3.1113 \(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\)
Optimal. Leaf size=192 \[ -\frac {(3 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {3 (5 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {(3 A+5 C) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(5 A+7 C) \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}+\frac {3 (5 A+7 C) \sin (c+d x)}{5 a d \sqrt {\cos (c+d x)}} \]
[Out]
-3/5*(5*A+7*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))/a/d-1/3*(
3*A+5*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/d+1/5*(5*A+7*
C)*sin(d*x+c)/a/d/cos(d*x+c)^(5/2)-1/3*(3*A+5*C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)-(A+C)*sin(d*x+c)/d/cos(d*x+c)
^(5/2)/(a+a*cos(d*x+c))+3/5*(5*A+7*C)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)
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Rubi [A] time = 0.27, antiderivative size = 192, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used =
{4114, 3042, 2748, 2636, 2639, 2641} \[ -\frac {(3 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {3 (5 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}-\frac {(3 A+5 C) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {(5 A+7 C) \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}+\frac {3 (5 A+7 C) \sin (c+d x)}{5 a d \sqrt {\cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x]
[Out]
(-3*(5*A + 7*C)*EllipticE[(c + d*x)/2, 2])/(5*a*d) - ((3*A + 5*C)*EllipticF[(c + d*x)/2, 2])/(3*a*d) + ((5*A +
7*C)*Sin[c + d*x])/(5*a*d*Cos[c + d*x]^(5/2)) - ((3*A + 5*C)*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)) + (3*(5
*A + 7*C)*Sin[c + d*x])/(5*a*d*Sqrt[Cos[c + d*x]]) - ((A + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(5/2)*(a + a*Cos[c
+ d*x]))
Rule 2636
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 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 2748
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 3042
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]
Rule 4114
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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 + A
*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Rubi steps
\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx &=\int \frac {C+A \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))} \, dx\\ &=-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} a (5 A+7 C)-\frac {1}{2} a (3 A+5 C) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))}-\frac {(3 A+5 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{2 a}+\frac {(5 A+7 C) \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{2 a}\\ &=\frac {(5 A+7 C) \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {(3 A+5 C) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))}-\frac {(3 A+5 C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}+\frac {(3 (5 A+7 C)) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{10 a}\\ &=-\frac {(3 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {(5 A+7 C) \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {(3 A+5 C) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {3 (5 A+7 C) \sin (c+d x)}{5 a d \sqrt {\cos (c+d x)}}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))}-\frac {(3 (5 A+7 C)) \int \sqrt {\cos (c+d x)} \, dx}{10 a}\\ &=-\frac {3 (5 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {(3 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {(5 A+7 C) \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {(3 A+5 C) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {3 (5 A+7 C) \sin (c+d x)}{5 a d \sqrt {\cos (c+d x)}}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))}\\ \end {align*}
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Mathematica [C] time = 7.41, size = 1382, normalized size = 7.20 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x]
[Out]
(((-3*I)/2)*A*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hyp
ergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*
I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)
*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((
2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I
*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1
+ E^((2*I)*d*x))*Sin[c])))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) - (((21*I)/10)*C*Cos[c/2 + (
d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4,
7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*S
in[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Co
s[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin
[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*
d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])
))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]^(3/2)*(A + C*Sec
[c + d*x]^2)*((2*(10*A + 16*C + 5*A*Cos[c] + 5*C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/(5*d) + (4*Sec[c/2]*Sec[c/2
+ (d*x)/2]*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d + (8*C*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(5*d) - (8*Sec[c]*Sec[
c + d*x]*(5*C*Sin[c] - 15*A*Sin[d*x] - 24*C*Sin[d*x]))/(15*d) + (8*Sec[c]*Sec[c + d*x]^2*(3*C*Sin[c] - 5*C*Sin
[d*x]))/(15*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (2*A*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x
]*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*S
ec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan
[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*S
ec[c + d*x])) + (10*C*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x
- ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c
]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A
+ 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x]))
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fricas [F] time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {\cos \left (d x + c\right )}}{a \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + a \cos \left (d x + c\right )^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")
[Out]
integral((C*sec(d*x + c)^2 + A)*sqrt(cos(d*x + c))/(a*cos(d*x + c)^3*sec(d*x + c) + a*cos(d*x + c)^3), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + A)/((a*sec(d*x + c) + a)*cos(d*x + c)^(5/2)), x)
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maple [B] time = 17.34, size = 803, normalized size = 4.18 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*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)/a*(-2/5*C/(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*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)^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*(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))-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+2*C)*(-(-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)-2*C*(-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)))+(-A-C)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1
/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(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)/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))/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+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))),x)
[Out]
int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/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+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+a*sec(d*x+c)),x)
[Out]
Timed out
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