\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [1185]
Optimal result
Integrand size = 31, antiderivative size = 123 \[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 B \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}
\]
[Out]
-2/5*(5*A+3*C)*(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))/d+2/3*B*(
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))/d+2/5*C*sin(d*x+c)/d/cos(
d*x+c)^(5/2)+2/3*B*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*(5*A+3*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
Rubi [A] (verified)
Time = 0.18 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4149, 3100, 2827, 2716, 2720,
2719} \[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 B \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}
\]
[In]
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/Cos[c + d*x]^(3/2),x]
[Out]
(-2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*B*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*C*Sin[c + d*x])/
(5*d*Cos[c + d*x]^(5/2)) + (2*B*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*(5*A + 3*C)*Sin[c + d*x])/(5*d*Sqr
t[Cos[c + d*x]])
Rule 2716
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]
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 2827
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 3100
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(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)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[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))*Sin[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 4149
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)
]^2), x_Symbol] :> Dist[b^2, Int[(b*Cos[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; F
reeQ[{b, e, f, A, B, C, m}, x] && !IntegerQ[m]
Rubi steps \begin{align*}
\text {integral}& = \int \frac {C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {\frac {5 B}{2}+\frac {1}{2} (5 A+3 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+B \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx+\frac {1}{5} (5 A+3 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 B \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {1}{3} B \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} (-5 A-3 C) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 B \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 A+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.57 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.91
\[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-6 (5 A+3 C) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 B \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+10 B \sin (c+d x)+15 A \sin (2 (c+d x))+9 C \sin (2 (c+d x))+6 C \tan (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}
\]
[In]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/Cos[c + d*x]^(3/2),x]
[Out]
(-6*(5*A + 3*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*B*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2,
2] + 10*B*Sin[c + d*x] + 15*A*Sin[2*(c + d*x)] + 9*C*Sin[2*(c + d*x)] + 6*C*Tan[c + d*x])/(15*d*Cos[c + d*x]^(
3/2))
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(798\) vs. \(2(159)=318\).
Time = 3.60 (sec) , antiderivative size = 799, normalized size of antiderivative = 6.50
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\(799\) |
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[In]
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
[Out]
-2/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(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)^3*(120*A*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-60*A*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^
4-20*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(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)^4+72*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-36*C*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-120*A*sin(1/2*d*x+1/2*c)^4*cos(
1/2*d*x+1/2*c)+60*A*(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))*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+20*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
EllipticF(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)^2-72*C*cos(1/2*d*x+1
/2*c)*sin(1/2*d*x+1/2*c)^4+36*C*(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))*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+30*A*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-15*A*(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))+10*B*sin(1/2*d*x+1/2*c)^2*c
os(1/2*d*x+1/2*c)-5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))+24*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-9*C*(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*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*co
s(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Fricas [C] (verification not implemented)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.67
\[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-5 i \, \sqrt {2} B \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (5 i \, A + 3 i \, C\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (-5 i \, A - 3 i \, C\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, {\left (5 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, B \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{3}}
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
1/15*(-5*I*sqrt(2)*B*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*B*
cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(5*I*A + 3*I*C)*cos(d*x +
c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(-5*I*A -
3*I*C)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3
*(5*A + 3*C)*cos(d*x + c)^2 + 5*B*cos(d*x + c) + 3*C)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
Sympy [F]
\[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(3/2),x)
[Out]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)/cos(c + d*x)**(3/2), x)
Maxima [F]
\[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x }
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/cos(d*x + c)^(3/2), x)
Giac [F]
\[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x }
\]
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/cos(d*x + c)^(3/2), x)
Mupad [B] (verification not implemented)
Time = 18.51 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.88
\[
\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {6\,C\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+10\,B\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}
\]
[In]
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/cos(c + d*x)^(3/2),x)
[Out]
(6*C*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 10*B*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/
4, 1/2], 1/4, cos(c + d*x)^2) + 30*A*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/
(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2))