3.2.22 \(\int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [122]

Optimal. Leaf size=264 \[ \frac {(39 A-20 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

1/4*(39*A-20*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d-1/32*(219*A-115*B)*arctanh(1/2*si
n(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(A-B)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(
d*x+c))^(5/2)-1/16*(19*A-11*B)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)-7/16*(9*A-5*B)*tan(d*x+c)/a^2/
d/(a+a*cos(d*x+c))^(1/2)+1/16*(31*A-15*B)*sec(d*x+c)*tan(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.61, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3057, 3063, 3064, 2728, 212, 2852} \begin {gather*} \frac {(39 A-20 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(31 A-15 B) \tan (c+d x) \sec (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(19 A-11 B) \tan (c+d x) \sec (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

((39*A - 20*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*a^(5/2)*d) - ((219*A - 115*B)*ArcT
anh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (7*(9*A - 5*B)*Tan[c
+ d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c + d*x]*Tan[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5
/2)) - ((19*A - 11*B)*Sec[c + d*x]*Tan[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + ((31*A - 15*B)*Sec[c +
d*x]*Tan[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2852

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[-2*(
b/f), Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

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])

Rule 3063

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 3064

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\left (2 a (3 A-B)-\frac {7}{2} a (A-B) \cos (c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (a^2 (31 A-15 B)-\frac {5}{4} a^2 (19 A-11 B) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-7 a^3 (9 A-5 B)+\frac {3}{2} a^3 (31 A-15 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{16 a^5}\\ &=-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (2 a^4 (39 A-20 B)-\frac {7}{2} a^4 (9 A-5 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{16 a^6}\\ &=-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(219 A-115 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}+\frac {(39 A-20 B) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{8 a^3}\\ &=-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(219 A-115 B) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{16 a^2 d}-\frac {(39 A-20 B) \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^2 d}\\ &=\frac {(39 A-20 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {7 (9 A-5 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(19 A-11 B) \sec (c+d x) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(31 A-15 B) \sec (c+d x) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 6.26, size = 656, normalized size = 2.48 \begin {gather*} \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (8 (219 A-115 B) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )+8 (-219 A+115 B) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \log \left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )+16 \sqrt {2} (39 A-20 B) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \log \left (\sqrt {2}+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {16 i (39 A-20 B) \text {ArcTan}\left (\frac {\cos \left (\frac {1}{4} (c+d x)\right )-\left (-1+\sqrt {2}\right ) \sin \left (\frac {1}{4} (c+d x)\right )}{\left (1+\sqrt {2}\right ) \cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2}-2 \sin \left (\frac {c}{2}\right )\right )}{-1+\sqrt {2} \sin \left (\frac {c}{2}\right )}+\frac {16 i (39 A-20 B) \text {ArcTan}\left (\frac {\cos \left (\frac {1}{4} (c+d x)\right )-\left (1+\sqrt {2}\right ) \sin \left (\frac {1}{4} (c+d x)\right )}{\left (-1+\sqrt {2}\right ) \cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2}-2 \sin \left (\frac {c}{2}\right )\right )}{-1+\sqrt {2} \sin \left (\frac {c}{2}\right )}+\frac {8 (39 A-20 B) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \log \left (2-\sqrt {2} \cos \left (\frac {1}{2} (c+d x)\right )-\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sqrt {2}-2 \sin \left (\frac {c}{2}\right )\right )}{-1+\sqrt {2} \sin \left (\frac {c}{2}\right )}+\frac {8 (39 A-20 B) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \log \left (2+\sqrt {2} \cos \left (\frac {1}{2} (c+d x)\right )-\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sqrt {2}-2 \sin \left (\frac {c}{2}\right )\right )}{-1+\sqrt {2} \sin \left (\frac {c}{2}\right )}-(158 A-110 B+(269 A-169 B) \cos (c+d x)+10 (19 A-11 B) \cos (2 (c+d x))+63 A \cos (3 (c+d x))-35 B \cos (3 (c+d x))) \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{32 d (a (1+\cos (c+d x)))^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(Cos[(c + d*x)/2]*(8*(219*A - 115*B)*Cos[(c + d*x)/2]^4*Log[Cos[(c + d*x)/4] - Sin[(c + d*x)/4]] + 8*(-219*A +
115*B)*Cos[(c + d*x)/2]^4*Log[Cos[(c + d*x)/4] + Sin[(c + d*x)/4]] + 16*Sqrt[2]*(39*A - 20*B)*Cos[(c + d*x)/2
]^4*Log[Sqrt[2] + 2*Sin[(c + d*x)/2]] + ((16*I)*(39*A - 20*B)*ArcTan[(Cos[(c + d*x)/4] - (-1 + Sqrt[2])*Sin[(c
+ d*x)/4])/((1 + Sqrt[2])*Cos[(c + d*x)/4] - Sin[(c + d*x)/4])]*Cos[(c + d*x)/2]^4*(Sqrt[2] - 2*Sin[c/2]))/(-
1 + Sqrt[2]*Sin[c/2]) + ((16*I)*(39*A - 20*B)*ArcTan[(Cos[(c + d*x)/4] - (1 + Sqrt[2])*Sin[(c + d*x)/4])/((-1
+ Sqrt[2])*Cos[(c + d*x)/4] - Sin[(c + d*x)/4])]*Cos[(c + d*x)/2]^4*(Sqrt[2] - 2*Sin[c/2]))/(-1 + Sqrt[2]*Sin[
c/2]) + (8*(39*A - 20*B)*Cos[(c + d*x)/2]^4*Log[2 - Sqrt[2]*Cos[(c + d*x)/2] - Sqrt[2]*Sin[(c + d*x)/2]]*(Sqrt
[2] - 2*Sin[c/2]))/(-1 + Sqrt[2]*Sin[c/2]) + (8*(39*A - 20*B)*Cos[(c + d*x)/2]^4*Log[2 + Sqrt[2]*Cos[(c + d*x)
/2] - Sqrt[2]*Sin[(c + d*x)/2]]*(Sqrt[2] - 2*Sin[c/2]))/(-1 + Sqrt[2]*Sin[c/2]) - (158*A - 110*B + (269*A - 16
9*B)*Cos[c + d*x] + 10*(19*A - 11*B)*Cos[2*(c + d*x)] + 63*A*Cos[3*(c + d*x)] - 35*B*Cos[3*(c + d*x)])*Sec[c +
d*x]^2*Sin[(c + d*x)/2]))/(32*d*(a*(1 + Cos[c + d*x]))^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1609\) vs. \(2(229)=458\).
time = 0.57, size = 1610, normalized size = 6.10

method	result	size
default	\(\text {Expression too large to display}\)	\(1610\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/8*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(-624*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a
^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^6*a-624*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2
^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^
6*a+320*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c
)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^6*a+320*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/
2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^6*a-876*A*ln(2*(2*a^(1/2)*(sin(1/
2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^8*a+460*B*ln(2*(2*a^(1/2)*(sin(1/2
*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^8*a-252*A*2^(1/2)*(sin(1/2*d*x+1/2*
c)^2*a)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6+140*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*cos(1/2*d*x+1/
2*c)^6-320*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1
/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^8*a+624*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x
+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^8*a-320*B*ln(4/(2*cos(1/2*d*x+
1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x
+1/2*c)^8*a+624*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*
d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^8*a-219*A*2^(1/2)*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2
)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a+115*B*2^(1/2)*ln(2*(2*a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)
+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a-2*A*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-19*A*a^(1/
2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*cos(1/2*d*x+1/2*c)^2+11*B*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(
1/2)*cos(1/2*d*x+1/2*c)^2-100*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4+156*A*ln(-
4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-
2*a))*cos(1/2*d*x+1/2*c)^4*a+156*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2
^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a-80*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(
a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^4*a-80*B*
ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/
2)+2*a))*cos(1/2*d*x+1/2*c)^4*a+2*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)+876*A*ln(2*(2*a^(1/2)*(sin(
1/2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^6*a-460*B*ln(2*(2*a^(1/2)*(sin(1
/2*d*x+1/2*c)^2*a)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^6*a+188*A*2^(1/2)*(sin(1/2*d*x+1/
2*c)^2*a)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4)/a^(7/2)/cos(1/2*d*x+1/2*c)^3/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^2/(2
*cos(1/2*d*x+1/2*c)+2^(1/2))^2/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(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*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 0.52, size = 428, normalized size = 1.62 \begin {gather*} -\frac {\sqrt {2} {\left ({\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (7 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (19 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (5 \, A - 4 \, B\right )} \cos \left (d x + c\right ) - 8 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/64*(sqrt(2)*((219*A - 115*B)*cos(d*x + c)^5 + 3*(219*A - 115*B)*cos(d*x + c)^4 + 3*(219*A - 115*B)*cos(d*x
+ c)^3 + (219*A - 115*B)*cos(d*x + c)^2)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*s
qrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((39*A - 20*B)*cos(d*
x + c)^5 + 3*(39*A - 20*B)*cos(d*x + c)^4 + 3*(39*A - 20*B)*cos(d*x + c)^3 + (39*A - 20*B)*cos(d*x + c)^2)*sqr
t(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*
x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(7*(9*A - 5*B)*cos(d*x + c)^3 + 5*(19*A - 11*B)*cos(d*x +
c)^2 + 4*(5*A - 4*B)*cos(d*x + c) - 8*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3
*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

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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*cos(d*x+c))*sec(d*x+c)**3/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 0.86, size = 374, normalized size = 1.42 \begin {gather*} -\frac {\frac {\sqrt {2} {\left (219 \, A \sqrt {a} - 115 \, B \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (219 \, A \sqrt {a} - 115 \, B \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {8 \, {\left (39 \, A \sqrt {a} - 20 \, B \sqrt {a}\right )} \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {8 \, {\left (39 \, A \sqrt {a} - 20 \, B \sqrt {a}\right )} \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (252 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 140 \, B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 568 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 320 \, B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 399 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 231 \, B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 85 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 53 \, B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/64*(sqrt(2)*(219*A*sqrt(a) - 115*B*sqrt(a))*log(sin(1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(1/2*d*x + 1/2*c))) -
sqrt(2)*(219*A*sqrt(a) - 115*B*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(1/2*d*x + 1/2*c))) - 8*(3
9*A*sqrt(a) - 20*B*sqrt(a))*log(abs(1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a^3*sgn(cos(1/2*d*x + 1/2*c))) + 8*(
39*A*sqrt(a) - 20*B*sqrt(a))*log(abs(-1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a^3*sgn(cos(1/2*d*x + 1/2*c))) - 2
*sqrt(2)*(252*A*sqrt(a)*sin(1/2*d*x + 1/2*c)^7 - 140*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^7 - 568*A*sqrt(a)*sin(1/2*
d*x + 1/2*c)^5 + 320*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^5 + 399*A*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 231*B*sqrt(a)*s
in(1/2*d*x + 1/2*c)^3 - 85*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + 53*B*sqrt(a)*sin(1/2*d*x + 1/2*c))/((2*sin(1/2*d*x
+ 1/2*c)^4 - 3*sin(1/2*d*x + 1/2*c)^2 + 1)^2*a^3*sgn(cos(1/2*d*x + 1/2*c))))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,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)^3*(a + a*cos(c + d*x))^(5/2)),x)

[Out]

int((A + B*cos(c + d*x))/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(5/2)), x)

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