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

Optimal. Leaf size=221 \[ \frac {(13 A+7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}-\frac {(5 A-17 B) \sin (c+d x)}{192 a^2 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}+\frac {(A+3 B) \sin (c+d x)}{16 a d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}+\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{7/2}} \]

[Out]

1/6*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(1/2)+1/16*(A+3*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(5
/2)/sec(d*x+c)^(1/2)-1/192*(5*A-17*B)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2)+1/128*(13*A+7*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)

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Rubi [A]  time = 0.73, antiderivative size = 221, normalized size of antiderivative = 1.00, 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} \[ -\frac {(5 A-17 B) \sin (c+d x)}{192 a^2 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}+\frac {(13 A+7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {(A+3 B) \sin (c+d x)}{16 a d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}+\frac {(A-B) \sin (c+d x)}{6 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((13*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]])/(64*Sqrt[2]*a^(7/2)*d) + ((A - B)*Sin[c + d*x])/(6*d*(a + a*Cos[c + d*x])^(7/2)*Sq
rt[Sec[c + d*x]]) + ((A + 3*B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]) - ((5*A -
17*B)*Sin[c + d*x])/(192*a^2*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[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 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])

Rubi steps

\begin {align*} \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (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} \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (A-B)+2 a (A+2 B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (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} \sqrt {\sec (c+d x)}}+\frac {(A+3 B) \sin (c+d x)}{16 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 (11 A+B)+\frac {3}{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} \sqrt {\sec (c+d x)}}+\frac {(A+3 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {(5 A-17 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 (13 A+7 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} \sqrt {\sec (c+d x)}}+\frac {(A+3 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {(5 A-17 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {\left ((13 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}{128 a^3}\\ &=\frac {(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}}+\frac {(A+3 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {(5 A-17 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {\left ((13 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 )}{64 a^2 d}\\ &=\frac {(13 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)}}{64 \sqrt {2} a^{7/2} d}+\frac {(A-B) \sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}}+\frac {(A+3 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {(5 A-17 B) \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 2.95, size = 233, normalized size = 1.05 \[ \frac {\cos ^7\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right ) \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} (4 (A+35 B) \cos (c+d x)+(17 B-5 A) \cos (2 (c+d x))+73 A+59 B)}{48 d}+\frac {i (13 A+7 B) e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{d}\right )}{8 (a (\cos (c+d x)+1))^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[(c + d*x)/2]^7*((I*(13*A + 7*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))) - ((73*A
+ 59*B + 4*(A + 35*B)*Cos[c + d*x] + (-5*A + 17*B)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^6*Sqrt[Sec[c + d*x]]*(Si
n[(c + d*x)/2] - Sin[(3*(c + d*x))/2]))/(48*d)))/(8*(a*(1 + Cos[c + d*x]))^(7/2))

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fricas [A]  time = 2.39, size = 255, normalized size = 1.15 \[ -\frac {3 \, \sqrt {2} {\left ({\left (13 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (13 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (13 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (13 \, A + 7 \, B\right )} \cos \left (d x + c\right ) + 13 \, A + 7 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (5 \, A - 17 \, B\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (A + 35 \, B\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (13 \, A + 7 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

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)^(1/2),x, algorithm="fricas")

[Out]

-1/384*(3*sqrt(2)*((13*A + 7*B)*cos(d*x + c)^4 + 4*(13*A + 7*B)*cos(d*x + c)^3 + 6*(13*A + 7*B)*cos(d*x + c)^2
+ 4*(13*A + 7*B)*cos(d*x + c) + 13*A + 7*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c)
)/(sqrt(a)*sin(d*x + c))) + 2*((5*A - 17*B)*cos(d*x + c)^3 - 2*(A + 35*B)*cos(d*x + c)^2 - 3*(13*A + 7*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \sqrt {\sec \left (d x + c\right )}}\,{d x} \]

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.39, size = 512, normalized size = 2.32 \[ -\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \cos \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right )^{4} \left (-5 A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right )+17 B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right )+39 A \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )+7 A \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+21 B \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )+53 B \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+78 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 )+37 A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+42 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 )-49 B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+39 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-39 A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+21 B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-21 B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {2}}{384 d \sqrt {\frac {1}{\cos \left (d x +c \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 )^{9} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384/d*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^4*(-5*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*cos(d*x+c)^3+17*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3+39*A*cos(d*x+c)^2*arcsin((-1+cos(d*x
+c))/sin(d*x+c))*sin(d*x+c)+7*A*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+21*B*cos(d*x+c)^2*arcsi
n((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)+53*B*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+78*A*arcs
in((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)+37*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)
+42*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)-49*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
cos(d*x+c)+39*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)-39*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+2
1*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)-21*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))/(1/cos(d*x+c
))^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)/sin(d*x+c)^9*2^(1/2)/a^4

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

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)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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