\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx\) [1331]
Optimal result
Integrand size = 43, antiderivative size = 423 \[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx=-\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 b \left (a^2-b^2\right )^2 d}+\frac {\left (3 A b^4-7 a^3 b B+a b^3 B-a^2 b^2 (5 A-3 C)+a^4 (8 A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (3 A b^6-3 a^5 b B-10 a^3 b^3 B+a b^5 B-3 a^2 b^4 (2 A-C)-a^6 C+5 a^4 b^2 (3 A+2 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^3 (a-b)^2 b (a+b)^3 d}+\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}
\]
[Out]
-1/4*(3*A*b^4+5*B*a^3*b+B*a*b^3-a^4*C-a^2*b^2*(9*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip
ticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/b/(a^2-b^2)^2/d+1/4*(3*A*b^4-7*B*a^3*b+B*a*b^3-a^2*b^2*(5*A-3*C)+a^4*(8*A
+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/(a^2-b^2)^2/d
-1/4*(3*A*b^6-3*a^5*b*B-10*a^3*b^3*B+a*b^5*B-3*a^2*b^4*(2*A-C)-a^6*C+5*a^4*b^2*(3*A+2*C))*(cos(1/2*d*x+1/2*c)^
2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^3/(a-b)^2/b/(a+b)^3/d+1/2*(A*b^
2-a*(B*b-C*a))*sin(d*x+c)*cos(d*x+c)^(1/2)/a/(a^2-b^2)/d/(b+a*cos(d*x+c))^2+1/4*(3*A*b^4+5*B*a^3*b+B*a*b^3-a^4
*C-a^2*b^2*(9*A+5*C))*sin(d*x+c)*cos(d*x+c)^(1/2)/a/b/(a^2-b^2)^2/d/(b+a*cos(d*x+c))
Rubi [A] (verified)
Time = 1.49 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.00, number of
steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4197, 3126, 3134, 3138, 2719,
3081, 2720, 2884} \[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx=\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^4 (8 A+3 C)-7 a^3 b B-a^2 b^2 (5 A-3 C)+a b^3 B+3 A b^4\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{4 a^2 b d \left (a^2-b^2\right )^2}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{4 a b d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac {\left (a^6 (-C)-3 a^5 b B+5 a^4 b^2 (3 A+2 C)-10 a^3 b^3 B-3 a^2 b^4 (2 A-C)+a b^5 B+3 A b^6\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^3 b d (a-b)^2 (a+b)^3}
\]
[In]
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^3),x]
[Out]
-1/4*((3*A*b^4 + 5*a^3*b*B + a*b^3*B - a^4*C - a^2*b^2*(9*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(a^2*b*(a^2 - b
^2)^2*d) + ((3*A*b^4 - 7*a^3*b*B + a*b^3*B - a^2*b^2*(5*A - 3*C) + a^4*(8*A + 3*C))*EllipticF[(c + d*x)/2, 2])
/(4*a^3*(a^2 - b^2)^2*d) - ((3*A*b^6 - 3*a^5*b*B - 10*a^3*b^3*B + a*b^5*B - 3*a^2*b^4*(2*A - C) - a^6*C + 5*a^
4*b^2*(3*A + 2*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^3*(a - b)^2*b*(a + b)^3*d) + ((A*b^2 - a*(b
*B - a*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + ((3*A*b^4 + 5*a^3*b*B
+ a*b^3*B - a^4*C - a^2*b^2*(9*A + 5*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*a*b*(a^2 - b^2)^2*d*(b + a*Cos[c
+ d*x]))
Rule 2719
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]
Rule 2720
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]
Rule 2884
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Rule 3081
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3126
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) +
(c*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*
c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n +
1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Rule 3134
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&& !IntegerQ[m]) || EqQ[a, 0])))
Rule 3138
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 4197
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] && !IntegerQ[n] && IntegerQ[m]
Rubi steps \begin{align*}
\text {integral}& = \int \frac {\sqrt {\cos (c+d x)} \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{(b+a \cos (c+d x))^3} \, dx \\ & = \frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} \left (A b^2-a (b B-a C)\right )-2 a (A b-a B+b C) \cos (c+d x)-\frac {1}{2} \left (3 A b^2+a b B-a^2 (4 A+C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )} \\ & = \frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {\int \frac {\frac {1}{4} \left (-A b^4-3 a^3 b B-3 a b^3 B-a^4 C+7 a^2 b^2 (A+C)\right )+a b \left (3 a b B-a^2 (2 A+C)-b^2 (A+2 C)\right ) \cos (c+d x)+\frac {1}{4} \left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a b \left (a^2-b^2\right )^2} \\ & = \frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} a \left (A b^4+3 a^3 b B+3 a b^3 B+a^4 C-7 a^2 b^2 (A+C)\right )+\frac {1}{4} b \left (3 A b^4-7 a^3 b B+a b^3 B-a^2 b^2 (5 A-3 C)+a^4 (8 A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^2 b \left (a^2-b^2\right )^2}-\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^2 b \left (a^2-b^2\right )^2} \\ & = -\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 b \left (a^2-b^2\right )^2 d}+\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {\left (3 A b^6-3 a^5 b B-10 a^3 b^3 B+a b^5 B-3 a^2 b^4 (2 A-C)-a^6 C+5 a^4 b^2 (3 A+2 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^3 b \left (a^2-b^2\right )^2}+\frac {\left (3 A b^4-7 a^3 b B+a b^3 B-a^2 b^2 (5 A-3 C)+a^4 (8 A+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^3 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 b \left (a^2-b^2\right )^2 d}+\frac {\left (3 A b^4-7 a^3 b B+a b^3 B-a^2 b^2 (5 A-3 C)+a^4 (8 A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (3 A b^6-3 a^5 b B-10 a^3 b^3 B+a b^5 B-3 a^2 b^4 (2 A-C)-a^6 C+5 a^4 b^2 (3 A+2 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^3 (a-b)^2 b (a+b)^3 d}+\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \\
\end{align*}
Mathematica [A] (verified)
Time = 9.31 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.00
\[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx=\frac {-\frac {2 \sqrt {\cos (c+d x)} \left (-b \left (A b^4+3 a^3 b B+3 a b^3 B+a^4 C-7 a^2 b^2 (A+C)\right )+a \left (-3 A b^4-5 a^3 b B-a b^3 B+a^4 C+a^2 b^2 (9 A+5 C)\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+\frac {\frac {\left (-A b^4+a^3 b B+5 a b^3 B+3 a^4 C-a^2 b^2 (5 A+9 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 b \left (-3 a b B+a^2 (2 A+C)+b^2 (A+2 C)\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {\left (-3 A b^4-5 a^3 b B-a b^3 B+a^4 C+a^2 b^2 (9 A+5 C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 a b d}
\]
[In]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^3),x]
[Out]
((-2*Sqrt[Cos[c + d*x]]*(-(b*(A*b^4 + 3*a^3*b*B + 3*a*b^3*B + a^4*C - 7*a^2*b^2*(A + C))) + a*(-3*A*b^4 - 5*a^
3*b*B - a*b^3*B + a^4*C + a^2*b^2*(9*A + 5*C))*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(b + a*Cos[c + d*x])
^2) + (((-(A*b^4) + a^3*b*B + 5*a*b^3*B + 3*a^4*C - a^2*b^2*(5*A + 9*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2
, 2])/(a + b) + (8*b*(-3*a*b*B + a^2*(2*A + C) + b^2*(A + 2*C))*((a + b)*EllipticF[(c + d*x)/2, 2] - b*Ellipti
cPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) + ((-3*A*b^4 - 5*a^3*b*B - a*b^3*B + a^4*C + a^2*b^2*(9*A + 5*C))
*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (
a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/(
(a - b)^2*(a + b)^2))/(8*a*b*d)
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(1971\) vs. \(2(487)=974\).
Time = 4.38 (sec) , antiderivative size = 1972, normalized size of antiderivative =
4.66
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method | result | size |
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default |
\(\text {Expression too large to display}\) |
\(1972\) |
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[In]
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)-2*(-3*A*b+B*a)/a^2/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2/a^3*(3*A*b^2-2*B*a*b+C
*a^2)*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+
1/2*c)^2*a-a+b)-1/2/(a+b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2
*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(
1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b
)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(
1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))-2*b*(A*b^2-B*a*b+C*a^2)/a^3*(1/2*a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)^2+3/4*a^2*(a^2-3*b^2)/b^2/(a^2-b^
2)^2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-3/
8/(a+b)/(a^2-b^2)/b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+
sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-1/4/(a+b)/(a^2-b^2)/b*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))*a+7/8/(a+b)/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)
/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/8*a^3/b^2/(a^2-b
^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9/8*a/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)-3/8*a^3/b^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c
)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+9/8*a/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/
2*d*x+1/2*c),2^(1/2))-3/8/(a-b)/(a+b)/(a^2-b^2)/b^2/(a^2-a*b)*a^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x
+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b
),2^(1/2))+3/4/(a-b)/(a+b)/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1
/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-15/8
/(a-b)/(a+b)/(a^2-b^2)*b^2/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))))/sin(1/2*d*x+1/
2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Fricas [F(-1)]
Timed out. \[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
Timed out
Sympy [F(-1)]
Timed out. \[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3/cos(d*x+c)**(1/2),x)
[Out]
Timed out
Maxima [F(-1)]
Timed out. \[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
Timed out
Giac [F]
\[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3} \sqrt {\cos \left (d x + c\right )}} \,d x }
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^3*sqrt(cos(d*x + c))), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x
\]
[In]
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + b/cos(c + d*x))^3),x)
[Out]
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + b/cos(c + d*x))^3), x)