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

Optimal. Leaf size=207 \[ -\frac {(5 A-2 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(115 A-43 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-(5*A-2*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d+1/32*(115*A-43*B)*arctanh(1/2*sin(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)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-1/1
6*(15*A-7*B)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+1/16*(35*A-11*B)*tan(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.71, antiderivative size = 207, 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 = {2978, 2984, 2985, 2649, 206, 2773} \[ \frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(5 A-2 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(115 A-43 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((5*A - 2*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d)) + ((115*A - 43*B)*ArcTan
h[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B)*Tan[c + d*x])/
(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((15*A - 7*B)*Tan[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + ((35*A -
11*B)*Tan[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])

Rule 206

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

Rule 2649

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 2773

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 2978

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 2984

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 2985

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 ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\left (a (5 A-B)-\frac {5}{2} a (A-B) \cos (c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (\frac {1}{2} a^2 (35 A-11 B)-\frac {3}{4} a^2 (15 A-7 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-4 a^3 (5 A-2 B)+\frac {1}{4} a^3 (35 A-11 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=-\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(115 A-43 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}-\frac {(5 A-2 B) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{2 a^3}\\ &=-\frac {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(115 A-43 B) \operatorname {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 {(5 A-2 B) \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^2 d}\\ &=-\frac {(5 A-2 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {(115 A-43 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 {(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 3.47, size = 142, normalized size = 0.69 \[ \frac {\tan (c+d x) (10 (11 A-3 B) \cos (c+d x)+(35 A-11 B) \cos (2 (c+d x))+67 A-11 B)+8 (115 A-43 B) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-128 \sqrt {2} (5 A-2 B) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{32 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(115*A - 43*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 - 128*Sqrt[2]*(5*A - 2*B)*ArcTanh[Sqrt[2]*Sin[(
c + d*x)/2]]*Cos[(c + d*x)/2]^5 + (67*A - 11*B + 10*(11*A - 3*B)*Cos[c + d*x] + (35*A - 11*B)*Cos[2*(c + d*x)]
)*Tan[c + d*x])/(32*d*(a*(1 + Cos[c + d*x]))^(5/2))

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fricas [B]  time = 1.48, size = 404, normalized size = 1.95 \[ -\frac {\sqrt {2} {\left ({\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 16 \, {\left ({\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, {\left ({\left (35 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (11 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(sqrt(2)*((115*A - 43*B)*cos(d*x + c)^4 + 3*(115*A - 43*B)*cos(d*x + c)^3 + 3*(115*A - 43*B)*cos(d*x + c
)^2 + (115*A - 43*B)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)
*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 16*((5*A - 2*B)*cos(d*x + c)^
4 + 3*(5*A - 2*B)*cos(d*x + c)^3 + 3*(5*A - 2*B)*cos(d*x + c)^2 + (5*A - 2*B)*cos(d*x + c))*sqrt(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*((35*A - 11*B)*cos(d*x + c)^2 + 5*(11*A - 3*B)*cos(d*x + c) + 16*A)*sqrt
(a*cos(d*x + c) + a)*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*cos(d*x + c))

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giac [B]  time = 3.95, size = 409, normalized size = 1.98 \[ \frac {2 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} - B a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} + \frac {\sqrt {2} {\left (21 \, A a^{5} - 13 \, B a^{5}\right )}}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {\sqrt {2} {\left (115 \, A \sqrt {a} - 43 \, B \sqrt {a}\right )} \log \left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{a^{3}} - \frac {32 \, {\left (5 \, A \sqrt {a} - 2 \, B \sqrt {a}\right )} \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{a^{3}} + \frac {32 \, {\left (5 \, A \sqrt {a} - 2 \, B \sqrt {a}\right )} \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{a^{3}} + \frac {128 \, \sqrt {2} {\left (3 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} A \sqrt {a} - A a^{\frac {3}{2}}\right )}}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )} a^{2}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(2*sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5)*tan(1/2*d*x + 1/2*c)^2/a^8 + sqrt(2)*(21
*A*a^5 - 13*B*a^5)/a^8)*tan(1/2*d*x + 1/2*c) - sqrt(2)*(115*A*sqrt(a) - 43*B*sqrt(a))*log((sqrt(a)*tan(1/2*d*x
+ 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/a^3 - 32*(5*A*sqrt(a) - 2*B*sqrt(a))*log(abs((sqrt(a)*tan(1
/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3)))/a^3 + 32*(5*A*sqrt(a) - 2*B*sqrt
(a))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3)))/a^3 +
128*sqrt(2)*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(a) - A*a^(3/2))/(
((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqr
t(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)*a^2))/d

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maple [B]  time = 1.71, size = 1122, normalized size = 5.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(230*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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-86*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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-160*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^
(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^6*a-160*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2
*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^6*a
+64*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*
x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^6*a+64*B*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^6*a-115*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d
*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a+43*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x
+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^4*a+70*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4+80*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^4*a+80*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2
*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^4*a
-22*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-32*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1
/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*cos(1/2*d*x+1/2*c)^4*a
-32*B*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a)/(2*cos(1/2*d*x+1
/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^4*a-15*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^
2+7*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2-2*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(1/2)*a^(1/2)+2*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/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*cos(1/2*d*x+1/2*c)-2^(1/2))/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(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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