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

Optimal. Leaf size=254 \[ -\frac{5 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac{35 \sin (c+d x)}{16 a^2 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{115 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{15 \sin (c+d x)}{16 a d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{\sin (c+d x)}{4 d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

[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]])

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Rubi [A]  time = 0.666436, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {4222, 2765, 2977, 2983, 2982, 2782, 205, 2774, 216} \[ -\frac{5 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac{35 \sin (c+d x)}{16 a^2 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{115 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{15 \sin (c+d x)}{16 a d \sec ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac{\sin (c+d x)}{4 d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \]

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 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]

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 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 2983

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] && (I
ntegerQ[n] || EqQ[m + 1/2, 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 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]

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 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]

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 ) \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^3 d}-\frac{\left (115 \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 )}{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*}

Mathematica [C]  time = 3.17602, size = 412, normalized size = 1.62 \[ \frac{e^{-\frac{1}{2} i (c+d x)} \left (40 i \sqrt{2} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \cos ^5\left (\frac{1}{2} (c+d x)\right ) \sinh ^{-1}\left (e^{i (c+d x)}\right )+115 i \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \cos ^5\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-\frac{1}{16} i e^{-3 i (c+d x)} \cos \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \left (-47 e^{i (c+d x)}-39 e^{2 i (c+d x)}-16 e^{3 i (c+d x)}+16 e^{4 i (c+d x)}+39 e^{5 i (c+d x)}+47 e^{6 i (c+d x)}+8 e^{7 i (c+d x)}+40 e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^4 \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-8\right )\right )}{4 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((40*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c
+ d*x))]*Cos[(c + d*x)/2]^5 + (115*I)*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 + d*x)/2]^5 - ((I/16)*(-8
- 47*E^(I*(c + d*x)) - 39*E^((2*I)*(c + d*x)) - 16*E^((3*I)*(c + d*x)) + 16*E^((4*I)*(c + d*x)) + 39*E^((5*I)*
(c + d*x)) + 47*E^((6*I)*(c + d*x)) + 8*E^((7*I)*(c + d*x)) + 40*E^(I*(c + d*x))*(1 + E^(I*(c + d*x)))^4*Sqrt[
1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Cos[(c + d*x)/2]*Sqrt[Sec[c + d*x]])/E^((3*I)
*(c + d*x)))/(4*d*E^((I/2)*(c + d*x))*(a*(1 + Cos[c + d*x]))^(5/2))

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Maple [A]  time = 0.463, size = 352, normalized size = 1.4 \begin{align*} -{\frac{\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) ^{6}\cos \left ( dx+c \right ) }{32\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{13}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( 16\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+39\,\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+80\,\cos \left ( dx+c \right ) \sqrt{2}\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sin \left ( dx+c \right ) +115\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -20\,\sqrt{2}\cos \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+80\,\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sqrt{2}\sin \left ( dx+c \right ) +115\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) -35\,\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{-{\frac{7}{2}}} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

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 = 3.76728, size = 693, normalized size = 2.73 \begin{align*} -\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{align*}

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)]  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(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

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]

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