3.1246 \(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=315 \[ -\frac {(283 A+75 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(157 A+45 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{80 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(787 A+195 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{240 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(2671 A+735 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{240 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(21 A+5 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-1/4*(A+C)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-1/16*(21*A+5*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/
d/(a+a*cos(d*x+c))^(3/2)-1/240*(787*A+195*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)+1/80*(15
7*A+45*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)-1/32*(283*A+75*C)*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)+1/24
0*(2671*A+735*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a^2/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 1.09, antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {4221, 3042, 2978, 2984, 12, 2782, 205} \[ \frac {(157 A+45 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{80 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(787 A+195 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{240 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(2671 A+735 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{240 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(283 A+75 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(21 A+5 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((283*A + 75*C)*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) + ((2671*A + 735*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(24
0*a^2*d*Sqrt[a + a*Cos[c + d*x]]) - ((787*A + 195*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(240*a^2*d*Sqrt[a + a*Co
s[c + d*x]]) - ((A + C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((21*A + 5*C)*Sec[
c + d*x]^(5/2)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + ((157*A + 45*C)*Sec[c + d*x]^(5/2)*Sin[c +
d*x])/(80*a^2*d*Sqrt[a + a*Cos[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 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 2984

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

Rule 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 4221

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

Rubi steps

\begin {align*} \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\\ &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (13 A+5 C)-4 a A \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a^2 (157 A+45 C)-\frac {3}{2} a^2 (21 A+5 C) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A+45 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a^3 (787 A+195 C)+\frac {1}{2} a^3 (157 A+45 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{20 a^5}\\ &=-\frac {(787 A+195 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A+45 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{16} a^4 (2671 A+735 C)-\frac {1}{8} a^4 (787 A+195 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{30 a^6}\\ &=\frac {(2671 A+735 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(787 A+195 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A+45 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int -\frac {15 a^5 (283 A+75 C)}{32 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^7}\\ &=\frac {(2671 A+735 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(787 A+195 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A+45 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\left ((283 A+75 C) \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 {(2671 A+735 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(787 A+195 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A+45 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\left ((283 A+75 C) \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 {(283 A+75 C) \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 {(2671 A+735 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(787 A+195 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A+45 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 8.09, size = 261, normalized size = 0.83 \[ \frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) (50 (521 A+153 C) \cos (c+d x)+108 (157 A+45 C) \cos (2 (c+d x))+9110 A \cos (3 (c+d x))+2671 A \cos (4 (c+d x))+15053 A+2550 C \cos (3 (c+d x))+735 C \cos (4 (c+d x))+4125 C)-240 i (283 A+75 C) e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )\right )}{960 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*(((-240*I)*(283*A + 75*C)*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))])])/E^((I/2)*(c + d*x)) + (1
5053*A + 4125*C + 50*(521*A + 153*C)*Cos[c + d*x] + 108*(157*A + 45*C)*Cos[2*(c + d*x)] + 9110*A*Cos[3*(c + d*
x)] + 2550*C*Cos[3*(c + d*x)] + 2671*A*Cos[4*(c + d*x)] + 735*C*Cos[4*(c + d*x)])*Sec[(c + d*x)/2]^3*Sec[c + d
*x]^(5/2)*Tan[(c + d*x)/2]))/(960*d*(a*(1 + Cos[c + d*x]))^(5/2))

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fricas [A]  time = 0.51, size = 262, normalized size = 0.83 \[ \frac {15 \, \sqrt {2} {\left ({\left (283 \, A + 75 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (283 \, A + 75 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (283 \, A + 75 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (283 \, A + 75 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (2671 \, A + 735 \, C\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (911 \, A + 255 \, C\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (49 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2} - 160 \, A \cos \left (d x + c\right ) + 96 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{480 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(15*sqrt(2)*((283*A + 75*C)*cos(d*x + c)^5 + 3*(283*A + 75*C)*cos(d*x + c)^4 + 3*(283*A + 75*C)*cos(d*x
+ c)^3 + (283*A + 75*C)*cos(d*x + c)^2)*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*((2671*A + 735*C)*cos(d*x + c)^4 + 5*(911*A + 255*C)*cos(d*x + c)^3 + 32*(49*A + 15*C
)*cos(d*x + c)^2 - 160*A*cos(d*x + c) + 96*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d
*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.65, size = 717, normalized size = 2.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480/d*(-4245*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-
1125*C*cos(d*x+c)^4*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-16980*A*ar
csin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-4500*C*sin(d*x+c)*c
os(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-25470*A*arcsin((-1+cos(d*x+c)
)/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-6750*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*
x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-16980*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d
*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-4500*C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5
/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-4245*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos
(d*x+c)))^(5/2)-1125*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+2671*A*
2^(1/2)*cos(d*x+c)^5+735*C*2^(1/2)*cos(d*x+c)^5+1884*A*2^(1/2)*cos(d*x+c)^4+540*C*2^(1/2)*cos(d*x+c)^4-2987*A*
2^(1/2)*cos(d*x+c)^3-795*C*2^(1/2)*cos(d*x+c)^3-1728*A*2^(1/2)*cos(d*x+c)^2-480*C*2^(1/2)*cos(d*x+c)^2+256*A*2
^(1/2)*cos(d*x+c)-96*A*2^(1/2))*cos(d*x+c)*(1/cos(d*x+c))^(7/2)*(a*(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/(-1+cos(d*
x+c))/(1+cos(d*x+c))^3*2^(1/2)/a^3

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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