3.392 \(\int \frac {1}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=254 \[ \frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{7/2} d}-\frac {177 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}-\frac {49 \sin (c+d x)}{64 a^2 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac {17 \sin (c+d x)}{48 a d \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}-\frac {\sin (c+d x)}{6 d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]

[Out]

-1/6*sin(d*x+c)/d/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(5/2)-17/48*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+
c)^(3/2)-49/64*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2)+2*arcsin(sin(d*x+c)*a^(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-177/128*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 = 0.67, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4222, 2765, 2977, 2982, 2782, 205, 2774, 216} \[ \frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{7/2} d}-\frac {49 \sin (c+d x)}{64 a^2 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac {177 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}-\frac {17 \sin (c+d x)}{48 a d \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}-\frac {\sin (c+d x)}{6 d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(7/2)),x]

[Out]

(2*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(a^(7/2)*d)
- (177*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) - Sin[c + d*x]/(6*d*(a + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(5/2))
- (17*Sin[c + d*x])/(48*a*d*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(3/2)) - (49*Sin[c + d*x])/(64*a^2*d*(a +
a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(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 - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], 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] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, 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 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 2982

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*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]

Rule 4222

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps

\begin {align*} \int \frac {1}{(a+a \cos (c+d x))^{7/2} \sec ^{\frac {7}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{7/2}} \, dx\\ &=-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5 a}{2}-6 a \cos (c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {17 \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \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 {51 a^2}{4}-24 a^2 \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {17 \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}-\frac {49 \sin (c+d x)}{64 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 {\frac {147 a^3}{8}-48 a^3 \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {17 \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}-\frac {49 \sin (c+d x)}{64 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 {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{a^4}-\frac {\left (177 \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 {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {17 \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}-\frac {49 \sin (c+d x)}{64 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^4 d}+\frac {\left (177 \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 {2 \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{7/2} d}-\frac {177 \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 {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {17 \sin (c+d x)}{48 a d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {3}{2}}(c+d x)}-\frac {49 \sin (c+d x)}{64 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 = 6.79, size = 454, normalized size = 1.79 \[ \frac {\cos ^7\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\sec (c+d x)} \left (-\frac {247 \sin \left (\frac {c}{2}\right ) \cos \left (\frac {d x}{2}\right )}{12 d}-\frac {247 \cos \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )}{12 d}+\frac {\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d}+\frac {\tan \left (\frac {c}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d}-\frac {41 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{12 d}-\frac {41 \tan \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{12 d}+\frac {379 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{24 d}+\frac {379 \tan \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right )}{24 d}\right )}{(a (\cos (c+d x)+1))^{7/2}}-\frac {i e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^7\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (64 \sinh ^{-1}\left (e^{i (c+d x)}\right )+\frac {177 \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{\sqrt {2}}-64 \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{4 \sqrt {2} d (a (\cos (c+d x)+1))^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(7/2)),x]

[Out]

((-1/4*I)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(64*ArcSinh[E^(I*(c +
d*x))] + (177*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[2] - 64*ArcTanh[Sqr
t[1 + E^((2*I)*(c + d*x))]])*Cos[c/2 + (d*x)/2]^7)/(Sqrt[2]*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]]*((-247*Cos[(d*x)/2]*Sin[c/2])/(12*d) - (247*Cos[c/2]*Sin[(d*x)/2]
)/(12*d) + (379*Sec[c/2]*Sec[c/2 + (d*x)/2]^2*Sin[(d*x)/2])/(24*d) - (41*Sec[c/2]*Sec[c/2 + (d*x)/2]^4*Sin[(d*
x)/2])/(12*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]^6*Sin[(d*x)/2])/(3*d) + (379*Sec[c/2 + (d*x)/2]*Tan[c/2])/(24*d)
- (41*Sec[c/2 + (d*x)/2]^3*Tan[c/2])/(12*d) + (Sec[c/2 + (d*x)/2]^5*Tan[c/2])/(3*d)))/(a*(1 + Cos[c + d*x]))^(
7/2)

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fricas [A]  time = 2.29, size = 279, normalized size = 1.10 \[ \frac {531 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 768 \, {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (247 \, \cos \left (d x + c\right )^{3} + 362 \, \cos \left (d x + c\right )^{2} + 147 \, \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/384*(531*sqrt(2)*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*sqrt(a)*arctan(
sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 768*(cos(d*x + c)^4 + 4*cos(d*x
+ c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sq
rt(a)*sin(d*x + c))) - 2*(247*cos(d*x + c)^3 + 362*cos(d*x + c)^2 + 147*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 \[ \int \frac {1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \sec \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.26, size = 440, normalized size = 1.73 \[ -\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (-1+\cos \left (d x +c \right )\right )^{7} \cos \left (d x +c \right ) \left (384 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+247 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+768 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+115 \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 )}}+531 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+384 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \sqrt {2}\, \sin \left (d x +c \right )-215 \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+1062 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-147 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+531 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )\right ) \sqrt {2}}{384 d \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \sin \left (d x +c \right )^{15} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384/d*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^7*cos(d*x+c)*(384*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)/cos(d*x+c))*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+247*cos(d*x+c)^3*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)+768*cos(d*x+c)*sin(d*x+c)*2^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+115*cos(d
*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+531*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+
c)+384*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*2^(1/2)*sin(d*x+c)-215*cos(d*x+c)*2^(1/
2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+1062*cos(d*x+c)*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-147*2^(1/2)
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+531*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c))/(1/cos(d*x+c))^(7/2)/(co
s(d*x+c)/(1+cos(d*x+c)))^(9/2)/sin(d*x+c)^15*2^(1/2)/a^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \sec \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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