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

Optimal. Leaf size=223 \[ \frac{(11 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(7 A-3 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{6 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(19 A-15 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{6 a d \sqrt{a \cos (c+d x)+a}} \]

[Out]

((11*A - 7*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) - ((19*A - 15*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(6*a*d*Sqr
t[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((7*A -
3*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*a*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.701828, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{(11 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(7 A-3 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{6 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(19 A-15 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{6 a d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*A - 7*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) - ((19*A - 15*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(6*a*d*Sqr
t[a + a*Cos[c + d*x]]) - ((A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((7*A -
3*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*a*d*Sqrt[a + a*Cos[c + d*x]])

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])

Rule 12

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

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 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]

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))^{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{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\\ &=-\frac{(A-B) \sec ^{\frac{3}{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 (7 A-3 B)-2 a (A-B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 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 (19 A-15 B)+\frac{1}{2} a^2 (7 A-3 B) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{3 a^3}\\ &=-\frac{(19 A-15 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 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{3 a^3 (11 A-7 B)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{3 a^4}\\ &=-\frac{(19 A-15 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left ((11 A-7 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{(19 A-15 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}-\frac{\left ((11 A-7 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{(11 A-7 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{(19 A-15 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(7 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.84534, size = 981, normalized size = 4.4 \[ \frac{2 \cos ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\frac{1}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}} \sqrt{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )} \left (\frac{(A+3 B) \left (-12 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac{7}{2};1,\frac{9}{2};-\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}\right ) \sin ^8\left (\frac{c}{2}+\frac{d x}{2}\right )-12 \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};-\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}\right ) \left (3 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )-7 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )+4\right ) \sin ^8\left (\frac{c}{2}+\frac{d x}{2}\right )+7 \sqrt{-\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}} \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^3 \left (8 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )-20 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )+15\right ) \left (\left (3-7 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \sqrt{-\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}}-3 \tanh ^{-1}\left (\sqrt{-\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}}\right ) \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )\right )\right ) \csc ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{126 \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^{7/2}}-\frac{1}{2} (A-B) \left (5 \tan ^{-1}\left (\frac{1-2 \sin \left (\frac{c}{2}+\frac{d x}{2}\right )}{\sqrt{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}}\right )+\frac{3 \sqrt{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}}{1-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )}+\frac{\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+1}{\left (1-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \sqrt{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}}\right )+\frac{1}{2} (A-B) \left (5 \tan ^{-1}\left (\frac{2 \sin \left (\frac{c}{2}+\frac{d x}{2}\right )+1}{\sqrt{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}}\right )+\frac{3 \sqrt{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}}{\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+1}+\frac{1-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )}{\left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+1\right ) \sqrt{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}}\right )+\frac{(A-B) \left (2 \sin \left (\frac{c}{2}+\frac{d x}{2}\right )+1\right )}{12 \left (1-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^{3/2}}-\frac{(A-B) \left (1-2 \sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{12 \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+1\right ) \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^{3/2}}\right )}{d (a (\cos (c+d x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[c/2 + (d*x)/2]^3*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*(-((A - B)*(1
- 2*Sin[c/2 + (d*x)/2]))/(12*(1 + Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + ((A - B)*(1 + 2*S
in[c/2 + (d*x)/2]))/(12*(1 - Sin[c/2 + (d*x)/2])*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) - ((A - B)*(5*ArcTan[(1 -
2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (1 + Sin[c/2 + (d*x)/2])/((1 - Sin[c/2 + (d*x)/2])*
Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 - Sin[c/2 + (d*x)/2])))/2 + ((A -
B)*(5*ArcTan[(1 + 2*Sin[c/2 + (d*x)/2])/Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]] + (1 - Sin[c/2 + (d*x)/2])/((1 + Sin
[c/2 + (d*x)/2])*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (3*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2])/(1 + Sin[c/2 + (d*x)
/2])))/2 + ((A + 3*B)*Csc[c/2 + (d*x)/2]^5*(-12*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 7/2}, {1, 9/2}, -(
Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8 - 12*Hypergeometric2F1[2, 7/2, 9/2, -
(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/2]^8*(4 - 7*Sin[c/2 + (d*x)/2]^2 + 3*Sin[
c/2 + (d*x)/2]^4) + 7*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^
3*(15 - 20*Sin[c/2 + (d*x)/2]^2 + 8*Sin[c/2 + (d*x)/2]^4)*((3 - 7*Sin[c/2 + (d*x)/2]^2)*Sqrt[-(Sin[c/2 + (d*x)
/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))] - 3*ArcTanh[Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*(
1 - 2*Sin[c/2 + (d*x)/2]^2))))/(126*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2))))/(d*(a*(1 + Cos[c + d*x]))^(3/2))

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Maple [B]  time = 0.727, size = 457, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

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

[Out]

1/12/d*2^(1/2)/a^2*(33*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c))
)^(3/2)-21*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+66*A
*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-42*B*arcsin((-1+co
s(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+33*A*arcsin((-1+cos(d*x+c))/sin(
d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-21*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d
*x+c)/(1+cos(d*x+c)))^(3/2)-19*A*cos(d*x+c)^3*2^(1/2)+15*B*cos(d*x+c)^3*2^(1/2)+7*A*cos(d*x+c)^2*2^(1/2)-3*B*c
os(d*x+c)^2*2^(1/2)+16*A*cos(d*x+c)*2^(1/2)-12*B*cos(d*x+c)*2^(1/2)-4*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)/(-1+cos(d*x+c))/(1+cos(d*x+c))^2

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.17199, size = 531, normalized size = 2.38 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (11 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (11 \, A - 7 \, 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 (19 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \,{\left (A - B\right )} \cos \left (d x + c\right ) - 4 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{12 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \end{align*}

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))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(3*sqrt(2)*((11*A - 7*B)*cos(d*x + c)^3 + 2*(11*A - 7*B)*cos(d*x + c)^2 + (11*A - 7*B)*cos(d*x + c))*sqr
t(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((19*A - 15*B)*cos
(d*x + c)^2 + 12*(A - B)*cos(d*x + c) - 4*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*
cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

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

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))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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))^(3/2),x, algorithm="giac")

[Out]

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