3.4.85 \(\int \frac {1}{(a+a \cos (c+d x))^{5/2} \sec ^{\frac {7}{2}}(c+d x)} \, dx\) [385]

Optimal. Leaf size=254 \[ -\frac {5 \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{5/2} d}+\frac {115 \text {ArcTan}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {15 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {35 \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \]

[Out]

-1/4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(5/2)-15/16*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+
c)^(3/2)+35/16*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)-5*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^(5/2)/d+115/32*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^(5/2)/d*2^(1/2)

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Rubi [A]
time = 0.43, antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4307, 2844, 3056, 3062, 3061, 2861, 211, 2853, 222} \begin {gather*} -\frac {5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {115 \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 )}{16 \sqrt {2} a^{5/2} d}+\frac {35 \sin (c+d x)}{16 a^2 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {15 \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {\sin (c+d x)}{4 d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(7/2)),x]

[Out]

(-5*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^(5/2)*d)
+ (115*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]])/(16*Sqrt[2]*a^(5/2)*d) - Sin[c + d*x]/(4*d*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(5/2)
) - (15*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)) + (35*Sin[c + d*x])/(16*a^2*d*Sqr
t[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*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 222

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 2844

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 2853

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 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 3056

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 3061

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 3062

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*
(m + n + 1))), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c
*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*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] && GtQ[n, 0] &&
(IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 4307

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))^{5/2} \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))^{5/2}} \, dx\\ &=-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \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}-5 a \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {15 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \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 {45 a^2}{4}-\frac {35}{2} a^2 \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {15 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {35 \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {35 a^3}{4}+20 a^3 \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {15 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {35 \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {\left (5 \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}{2 a^3}+\frac {\left (115 \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}{32 a^2}\\ &=-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {15 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {35 \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {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^3 d}-\frac {\left (115 \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 )}{16 a d}\\ &=-\frac {5 \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^{5/2} d}+\frac {115 \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)}}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x)}-\frac {15 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)}+\frac {35 \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.03, size = 289, normalized size = 1.14 \begin {gather*} \frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (-5 i \sqrt {2} 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)}} \left (-16 i d x-16 \sinh ^{-1}\left (e^{i (c+d x)}\right )+23 \sqrt {2} \log \left (1+e^{i (c+d x)}\right )+16 \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )-23 \sqrt {2} \log \left (1-e^{i (c+d x)}+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )+\frac {1}{4} \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (16 \sin \left (\frac {1}{2} (c+d x)\right )+39 \sin \left (\frac {3}{2} (c+d x)\right )+47 \sin \left (\frac {5}{2} (c+d x)\right )+8 \sin \left (\frac {7}{2} (c+d x)\right )\right )\right )}{8 d (a (1+\cos (c+d x)))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(7/2)),x]

[Out]

(Cos[(c + d*x)/2]^5*(((-5*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*
x))]*((-16*I)*d*x - 16*ArcSinh[E^(I*(c + d*x))] + 23*Sqrt[2]*Log[1 + E^(I*(c + d*x))] + 16*Log[1 + Sqrt[1 + E^
((2*I)*(c + d*x))]] - 23*Sqrt[2]*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]]))/E^((I/2)*(
c + d*x)) + (Sec[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]]*(16*Sin[(c + d*x)/2] + 39*Sin[(3*(c + d*x))/2] + 47*Sin[(5*
(c + d*x))/2] + 8*Sin[(7*(c + d*x))/2]))/4))/(8*d*(a*(1 + Cos[c + d*x]))^(5/2))

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Maple [A]
time = 0.27, size = 352, normalized size = 1.39

method	result	size
default	\(-\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (-1+\cos \left (d x +c \right )\right )^{6} \cos \left (d x +c \right ) \left (16 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+80 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+39 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+80 \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )-20 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+115 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-35 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+115 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )\right ) \sqrt {2}}{32 d \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \sin \left (d x +c \right )^{13} a^{3}}\)	\(352\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/32/d*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^6*cos(d*x+c)*(16*cos(d*x+c)^3*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)+80*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))+39
*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+80*2^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)/cos(d*x+c))*sin(d*x+c)-20*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+115*arcsin((-1+cos(d*x
+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)-35*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+115*arcsin((-1+cos(d*x+c))
/sin(d*x+c))*sin(d*x+c))/(1/cos(d*x+c))^(7/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)/sin(d*x+c)^13*2^(1/2)/a^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate(1/((a*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^(7/2)), x)

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Fricas [A]
time = 0.70, size = 245, normalized size = 0.96 \begin {gather*} -\frac {115 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \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 ) - 160 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \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 (16 \, \cos \left (d x + c\right )^{3} + 55 \, \cos \left (d x + c\right )^{2} + 35 \, \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 )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/32*(115*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*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))) - 160*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x
+ c) + 1)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 2*(16*cos(d*x +
c)^3 + 55*cos(d*x + c)^2 + 35*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*
cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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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(1/(a+a*cos(d*x+c))**(5/2)/sec(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Evaluation time:
167Not invertible Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 )}^{5/2}} \,d x \end {gather*}

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))^(5/2)),x)

[Out]

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

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