3.1330 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx\)

Optimal. Leaf size=242 \[ \frac {2 a^{5/2} C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^3 (160 A+224 B+245 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (40 A+56 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{105 d}+\frac {2 a (5 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d} \]

[Out]

2/35*a*(5*A+7*B)*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/7*A*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^
(7/2)*sin(d*x+c)/d+2/105*a^2*(40*A+56*B+35*C)*sec(d*x+c)^(3/2)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+2*a^(5/2)*C
*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)/d+2/105*a^3*(160*A+224*B+
245*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.88, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4221, 3043, 2975, 2980, 2774, 216} \[ \frac {2 a^2 (40 A+56 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{105 d}+\frac {2 a^3 (160 A+224 B+245 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^{5/2} C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a (5 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{35 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2),x]

[Out]

(2*a^(5/2)*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d
+ (2*a^3*(160*A + 224*B + 245*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(4
0*A + 56*B + 35*C)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d) + (2*a*(5*A + 7*B)*(a + a
*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(35*d) + (2*A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(7
/2)*Sin[c + d*x])/(7*d)

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 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 2975

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

Rule 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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[n, -1]

Rule 3043

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

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 (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{5/2} \left (\frac {1}{2} a (5 A+7 B)+\frac {7}{2} a C \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac {2 a (5 A+7 B) (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{3/2} \left (\frac {1}{4} a^2 (40 A+56 B+35 C)+\frac {35}{4} a^2 C \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{35 a}\\ &=\frac {2 a^2 (40 A+56 B+35 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a (5 A+7 B) (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{8} a^3 (160 A+224 B+245 C)+\frac {105}{8} a^3 C \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{105 a}\\ &=\frac {2 a^3 (160 A+224 B+245 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (40 A+56 B+35 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a (5 A+7 B) (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\left (a^2 C \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\\ &=\frac {2 a^3 (160 A+224 B+245 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (40 A+56 B+35 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a (5 A+7 B) (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}-\frac {\left (2 a^2 C \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 )}{d}\\ &=\frac {2 a^{5/2} C \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)}}{d}+\frac {2 a^3 (160 A+224 B+245 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (40 A+56 B+35 C) \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac {2 a (5 A+7 B) (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 1.64, size = 172, normalized size = 0.71 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a (\cos (c+d x)+1)} \left (2 \sin \left (\frac {1}{2} (c+d x)\right ) ((930 A+987 B+840 C) \cos (c+d x)+2 (115 A+98 B+35 C) \cos (2 (c+d x))+230 A \cos (3 (c+d x))+290 A+301 B \cos (3 (c+d x))+196 B+280 C \cos (3 (c+d x))+70 C)+420 \sqrt {2} C \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {7}{2}}(c+d x)\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(7/2)*(420*Sqrt[2]*C*ArcSin[Sqrt[2]*Sin[(c + d*x
)/2]]*Cos[c + d*x]^(7/2) + 2*(290*A + 196*B + 70*C + (930*A + 987*B + 840*C)*Cos[c + d*x] + 2*(115*A + 98*B +
35*C)*Cos[2*(c + d*x)] + 230*A*Cos[3*(c + d*x)] + 301*B*Cos[3*(c + d*x)] + 280*C*Cos[3*(c + d*x)])*Sin[(c + d*
x)/2]))/(420*d)

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fricas [A]  time = 0.51, size = 188, normalized size = 0.78 \[ -\frac {2 \, {\left (105 \, {\left (C a^{2} \cos \left (d x + c\right )^{4} + C a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left ({\left (230 \, A + 301 \, B + 280 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (115 \, A + 98 \, B + 35 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (20 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

-2/105*(105*(C*a^2*cos(d*x + c)^4 + C*a^2*cos(d*x + c)^3)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x
+ c))/(sqrt(a)*sin(d*x + c))) - ((230*A + 301*B + 280*C)*a^2*cos(d*x + c)^3 + (115*A + 98*B + 35*C)*a^2*cos(d
*x + c)^2 + 3*(20*A + 7*B)*a^2*cos(d*x + c) + 15*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c
)))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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giac [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+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.52, size = 522, normalized size = 2.16 \[ -\frac {2 \left (105 C \left (\cos ^{4}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{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 )+420 C \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{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 )+630 C \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{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 )+420 C \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{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 )+105 C \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{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 )+230 A \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+301 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+280 C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+115 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+98 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+35 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+60 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+21 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+15 A \sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \left (\sin ^{6}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{105 d \left (-1+\cos \left (d x +c \right )\right )^{3} \left (1+\cos \left (d x +c \right )\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2),x)

[Out]

-2/105/d*(105*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)/cos(d*x+c))+420*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)/cos(d*x+c))+630*C*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)/cos(d*x+c))+420*C*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctan(sin(d*x+c)*(cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+105*C*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)/cos(d*x+c))+230*A*cos(d*x+c)^3*sin(d*x+c)+301*B*sin(d*x+c)*cos(d*x+c)^3+280*C*sin(d*x+c)*cos
(d*x+c)^3+115*A*cos(d*x+c)^2*sin(d*x+c)+98*B*sin(d*x+c)*cos(d*x+c)^2+35*C*sin(d*x+c)*cos(d*x+c)^2+60*A*cos(d*x
+c)*sin(d*x+c)+21*B*cos(d*x+c)*sin(d*x+c)+15*A*sin(d*x+c))*cos(d*x+c)*sin(d*x+c)^6*(1/cos(d*x+c))^(9/2)*(a*(1+
cos(d*x+c)))^(1/2)/(-1+cos(d*x+c))^3/(1+cos(d*x+c))^4*a^2

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maxima [B]  time = 0.94, size = 2584, normalized size = 10.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

1/210*(7*(6*(a^2*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 25*(a^2*cos(2*d*x + 2*c)^2 + a^2*s
in(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(c
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*sqrt(a) + 2*(cos(2*d*x + 2*c)^2 + sin(2
*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((15*a^2*sin(6*d*x + 6*c) + 50*a^2*sin(4*d*x + 4*c) + 58*a^2*sin
(2*d*x + 2*c) - 20*(3*a^2*sin(6*d*x + 6*c) + 10*a^2*sin(4*d*x + 4*c) + 11*a^2*sin(2*d*x + 2*c))*cos(7/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 20*(3*a^2*cos(6*d*x + 6*c) + 10*a^2*cos(4*d*x + 4*c) + 11*a^2*cos(2*d
*x + 2*c) + 4*a^2)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c) + 1)) - (15*a^2*cos(6*d*x + 6*c) + 50*a^2*cos(4*d*x + 4*c) + 58*a^2*cos(2*d*x + 2*c) + 23*a^2 +
20*(3*a^2*cos(6*d*x + 6*c) + 10*a^2*cos(4*d*x + 4*c) + 11*a^2*cos(2*d*x + 2*c) + 4*a^2)*cos(7/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c))) + 20*(3*a^2*sin(6*d*x + 6*c) + 10*a^2*sin(4*d*x + 4*c) + 11*a^2*sin(2*d*x + 2*c
))*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1)) + 25*(a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*sin(3/2*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 15*((a^2*cos(2*d*x + 2*c)^4 + a^2*sin(2*d*x + 2*c)^4 + 4*a^2*c
os(2*d*x + 2*c)^3 + 6*a^2*cos(2*d*x + 2*c)^2 + 4*a^2*cos(2*d*x + 2*c) + 2*(a^2*cos(2*d*x + 2*c)^2 + 2*a^2*cos(
2*d*x + 2*c) + a^2)*sin(2*d*x + 2*c)^2 + a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arct
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 1) - (a^2
*cos(2*d*x + 2*c)^4 + a^2*sin(2*d*x + 2*c)^4 + 4*a^2*cos(2*d*x + 2*c)^3 + 6*a^2*cos(2*d*x + 2*c)^2 + 4*a^2*cos
(2*d*x + 2*c) + 2*(a^2*cos(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*sin(2*d*x + 2*c)^2 + a^2)*arctan2((c
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), co
s(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x +
2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arc
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) - 1) - (a^2*cos(2*d*x + 2*c)^4 + a^2*sin(2*d*x + 2*c)^4 + 4*a^2*cos
(2*d*x + 2*c)^3 + 6*a^2*cos(2*d*x + 2*c)^2 + 4*a^2*cos(2*d*x + 2*c) + 2*(a^2*cos(2*d*x + 2*c)^2 + 2*a^2*cos(2*
d*x + 2*c) + a^2)*sin(2*d*x + 2*c)^2 + a^2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2
*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c
)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + (a^2*cos(2
*d*x + 2*c)^4 + a^2*sin(2*d*x + 2*c)^4 + 4*a^2*cos(2*d*x + 2*c)^3 + 6*a^2*cos(2*d*x + 2*c)^2 + 4*a^2*cos(2*d*x
+ 2*c) + 2*(a^2*cos(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*sin(2*d*x + 2*c)^2 + a^2)*arctan2((cos(2*d
*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c) + 1)) - 1))*sqrt(a))*C/(cos(2*d*x + 2*c)^4 + sin(2*d*x + 2*c)^4 + 4*cos(2*d*x + 2*c
)^3 + 2*(cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 6*cos(2*d*x + 2*c)^2 + 4*cos(2*d*x
+ 2*c) + 1) + 80*(21*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 56*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 + 63*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 36*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(
cos(d*x + c) + 1)^7 + 8*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*A*(sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 1)^2/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(2*sin(
d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + 112*(15*sqrt(2)*a^(5/2)*sin(d*x
+ c)/(cos(d*x + c) + 1) - 50*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sqrt(2)*a^(5/2)*sin(d*x
+ c)^5/(cos(d*x + c) + 1)^5 - 36*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 8*sqrt(2)*a^(5/2)*sin(d
*x + c)^9/(cos(d*x + c) + 1)^9)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x + c)/(cos(d*x + c) + 1
) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c
)^4/(cos(d*x + c) + 1)^4 + 1)))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

int((1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^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+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(9/2),x)

[Out]

Timed out

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