3.6.14 \(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\) [514]

Optimal. Leaf size=221 \[ \frac {(9 A-49 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A-13 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {(9 A-49 B) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}+\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(3 A-13 B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

1/10*(9*A-49*B)*(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^3/d+1/
6*(3*A-13*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))/a^3/d-1/10*
(9*A-49*B)*sin(d*x+c)/a^3/d/cos(d*x+c)^(1/2)+1/5*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3/cos(d*x+c)^(1/2)+1/15*(
3*A-8*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2/cos(d*x+c)^(1/2)+1/6*(3*A-13*B)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))/c
os(d*x+c)^(1/2)

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Rubi [A]
time = 0.39, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3033, 3057, 2827, 2716, 2719, 2720} \begin {gather*} \frac {(3 A-13 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {(9 A-49 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(9 A-49 B) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}+\frac {(3 A-13 B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(3 A-8 B) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}+\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

((9*A - 49*B)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((3*A - 13*B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) - ((9
*A - 49*B)*Sin[c + d*x])/(10*a^3*d*Sqrt[Cos[c + d*x]]) + ((A - B)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]*(a + a
*Cos[c + d*x])^3) + ((3*A - 8*B)*Sin[c + d*x])/(15*a*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2) + ((3*A - 13
*B)*Sin[c + d*x])/(6*d*Sqrt[Cos[c + d*x]]*(a^3 + a^3*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 3033

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx &=\int \frac {B+A \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx\\ &=\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {\int \frac {-\frac {1}{2} a (A-11 B)+\frac {5}{2} a (A-B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {\int \frac {-\frac {1}{2} a^2 (6 A-41 B)+\frac {3}{2} a^2 (3 A-8 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(3 A-13 B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {-\frac {3}{4} a^3 (9 A-49 B)+\frac {5}{4} a^3 (3 A-13 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(3 A-13 B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(9 A-49 B) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{20 a^3}+\frac {(3 A-13 B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}\\ &=\frac {(3 A-13 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {(9 A-49 B) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}+\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(3 A-13 B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(9 A-49 B) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac {(9 A-49 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(3 A-13 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac {(9 A-49 B) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}+\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(3 A-8 B) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(3 A-13 B) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.17, size = 1447, normalized size = 6.55 \begin {gather*} \frac {9 i A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) (A+B \sec (c+d x)) \left (\frac {2 e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{3 i d \left (1+e^{2 i d x}\right ) \cos (c)-3 d \left (-1+e^{2 i d x}\right ) \sin (c)}-\frac {2 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{-i d \left (1+e^{2 i d x}\right ) \cos (c)+d \left (-1+e^{2 i d x}\right ) \sin (c)}\right )}{10 (B+A \cos (c+d x)) (a+a \sec (c+d x))^3}-\frac {49 i B \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) (A+B \sec (c+d x)) \left (\frac {2 e^{2 i d x} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{3 i d \left (1+e^{2 i d x}\right ) \cos (c)-3 d \left (-1+e^{2 i d x}\right ) \sin (c)}-\frac {2 \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt {e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt {1+e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)}}{-i d \left (1+e^{2 i d x}\right ) \cos (c)+d \left (-1+e^{2 i d x}\right ) \sin (c)}\right )}{10 (B+A \cos (c+d x)) (a+a \sec (c+d x))^3}-\frac {2 A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) (A+B \sec (c+d x)) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))^3}+\frac {26 B \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec ^2(c+d x) (A+B \sec (c+d x)) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d (B+A \cos (c+d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))^3}+\frac {\cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (A+B \sec (c+d x)) \left (\frac {2 (20 B-9 A \cos (c)+29 B \cos (c)) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c)}{5 d}+\frac {2 \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-A \sin \left (\frac {d x}{2}\right )+B \sin \left (\frac {d x}{2}\right )\right )}{5 d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-6 A \sin \left (\frac {d x}{2}\right )+11 B \sin \left (\frac {d x}{2}\right )\right )}{15 d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-9 A \sin \left (\frac {d x}{2}\right )+29 B \sin \left (\frac {d x}{2}\right )\right )}{5 d}+\frac {16 B \sec (c) \sec (c+d x) \sin (d x)}{d}+\frac {4 (-6 A+11 B) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{15 d}+\frac {2 (-A+B) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{5 d}\right )}{\cos ^{\frac {3}{2}}(c+d x) (B+A \cos (c+d x)) (a+a \sec (c+d x))^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

(((9*I)/10)*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((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])))/((B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) - (((49*I)/10)*B*Cos[c/2 + (d*x)/2]^
6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((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))*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])))/((B +
A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) - (2*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5
/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*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]]]])/(d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^3) + (26*B*Cos[c/2 + (d*x)/
2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B
*Sec[c + d*x])*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*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*
(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*(A + B*Sec[c + d*x])*((2*(20*B - 9*A*Cos[c] + 29*B*Cos[c])*Csc
[c/2]*Sec[c/2]*Sec[c])/(5*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(5*d) +
(4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(-6*A*Sin[(d*x)/2] + 11*B*Sin[(d*x)/2]))/(15*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)
/2]*(-9*A*Sin[(d*x)/2] + 29*B*Sin[(d*x)/2]))/(5*d) + (16*B*Sec[c]*Sec[c + d*x]*Sin[d*x])/d + (4*(-6*A + 11*B)*
Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) + (2*(-A + B)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(Cos[c + d*x]^(3/2)
*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(684\) vs. \(2(253)=506\).
time = 3.14, size = 685, normalized size = 3.10

method	result	size
default	\(\frac {-2 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \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 )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (15 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-27 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-65 B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+147 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \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 )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (15 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-27 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-65 B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+147 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \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 )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (15 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-27 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-65 B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+147 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \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 )}\, \left (9 A -49 B \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \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 )}\, \left (147 A -817 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \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 )}\, \left (43 A -248 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )}\, \left (69 A -439 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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}\)	\(685\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/60*(-2*(2*sin(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)*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*(15*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-27*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*B*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*B*EllipticE(cos(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)^4+4*(2*sin(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)*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(15*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-27*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*B*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*B*EllipticE(cos(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*(2*sin(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)*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(15*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-27*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*B*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+12*(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(9*A-49*B)*sin(1/2*d*x+1/2*c)^8-2*(-2*sin(1/2*d*x+1/2*c)^4+sin
(1/2*d*x+1/2*c)^2)^(1/2)*(147*A-817*B)*sin(1/2*d*x+1/2*c)^6+6*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(
1/2)*(43*A-248*B)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(69*A-439*B)*sin(1
/2*d*x+1/2*c)^2)/a^3/cos(1/2*d*x+1/2*c)^5/(-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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.56, size = 521, normalized size = 2.36 \begin {gather*} -\frac {2 \, {\left (3 \, {\left (9 \, A - 49 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (33 \, A - 188 \, B\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, A - 59 \, B\right )} \cos \left (d x + c\right ) - 60 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (3 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (3 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (3 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (3 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (-3 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (-3 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-3 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-3 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-9 i \, A + 49 i \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (-9 i \, A + 49 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-9 i \, A + 49 i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-9 i \, A + 49 i \, B\right )} \cos \left (d x + c\right )\right )} {\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 \, {\left (\sqrt {2} {\left (9 i \, A - 49 i \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (9 i \, A - 49 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (9 i \, A - 49 i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (9 i \, A - 49 i \, B\right )} \cos \left (d x + c\right )\right )} {\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 )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(2*(3*(9*A - 49*B)*cos(d*x + c)^3 + 2*(33*A - 188*B)*cos(d*x + c)^2 + 5*(9*A - 59*B)*cos(d*x + c) - 60*B
)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(3*I*A - 13*I*B)*cos(d*x + c)^4 + 3*sqrt(2)*(3*I*A - 13*I*B)*co
s(d*x + c)^3 + 3*sqrt(2)*(3*I*A - 13*I*B)*cos(d*x + c)^2 + sqrt(2)*(3*I*A - 13*I*B)*cos(d*x + c))*weierstrassP
Inverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-3*I*A + 13*I*B)*cos(d*x + c)^4 + 3*sqrt(2)*(-3*I*
A + 13*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-3*I*A + 13*I*B)*cos(d*x + c)^2 + sqrt(2)*(-3*I*A + 13*I*B)*cos(d*x +
c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-9*I*A + 49*I*B)*cos(d*x + c)^4 +
3*sqrt(2)*(-9*I*A + 49*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-9*I*A + 49*I*B)*cos(d*x + c)^2 + sqrt(2)*(-9*I*A + 49
*I*B)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqr
t(2)*(9*I*A - 49*I*B)*cos(d*x + c)^4 + 3*sqrt(2)*(9*I*A - 49*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(9*I*A - 49*I*B)*
cos(d*x + c)^2 + sqrt(2)*(9*I*A - 49*I*B)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(
d*x + c) - I*sin(d*x + c))))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*c
os(d*x + c))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)**(7/2)/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^3*cos(d*x + c)^(7/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(7/2)*(a + a/cos(c + d*x))^3),x)

[Out]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(7/2)*(a + a/cos(c + d*x))^3), x)

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