3.530 \(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=270 \[ -\frac {(15 A-11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(9 A-5 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{10 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(39 A-35 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{30 a d \sqrt {a \cos (c+d x)+a}}+\frac {(147 A-95 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{30 a d \sqrt {a \cos (c+d x)+a}} \]

[Out]

-1/2*(A-B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-1/30*(39*A-35*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a
/d/(a+a*cos(d*x+c))^(1/2)+1/10*(9*A-5*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/4*(15*A-11*B
)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(
1/2)/a^(3/2)/d*2^(1/2)+1/30*(147*A-95*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.89, antiderivative size = 270, 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 = {2961, 2978, 2984, 12, 2782, 205} \[ -\frac {(15 A-11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(9 A-5 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{10 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(39 A-35 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{30 a d \sqrt {a \cos (c+d x)+a}}+\frac {(147 A-95 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{30 a d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*A - 11*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[
c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) + ((147*A - 95*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(30*a*d
*Sqrt[a + a*Cos[c + d*x]]) - ((39*A - 35*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(30*a*d*Sqrt[a + a*Cos[c + d*x]])
- ((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((9*A - 5*B)*Sec[c + d*x]^(5/2
)*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Cos[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*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 2961

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

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])

Rubi steps

\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\\ &=-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (9 A-5 B)-3 a (A-B) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-5 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{4} a^2 (39 A-35 B)+a^2 (9 A-5 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{5 a^3}\\ &=-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-5 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} a^3 (147 A-95 B)-\frac {1}{4} a^3 (39 A-35 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^4}\\ &=\frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-5 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int -\frac {15 a^4 (15 A-11 B)}{16 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^5}\\ &=\frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-5 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {\left ((15 A-11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=\frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-5 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}+\frac {\left ((15 A-11 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 d}\\ &=-\frac {(15 A-11 B) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}+\frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(39 A-35 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{30 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(9 A-5 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [A]  time = 2.02, size = 220, normalized size = 0.81 \[ \frac {15 \, \sqrt {2} {\left ({\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (147 \, A - 95 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 12 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*sqrt(2)*((15*A - 11*B)*cos(d*x + c)^4 + 2*(15*A - 11*B)*cos(d*x + c)^3 + (15*A - 11*B)*cos(d*x + c)^2
)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((147*A - 95*
B)*cos(d*x + c)^3 + 12*(9*A - 5*B)*cos(d*x + c)^2 - 4*(3*A - 5*B)*cos(d*x + c) + 12*A)*sqrt(a*cos(d*x + c) + a
)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.56, size = 595, normalized size = 2.20 \[ \frac {\left (225 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-165 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+675 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-495 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+675 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-495 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+225 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-165 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-147 A \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right )+95 B \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right )+39 A \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )-35 B \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )+120 A \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )-80 B \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )-24 A \sqrt {2}\, \cos \left (d x +c \right )+20 B \sqrt {2}\, \cos \left (d x +c \right )+12 A \sqrt {2}\right ) \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{60 d \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{3} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60/d*(225*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-165
*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+675*A*arcsin((
-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-495*B*arcsin((-1+cos(d*x+
c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+675*A*arcsin((-1+cos(d*x+c))/sin(d*x
+c))*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-495*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+
c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+225*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c
)/(1+cos(d*x+c)))^(5/2)-165*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-
147*A*2^(1/2)*cos(d*x+c)^4+95*B*2^(1/2)*cos(d*x+c)^4+39*A*2^(1/2)*cos(d*x+c)^3-35*B*2^(1/2)*cos(d*x+c)^3+120*A
*2^(1/2)*cos(d*x+c)^2-80*B*2^(1/2)*cos(d*x+c)^2-24*A*2^(1/2)*cos(d*x+c)+20*B*2^(1/2)*cos(d*x+c)+12*A*2^(1/2))*
cos(d*x+c)*sin(d*x+c)^3*(1/cos(d*x+c))^(7/2)*(a*(1+cos(d*x+c)))^(1/2)/(-1+cos(d*x+c))^2/(1+cos(d*x+c))^3*2^(1/
2)/a^2

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + a*cos(c + d*x))^(3/2),x)

[Out]

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + a*cos(c + d*x))^(3/2), 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)**(7/2)/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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