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

Optimal. Leaf size=214 \[ -\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}+\frac {(7 A-5 B+5 C) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(5 A-5 B+3 C) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {(5 A-5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d} \]

[Out]

1/5*(7*A-5*B+5*C)*sin(d*x+c)/a/d/sec(d*x+c)^(3/2)-(A-B+C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))-1/3*(
5*A-5*B+3*C)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)+3/5*(7*A-5*B+5*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d-1/3*(5*A-5*B+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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d

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Rubi [A]  time = 0.25, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4084, 3787, 3769, 3771, 2639, 2641} \[ -\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}+\frac {(7 A-5 B+5 C) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(5 A-5 B+3 C) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {(5 A-5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}+\frac {3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(7*A - 5*B + 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a*d) - ((5*A - 5*B +
3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) + ((7*A - 5*B + 5*C)*Sin[c + d*x
])/(5*a*d*Sec[c + d*x]^(3/2)) - ((5*A - 5*B + 3*C)*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]) - ((A - B + C)*Sin
[c + d*x])/(d*Sec[c + d*x]^(3/2)*(a + a*Sec[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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx &=-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}+\frac {\int \frac {\frac {1}{2} a (7 A-5 B+5 C)-\frac {1}{2} a (5 A-5 B+3 C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{a^2}\\ &=-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {(5 A-5 B+3 C) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{2 a}+\frac {(7 A-5 B+5 C) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{2 a}\\ &=\frac {(7 A-5 B+5 C) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(5 A-5 B+3 C) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {(5 A-5 B+3 C) \int \sqrt {\sec (c+d x)} \, dx}{6 a}+\frac {(3 (7 A-5 B+5 C)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{10 a}\\ &=\frac {(7 A-5 B+5 C) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(5 A-5 B+3 C) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac {\left ((5 A-5 B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}+\frac {\left (3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{10 a}\\ &=\frac {3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(5 A-5 B+3 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {(7 A-5 B+5 C) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(5 A-5 B+3 C) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 6.79, size = 1350, normalized size = 6.31 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-7*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]
^2*Cos[c + d*x]*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F
1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(5*d*E^(I*d*x)*(A +
2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((
2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*(-3*Sqrt[1 + E^((2*
I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[
c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(
a + a*Sec[c + d*x])) - (Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))
]*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*
c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/
(d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) - (10*A*Cos[c/2 + (d*x)/2
]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin
[c])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])) + (10*B*C
os[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec
[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + a*Sec[c +
d*x])) - (2*C*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*(A + B*Sec[c
+ d*x] + C*Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a
+ a*Sec[c + d*x])) + (Cos[c/2 + (d*x)/2]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-1/10*((51*A - 40*B + 40*
C + 33*A*Cos[2*c] - 20*B*Cos[2*c] + 20*C*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2])/d - (4*(A - B)*Cos[2*d*x]*Sin[2
*c])/(3*d) + (2*A*Cos[3*d*x]*Sin[3*c])/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]
+ C*Sin[(d*x)/2]))/d + (2*(33*A - 20*B + 20*C)*Cos[c]*Sin[d*x])/(5*d) - (4*(A - B)*Cos[2*c]*Sin[2*d*x])/(3*d)
+ (2*A*Cos[3*c]*Sin[3*d*x])/(5*d) + (4*(A - B + C)*Tan[c/2])/d))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2
*d*x])*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x]))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{a \sec \left (d x + c\right )^{4} + a \sec \left (d x + c\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(a*sec(d*x + c)^4 + a*sec(d*x + c)^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 5.04, size = 320, normalized size = 1.50 \[ -\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (25 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+63 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-25 B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-45 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+15 C \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+45 C \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+48 A \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-56 A -40 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-30 A +90 B -30 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (23 A -35 B +15 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{15 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-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)*(25*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+63*A*EllipticE(cos(1/2*d*x+1/2*
c),2^(1/2))-25*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-45*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*C*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))+45*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+48*A*sin(1/2*d*x+1/2*c)^8+(-56*A-40
*B)*sin(1/2*d*x+1/2*c)^6+(-30*A+90*B-30*C)*sin(1/2*d*x+1/2*c)^4+(23*A-35*B+15*C)*sin(1/2*d*x+1/2*c)^2)/a/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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))*(1/cos(c + d*x))^(5/2)),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))*(1/cos(c + d*x))^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A/(sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x) + Integral(B*sec(c + d*x)/(sec(c + d*x)**(7/2) + s
ec(c + d*x)**(5/2)), x) + Integral(C*sec(c + d*x)**2/(sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x))/a

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