3.1323 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\)

Optimal. Leaf size=236 \[ -\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}+\frac {2 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d \sqrt {\cos (c+d x)}}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}+\frac {2 (b B-a C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 d}+\frac {2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

-2/5*(5*A*b^2-5*B*a*b+5*C*a^2+3*C*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1
/2*c),2^(1/2))/b^3/d+2/3*(B*b-C*a)*(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))/b^2/d-2*a*(A*b^2-a*(B*b-C*a))*(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))/b^3/(a+b)/d+2/5*C*sin(d*x+c)/b/d/cos(d*x+c)^(5/2)+2/3*(B*b-C*a)*sin(d*x+c)/b^2/d/c
os(d*x+c)^(3/2)+2/5*(5*A*b^2-5*B*a*b+5*C*a^2+3*C*b^2)*sin(d*x+c)/b^3/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 1.28, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4112, 3055, 3059, 2639, 3002, 2641, 2805} \[ -\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}+\frac {2 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d \sqrt {\cos (c+d x)}}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}+\frac {2 (b B-a C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 d}+\frac {2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])),x]

[Out]

(-2*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*C)*EllipticE[(c + d*x)/2, 2])/(5*b^3*d) + (2*(b*B - a*C)*EllipticF[(c
+ d*x)/2, 2])/(3*b^2*d) - (2*a*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(b^3*(a + b
)*d) + (2*C*Sin[c + d*x])/(5*b*d*Cos[c + d*x]^(5/2)) + (2*(b*B - a*C)*Sin[c + d*x])/(3*b^2*d*Cos[c + d*x]^(3/2
)) + (2*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*C)*Sin[c + d*x])/(5*b^3*d*Sqrt[Cos[c + d*x]])

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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 3002

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 3055

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] + Dis
t[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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

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 4112

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*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx &=\int \frac {C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (b+a \cos (c+d x))} \, dx\\ &=\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {\frac {5}{2} (b B-a C)+\frac {1}{2} b (5 A+3 C) \cos (c+d x)+\frac {3}{2} a C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{5 b}\\ &=\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\frac {3}{4} \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )+\frac {1}{4} b (5 b B+4 a C) \cos (c+d x)+\frac {5}{4} a (b B-a C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{15 b^2}\\ &=\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 b^3 d \sqrt {\cos (c+d x)}}+\frac {8 \int \frac {\frac {5}{8} \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+C)\right )-\frac {1}{8} b \left (15 A b^2-20 a b B+20 a^2 C+9 b^2 C\right ) \cos (c+d x)-\frac {3}{8} a \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{15 b^3}\\ &=\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 b^3 d \sqrt {\cos (c+d x)}}-\frac {8 \int \frac {-\frac {5}{8} a \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+C)\right )-\frac {5}{8} a^2 b (b B-a C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{15 a b^3}-\frac {\left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^3}\\ &=-\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 b^3 d \sqrt {\cos (c+d x)}}+\frac {(b B-a C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 b^2}-\frac {\left (a \left (A b^2-a (b B-a C)\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b^3}\\ &=-\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac {2 (b B-a C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 (a+b) d}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 b^3 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 4.29, size = 332, normalized size = 1.41 \[ -\frac {\frac {2 b \left (20 a^2 C-20 a b B+15 A b^2+9 b^2 C\right ) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}-\frac {2 \left (3 \sin (2 (c+d x)) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )+10 b (b B-a C) \sin (c+d x)+6 b^2 C \tan (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \left (\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt {\sin ^2(c+d x)}}+\frac {2 \left (45 a^3 C-45 a^2 b B+a b^2 (45 A+19 C)-10 b^3 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}}{30 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])),x]

[Out]

-1/30*((2*(-45*a^2*b*B - 10*b^3*B + 45*a^3*C + a*b^2*(45*A + 19*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])
/(a + b) + (2*b*(15*A*b^2 - 20*a*b*B + 20*a^2*C + 9*b^2*C)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a
)/(a + b), (c + d*x)/2, 2])/(a + b)))/a + (6*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*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*b*Sqrt[Sin[c + d*x]^2]) - (2*(10*b*(b*B - a*C)*Sin[c +
d*x] + 3*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*C)*Sin[2*(c + d*x)] + 6*b^2*C*Tan[c + d*x]))/Cos[c + d*x]^(3/2)
)/(b^3*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)*cos(d*x + c)^(5/2)), x)

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maple [B]  time = 19.77, size = 800, normalized size = 3.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c)),x)

[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*b^2-B*a*b+C*a^2)*a^2/b^3/(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)*Elli
pticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2*(B*b-C*a)/b^2*(-1/6*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)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(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/
5*C/b/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2
*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c
)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-8*sin(1/2*d*
x+1/2*c)^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*(A*b^2-B*a*b+C*a^2)/b^3*
(-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)
^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*
x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*
d*x+1/2*c)^2-1)^(1/2)/d

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+b*sec(d*x+c)),x)

[Out]

Timed out

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