5*Sin[e] + 9*d^6*Sin[e] - 90*c^2*d^4*Sin[f*x] + 108*c*d^5*Sin[f*x]
- 40*d^6*Sin[f*x]))/(3*f*(d + c*Cos[e + f*x])^6*(a + a*Sec[e + f*x])^3) + (4*Cos[e/2 + (f*x)/2]^6*Cos[e + f*x
]*Sec[e]*(c + d*Sec[e + f*x])^6*(2*d^6*Sin[e] + 18*c*d^5*Sin[f*x] - 9*d^6*Sin[f*x]))/(3*f*(d + c*Cos[e + f*x])
^6*(a + a*Sec[e + f*x])^3)

________________________________________________________________________________________

Maple [B]  time = 0.119, size = 956, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))^6/(a+a*sec(f*x+e))^3,x)

[Out]

1/20/f*c^6/a^3*tan(1/2*f*x+1/2*e)^5-1/6/f*c^6/a^3*tan(1/2*f*x+1/2*e)^3+1/4/f*c^6/a^3*tan(1/2*f*x+1/2*e)-3/10/f
/a^3*tan(1/2*f*x+1/2*e)^5*c*d^5+5/2/f/a^3*tan(1/2*f*x+1/2*e)^3*c^4*d^2+15/2/f/a^3*tan(1/2*f*x+1/2*e)^3*c^2*d^4
+3/2/f/a^3*tan(1/2*f*x+1/2*e)*c^5*d+15/4/f/a^3*tan(1/2*f*x+1/2*e)*c^4*d^2+255/4/f/a^3*tan(1/2*f*x+1/2*e)*c^2*d
^4-93/2/f/a^3*tan(1/2*f*x+1/2*e)*c*d^5+3/f/a^3*d^5/(tan(1/2*f*x+1/2*e)-1)^2*c-23/2/f/a^3*ln(tan(1/2*f*x+1/2*e)
+1)*d^6+1/20/f/a^3*tan(1/2*f*x+1/2*e)^5*d^6+5/6/f/a^3*tan(1/2*f*x+1/2*e)^3*d^6+49/4/f/a^3*tan(1/2*f*x+1/2*e)*d
^6-1/3/f/a^3*d^6/(tan(1/2*f*x+1/2*e)+1)^3-1/3/f/a^3*d^6/(tan(1/2*f*x+1/2*e)-1)^3-17/2/f/a^3*d^6/(tan(1/2*f*x+1
/2*e)+1)+2/f/a^3*d^6/(tan(1/2*f*x+1/2*e)+1)^2-17/2/f/a^3*d^6/(tan(1/2*f*x+1/2*e)-1)-2/f/a^3*d^6/(tan(1/2*f*x+1
/2*e)-1)^2+23/2/f/a^3*ln(tan(1/2*f*x+1/2*e)-1)*d^6-15/f/a^3*d^4/(tan(1/2*f*x+1/2*e)+1)*c^2+21/f/a^3*d^5/(tan(1
/2*f*x+1/2*e)+1)*c-3/f/a^3*d^5/(tan(1/2*f*x+1/2*e)+1)^2*c-1/f/a^3*tan(1/2*f*x+1/2*e)^5*c^3*d^3-20/3/f/a^3*tan(
1/2*f*x+1/2*e)^3*c^3*d^3-4/f/a^3*tan(1/2*f*x+1/2*e)^3*c*d^5-35/f/a^3*c^3*d^3*tan(1/2*f*x+1/2*e)-15/f/a^3*d^4/(
tan(1/2*f*x+1/2*e)-1)*c^2+21/f/a^3*d^5/(tan(1/2*f*x+1/2*e)-1)*c+20/f/a^3*ln(tan(1/2*f*x+1/2*e)+1)*c^3*d^3-45/f
/a^3*ln(tan(1/2*f*x+1/2*e)+1)*c^2*d^4+39/f/a^3*ln(tan(1/2*f*x+1/2*e)+1)*c*d^5-20/f/a^3*ln(tan(1/2*f*x+1/2*e)-1
)*c^3*d^3+45/f/a^3*ln(tan(1/2*f*x+1/2*e)-1)*c^2*d^4-39/f/a^3*ln(tan(1/2*f*x+1/2*e)-1)*c*d^5-3/10/f/a^3*tan(1/2
*f*x+1/2*e)^5*c^5*d+3/4/f/a^3*tan(1/2*f*x+1/2*e)^5*c^4*d^2+3/4/f/a^3*tan(1/2*f*x+1/2*e)^5*c^2*d^4

________________________________________________________________________________________

Maxima [B]  time = 1.12129, size = 1277, normalized size = 3.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^6/(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/60*(d^6*(20*(33*sin(f*x + e)/(cos(f*x + e) + 1) - 76*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 51*sin(f*x + e)^5
/(cos(f*x + e) + 1)^5)/(a^3 - 3*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*a^3*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 - a^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6) + (735*sin(f*x + e)/(cos(f*x + e) + 1) + 50*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 690*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)
/a^3 + 690*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^3) - 6*c*d^5*(60*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 7*
sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^3 - 2*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^3*sin(f*x + e)^4/(co
s(f*x + e) + 1)^4) + (465*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x
+ e)^5/(cos(f*x + e) + 1)^5)/a^3 - 390*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^3 + 390*log(sin(f*x + e)/(c
os(f*x + e) + 1) - 1)/a^3) + 45*c^2*d^4*(40*sin(f*x + e)/((a^3 - a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos
(f*x + e) + 1)) + (85*sin(f*x + e)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + sin(f*x + e)^
5/(cos(f*x + e) + 1)^5)/a^3 - 60*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^3 + 60*log(sin(f*x + e)/(cos(f*x +
e) + 1) - 1)/a^3) - 20*c^3*d^3*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 60*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^3 + 60*log(sin(f
*x + e)/(cos(f*x + e) + 1) - 1)/a^3) + 15*c^4*d^2*(15*sin(f*x + e)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^3/(cos
(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 + c^6*(15*sin(f*x + e)/(cos(f*x + e) + 1) - 10*s
in(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 + 18*c^5*d*(5*sin(f*x + e)/(co
s(f*x + e) + 1) - sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3)/f

________________________________________________________________________________________

Fricas [A]  time = 0.629447, size = 1470, normalized size = 4.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^6/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/60*(15*((40*c^3*d^3 - 90*c^2*d^4 + 78*c*d^5 - 23*d^6)*cos(f*x + e)^6 + 3*(40*c^3*d^3 - 90*c^2*d^4 + 78*c*d^5
- 23*d^6)*cos(f*x + e)^5 + 3*(40*c^3*d^3 - 90*c^2*d^4 + 78*c*d^5 - 23*d^6)*cos(f*x + e)^4 + (40*c^3*d^3 - 90*
c^2*d^4 + 78*c*d^5 - 23*d^6)*cos(f*x + e)^3)*log(sin(f*x + e) + 1) - 15*((40*c^3*d^3 - 90*c^2*d^4 + 78*c*d^5 -
23*d^6)*cos(f*x + e)^6 + 3*(40*c^3*d^3 - 90*c^2*d^4 + 78*c*d^5 - 23*d^6)*cos(f*x + e)^5 + 3*(40*c^3*d^3 - 90*
c^2*d^4 + 78*c*d^5 - 23*d^6)*cos(f*x + e)^4 + (40*c^3*d^3 - 90*c^2*d^4 + 78*c*d^5 - 23*d^6)*cos(f*x + e)^3)*lo
g(-sin(f*x + e) + 1) + 2*(10*d^6 + 2*(7*c^6 + 18*c^5*d + 30*c^4*d^2 - 440*c^3*d^3 + 1080*c^2*d^4 - 912*c*d^5 +
272*d^6)*cos(f*x + e)^5 + 3*(4*c^6 + 36*c^5*d + 60*c^4*d^2 - 680*c^3*d^3 + 1710*c^2*d^4 - 1434*c*d^5 + 429*d^
6)*cos(f*x + e)^4 + (4*c^6 + 36*c^5*d + 210*c^4*d^2 - 1280*c^3*d^3 + 3510*c^2*d^4 - 2874*c*d^5 + 869*d^6)*cos(
f*x + e)^3 + 5*(90*c^2*d^4 - 54*c*d^5 + 19*d^6)*cos(f*x + e)^2 + 15*(6*c*d^5 - d^6)*cos(f*x + e))*sin(f*x + e)
)/(a^3*f*cos(f*x + e)^6 + 3*a^3*f*cos(f*x + e)^5 + 3*a^3*f*cos(f*x + e)^4 + a^3*f*cos(f*x + e)^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{6} \sec{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{6} \sec ^{7}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{6 c d^{5} \sec ^{6}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{15 c^{2} d^{4} \sec ^{5}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{20 c^{3} d^{3} \sec ^{4}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{15 c^{4} d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{6 c^{5} d \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))**6/(a+a*sec(f*x+e))**3,x)

[Out]

(Integral(c**6*sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(d**6*sec
(e + f*x)**7/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(6*c*d**5*sec(e + f*x)**
6/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(15*c**2*d**4*sec(e + f*x)**5/(sec(
e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(20*c**3*d**3*sec(e + f*x)**4/(sec(e + f*x
)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(15*c**4*d**2*sec(e + f*x)**3/(sec(e + f*x)**3 +
3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(6*c**5*d*sec(e + f*x)**2/(sec(e + f*x)**3 + 3*sec(e + f
*x)**2 + 3*sec(e + f*x) + 1), x))/a**3

________________________________________________________________________________________

Giac [B]  time = 1.47798, size = 950, normalized size = 2.62 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^6/(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/60*(30*(40*c^3*d^3 - 90*c^2*d^4 + 78*c*d^5 - 23*d^6)*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^3 - 30*(40*c^3*d^3
- 90*c^2*d^4 + 78*c*d^5 - 23*d^6)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^3 - 20*(90*c^2*d^4*tan(1/2*f*x + 1/2*e
)^5 - 126*c*d^5*tan(1/2*f*x + 1/2*e)^5 + 51*d^6*tan(1/2*f*x + 1/2*e)^5 - 180*c^2*d^4*tan(1/2*f*x + 1/2*e)^3 +
216*c*d^5*tan(1/2*f*x + 1/2*e)^3 - 76*d^6*tan(1/2*f*x + 1/2*e)^3 + 90*c^2*d^4*tan(1/2*f*x + 1/2*e) - 90*c*d^5*
tan(1/2*f*x + 1/2*e) + 33*d^6*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*a^3) + (3*a^12*c^6*tan(1/2
*f*x + 1/2*e)^5 - 18*a^12*c^5*d*tan(1/2*f*x + 1/2*e)^5 + 45*a^12*c^4*d^2*tan(1/2*f*x + 1/2*e)^5 - 60*a^12*c^3*
d^3*tan(1/2*f*x + 1/2*e)^5 + 45*a^12*c^2*d^4*tan(1/2*f*x + 1/2*e)^5 - 18*a^12*c*d^5*tan(1/2*f*x + 1/2*e)^5 + 3
*a^12*d^6*tan(1/2*f*x + 1/2*e)^5 - 10*a^12*c^6*tan(1/2*f*x + 1/2*e)^3 + 150*a^12*c^4*d^2*tan(1/2*f*x + 1/2*e)^
3 - 400*a^12*c^3*d^3*tan(1/2*f*x + 1/2*e)^3 + 450*a^12*c^2*d^4*tan(1/2*f*x + 1/2*e)^3 - 240*a^12*c*d^5*tan(1/2
*f*x + 1/2*e)^3 + 50*a^12*d^6*tan(1/2*f*x + 1/2*e)^3 + 15*a^12*c^6*tan(1/2*f*x + 1/2*e) + 90*a^12*c^5*d*tan(1/
2*f*x + 1/2*e) + 225*a^12*c^4*d^2*tan(1/2*f*x + 1/2*e) - 2100*a^12*c^3*d^3*tan(1/2*f*x + 1/2*e) + 3825*a^12*c^
2*d^4*tan(1/2*f*x + 1/2*e) - 2790*a^12*c*d^5*tan(1/2*f*x + 1/2*e) + 735*a^12*d^6*tan(1/2*f*x + 1/2*e))/a^15)/f