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

Optimal. Leaf size=221 \[ \frac{(17 A+67 B) \sin (c+d x)}{192 a^2 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}+\frac{(7 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{(A-B) \sin (c+d x)}{6 d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}}+\frac{(A-13 B) \sin (c+d x)}{48 a d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

[Out]

((7*A + 5*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]])/(64*Sqrt[2]*a^(7/2)*d) + ((A - B)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)*Sec
[c + d*x]^(3/2)) + ((A - 13*B)*Sin[c + d*x])/(48*a*d*(a + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]) + ((17*A +
67*B)*Sin[c + d*x])/(192*a^2*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.723135, antiderivative size = 221, 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, 2977, 2978, 12, 2782, 205} \[ \frac{(17 A+67 B) \sin (c+d x)}{192 a^2 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}+\frac{(7 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{64 \sqrt{2} a^{7/2} d}+\frac{(A-B) \sin (c+d x)}{6 d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}}+\frac{(A-13 B) \sin (c+d x)}{48 a d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*A + 5*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]])/(64*Sqrt[2]*a^(7/2)*d) + ((A - B)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)*Sec
[c + d*x]^(3/2)) + ((A - 13*B)*Sin[c + d*x])/(48*a*d*(a + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]) + ((17*A +
67*B)*Sin[c + d*x])/(192*a^2*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[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 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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 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)}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac{3}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{7/2}} \, dx\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{2} a (A-B)+a (A+5 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(A-13 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a^2 (A-13 B)+\frac{9}{2} a^2 (A+3 B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(A-13 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}+\frac{(17 A+67 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{3 a^3 (7 A+5 B)}{8 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=\frac{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(A-13 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}+\frac{(17 A+67 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}+\frac{\left ((7 A+5 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{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(A-13 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}+\frac{(17 A+67 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}-\frac{\left ((7 A+5 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 )}{64 a^2 d}\\ &=\frac{(7 A+5 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{(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac{3}{2}}(c+d x)}+\frac{(A-13 B) \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}+\frac{(17 A+67 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 7.16513, size = 488, normalized size = 2.21 \[ \frac{\cos ^7\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} \left (\frac{(17 A+67 B) \sin \left (\frac{c}{2}\right ) \cos \left (\frac{d x}{2}\right )}{12 d}+\frac{(17 A+67 B) \cos \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )}{12 d}+\frac{\sec \left (\frac{c}{2}\right ) \sec ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )-B \sin \left (\frac{d x}{2}\right )\right )}{3 d}+\frac{\sec \left (\frac{c}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (29 B \sin \left (\frac{d x}{2}\right )-17 A \sin \left (\frac{d x}{2}\right )\right )}{12 d}+\frac{\sec \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (19 A \sin \left (\frac{d x}{2}\right )-151 B \sin \left (\frac{d x}{2}\right )\right )}{24 d}+\frac{(A-B) \tan \left (\frac{c}{2}\right ) \sec ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{(17 A-29 B) \tan \left (\frac{c}{2}\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{12 d}+\frac{(19 A-151 B) \tan \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{24 d}\right )}{(a (\cos (c+d x)+1))^{7/2}}+\frac{i (7 A+5 B) e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \cos ^7\left (\frac{c}{2}+\frac{d x}{2}\right ) \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{8 d (a (\cos (c+d x)+1))^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((I/8)*(7*A + 5*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))])]*Cos[c/2 + (d*x)/2]^7)/(d*E^((I/2)*(c + d*x))*(a*(1 +
Cos[c + d*x]))^(7/2)) + (Cos[c/2 + (d*x)/2]^7*Sqrt[Sec[c + d*x]]*(((17*A + 67*B)*Cos[(d*x)/2]*Sin[c/2])/(12*d
) + ((17*A + 67*B)*Cos[c/2]*Sin[(d*x)/2])/(12*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]^2*(19*A*Sin[(d*x)/2] - 151*B*S
in[(d*x)/2]))/(24*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]^6*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/(3*d) + (Sec[c/2]*Sec
[c/2 + (d*x)/2]^4*(-17*A*Sin[(d*x)/2] + 29*B*Sin[(d*x)/2]))/(12*d) + ((19*A - 151*B)*Sec[c/2 + (d*x)/2]*Tan[c/
2])/(24*d) - ((17*A - 29*B)*Sec[c/2 + (d*x)/2]^3*Tan[c/2])/(12*d) + ((A - B)*Sec[c/2 + (d*x)/2]^5*Tan[c/2])/(3
*d)))/(a*(1 + Cos[c + d*x]))^(7/2)

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Maple [B]  time = 0.598, size = 512, normalized size = 2.3 \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))/(a+cos(d*x+c)*a)^(7/2)/sec(d*x+c)^(3/2),x)

[Out]

1/384/d*2^(1/2)/a^4*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^5*cos(d*x+c)*(17*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*cos(d*x+c)^3+67*B*cos(d*x+c)^3*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+53*A*2^(1/2)*(cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+21*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)-17*B*2^(1/2
)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+15*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+
c)-49*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+42*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c
)*cos(d*x+c)-35*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+30*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))
*sin(d*x+c)*cos(d*x+c)-21*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+21*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*
sin(d*x+c)-15*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+15*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c))/
(1/cos(d*x+c))^(3/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)/sin(d*x+c)^11

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \sec \left (d x + c\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))/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.84914, size = 675, normalized size = 3.05 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 7 \, A + 5 \, B\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 (17 \, A + 67 \, B\right )} \cos \left (d x + c\right )^{3} + 10 \,{\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (7 \, A + 5 \, B\right )} \cos \left (d x + c\right )\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 )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(3*sqrt(2)*((7*A + 5*B)*cos(d*x + c)^4 + 4*(7*A + 5*B)*cos(d*x + c)^3 + 6*(7*A + 5*B)*cos(d*x + c)^2 +
4*(7*A + 5*B)*cos(d*x + c) + 7*A + 5*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sq
rt(a)*sin(d*x + c))) - 2*((17*A + 67*B)*cos(d*x + c)^3 + 10*(7*A + 5*B)*cos(d*x + c)^2 + 3*(7*A + 5*B)*cos(d*x
+ c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^
3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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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))/(a+a*cos(d*x+c))**(7/2)/sec(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{B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \sec \left (d x + c\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))/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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