3.6.43 \(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx\) [543]

Optimal. Leaf size=317 \[ \frac {(1015 A-363 B) \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 \sqrt {2} a^{7/2} d}-\frac {(1887 A-691 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-1/6*(A-B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(7/2)-1/48*(23*A-11*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a
/d/(a+a*cos(d*x+c))^(5/2)-1/64*(109*A-41*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(3/2)+1/192*(57
9*A-199*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^3/d/(a+a*cos(d*x+c))^(1/2)+1/128*(1015*A-363*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^(7/2)/d*2^(1/2)-
1/192*(1887*A-691*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/a^3/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.72, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3040, 3057, 3063, 12, 2861, 211} \begin {gather*} \frac {(1015 A-363 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {(579 A-199 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{192 a^3 d \sqrt {a \cos (c+d x)+a}}-\frac {(1887 A-691 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{192 a^3 d \sqrt {a \cos (c+d x)+a}}-\frac {(109 A-41 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{64 a^2 d (a \cos (c+d x)+a)^{3/2}}-\frac {(23 A-11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{48 a d (a \cos (c+d x)+a)^{5/2}}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{6 d (a \cos (c+d x)+a)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x])^(7/2),x]

[Out]

((1015*A - 363*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])/(64*Sqrt[2]*a^(7/2)*d) - ((1887*A - 691*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(1
92*a^3*d*Sqrt[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)
) - ((23*A - 11*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(48*a*d*(a + a*Cos[c + d*x])^(5/2)) - ((109*A - 41*B)*Sec[
c + d*x]^(3/2)*Sin[c + d*x])/(64*a^2*d*(a + a*Cos[c + d*x])^(3/2)) + ((579*A - 199*B)*Sec[c + d*x]^(3/2)*Sin[c
+ d*x])/(192*a^3*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 211

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

Rule 2861

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 3040

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 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])

Rubi steps

\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx\\ &=-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{2} a (5 A-B)-4 a (A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{4} a^2 (63 A-19 B)-\frac {3}{2} a^2 (23 A-11 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{64 a^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 {3}{8} a^3 (579 A-199 B)-\frac {3}{2} a^3 (109 A-41 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3}{16} a^4 (1887 A-691 B)+\frac {3}{8} a^4 (579 A-199 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{72 a^7}\\ &=-\frac {(1887 A-691 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {9 a^5 (1015 A-363 B)}{32 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{36 a^8}\\ &=-\frac {(1887 A-691 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}}+\frac {\left ((1015 A-363 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}{128 a^3}\\ &=-\frac {(1887 A-691 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}}-\frac {\left ((1015 A-363 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {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 )}{64 a^2 d}\\ &=\frac {(1015 A-363 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)}}{64 \sqrt {2} a^{7/2} d}-\frac {(1887 A-691 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{192 a^3 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 5.03, size = 267, normalized size = 0.84 \begin {gather*} \frac {\cos ^7\left (\frac {1}{2} (c+d x)\right ) \left (\frac {i (1015 A-363 B) e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{d}-\frac {(21641 A-8469 B+4 (9415 A-3579 B) \cos (c+d x)+8 (3069 A-1145 B) \cos (2 (c+d x))+10164 A \cos (3 (c+d x))-3748 B \cos (3 (c+d x))+1887 A \cos (4 (c+d x))-691 B \cos (4 (c+d x))) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{96 d}\right )}{8 (a (1+\cos (c+d x)))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x])^(7/2),x]

[Out]

(Cos[(c + d*x)/2]^7*((I*(1015*A - 363*B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c
+ d*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/(d*E^((I/2)*(c + d*x))) - ((2
1641*A - 8469*B + 4*(9415*A - 3579*B)*Cos[c + d*x] + 8*(3069*A - 1145*B)*Cos[2*(c + d*x)] + 10164*A*Cos[3*(c +
d*x)] - 3748*B*Cos[3*(c + d*x)] + 1887*A*Cos[4*(c + d*x)] - 691*B*Cos[4*(c + d*x)])*Sec[(c + d*x)/2]^5*Sec[c
+ d*x]^(3/2)*Tan[(c + d*x)/2])/(96*d)))/(8*(a*(1 + Cos[c + d*x]))^(7/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(728\) vs. \(2(270)=540\).
time = 0.51, size = 729, normalized size = 2.30

method	result	size
default	\(\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (3045 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-1089 B \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+12180 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4356 B \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+18270 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-6534 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+12180 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \sin \left (d x +c \right )-1887 A \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {2}-4356 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \sin \left (d x +c \right )+691 B \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {2}+3045 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )-3195 A \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}-1089 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )+1183 B \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}+831 A \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}-275 B \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+3355 A \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-1215 B \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+1024 A \cos \left (d x +c \right ) \sqrt {2}-384 B \cos \left (d x +c \right ) \sqrt {2}-128 A \sqrt {2}\right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{384 d \sin \left (d x +c \right )^{3} \left (1+\cos \left (d x +c \right )\right )^{2} a^{4}}\)	\(729\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/384/d*(-1+cos(d*x+c))*(3045*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^4*sin(d*x+c)*arcsin((-1+cos(d*x+c
))/sin(d*x+c))-1089*B*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^4*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x
+c))+12180*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*sin(d*x+c)-4356
*B*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*sin(d*x+c)+18270*A*arcsin
((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2*sin(d*x+c)-6534*B*arcsin((-1+cos(d
*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2*sin(d*x+c)+12180*A*arcsin((-1+cos(d*x+c))/si
n(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)*sin(d*x+c)-1887*A*cos(d*x+c)^5*2^(1/2)-4356*B*arcsin((-
1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)*sin(d*x+c)+691*B*cos(d*x+c)^5*2^(1/2)+3
045*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)-3195*A*cos(d*x+c)^4*2^(1
/2)-1089*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)+1183*B*cos(d*x+c)^4
*2^(1/2)+831*A*cos(d*x+c)^3*2^(1/2)-275*B*cos(d*x+c)^3*2^(1/2)+3355*A*cos(d*x+c)^2*2^(1/2)-1215*B*cos(d*x+c)^2
*2^(1/2)+1024*A*cos(d*x+c)*2^(1/2)-384*B*cos(d*x+c)*2^(1/2)-128*A*2^(1/2))*cos(d*x+c)*(1/cos(d*x+c))^(5/2)*(a*
(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^3/(1+cos(d*x+c))^2*2^(1/2)/a^4

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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)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 0.45, size = 295, normalized size = 0.93 \begin {gather*} -\frac {3 \, \sqrt {2} {\left ({\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )\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 (1887 \, A - 691 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2541 \, A - 937 \, B\right )} \cos \left (d x + c\right )^{3} + 39 \, {\left (109 \, A - 41 \, B\right )} \cos \left (d x + c\right )^{2} + 128 \, {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 128 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\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)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/384*(3*sqrt(2)*((1015*A - 363*B)*cos(d*x + c)^5 + 4*(1015*A - 363*B)*cos(d*x + c)^4 + 6*(1015*A - 363*B)*co
s(d*x + c)^3 + 4*(1015*A - 363*B)*cos(d*x + c)^2 + (1015*A - 363*B)*cos(d*x + c))*sqrt(a)*arctan(sqrt(2)*sqrt(
a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((1887*A - 691*B)*cos(d*x + c)^4 + 2*(2541*
A - 937*B)*cos(d*x + c)^3 + 39*(109*A - 41*B)*cos(d*x + c)^2 + 128*(7*A - 3*B)*cos(d*x + c) - 128*A)*sqrt(a*co
s(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^4*d*cos(d*x + c)^5 + 4*a^4*d*cos(d*x + c)^4 + 6*a^4*d*cos(
d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(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*cos(d*x+c))*sec(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [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)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(5/2))/(a + a*cos(c + d*x))^(7/2),x)

[Out]

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(5/2))/(a + a*cos(c + d*x))^(7/2), x)

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