\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\) [322]
Optimal result
Integrand size = 43, antiderivative size = 193 \[
\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {(3 A+4 C) \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{8 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(3 A+4 C) \sin (c+d x)}{8 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {B \sin ^3(c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}
\]
[Out]
1/4*A*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2)+1/8*(3*A+4*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(b*cos(d*
x+c))^(1/2)+1/3*B*sin(d*x+c)^3/d/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2)+B*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(b*cos(
d*x+c))^(1/2)+1/8*(3*A+4*C)*arctanh(sin(d*x+c))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)
Rubi [A] (verified)
Time = 0.13 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {18, 3100, 2827, 3852, 3853,
3855} \[
\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {(3 A+4 C) \sqrt {\cos (c+d x)} \text {arctanh}(\sin (c+d x))}{8 d \sqrt {b \cos (c+d x)}}+\frac {(3 A+4 C) \sin (c+d x)}{8 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin ^3(c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}
\]
[In]
Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*Sqrt[b*Cos[c + d*x]]),x]
[Out]
((3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(8*d*Sqrt[b*Cos[c + d*x]]) + (A*Sin[c + d*x])/(4*d*Cos[
c + d*x]^(7/2)*Sqrt[b*Cos[c + d*x]]) + ((3*A + 4*C)*Sin[c + d*x])/(8*d*Cos[c + d*x]^(3/2)*Sqrt[b*Cos[c + d*x]]
) + (B*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]) + (B*Sin[c + d*x]^3)/(3*d*Cos[c + d*x]^(5/2)*
Sqrt[b*Cos[c + d*x]])
Rule 18
Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[a^(m - 1/2)*b^(n + 1/2)*(Sqrt[a*v]/Sqrt[b*v])
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] && !IntegerQ[m] && ILtQ[n - 1/2, 0] && IntegerQ[m + n]
Rule 2827
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 3100
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Rule 3852
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Rule 3853
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]
Rule 3855
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps \begin{align*}
\text {integral}& = \frac {\sqrt {\cos (c+d x)} \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx}{\sqrt {b \cos (c+d x)}} \\ & = \frac {A \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \int (4 B+(3 A+4 C) \cos (c+d x)) \sec ^4(c+d x) \, dx}{4 \sqrt {b \cos (c+d x)}} \\ & = \frac {A \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\left (B \sqrt {\cos (c+d x)}\right ) \int \sec ^4(c+d x) \, dx}{\sqrt {b \cos (c+d x)}}+\frac {\left ((3 A+4 C) \sqrt {\cos (c+d x)}\right ) \int \sec ^3(c+d x) \, dx}{4 \sqrt {b \cos (c+d x)}} \\ & = \frac {A \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(3 A+4 C) \sin (c+d x)}{8 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\left ((3 A+4 C) \sqrt {\cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{8 \sqrt {b \cos (c+d x)}}-\frac {\left (B \sqrt {\cos (c+d x)}\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d \sqrt {b \cos (c+d x)}} \\ & = \frac {(3 A+4 C) \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{8 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(3 A+4 C) \sin (c+d x)}{8 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {B \sin ^3(c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.57
\[
\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\frac {3 (3 A+4 C) \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+\sin (c+d x) \left (6 A+3 (3 A+4 C) \cos ^2(c+d x)+24 B \cos ^3(c+d x)+8 B \cos (c+d x) \sin ^2(c+d x)\right )}{24 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}
\]
[In]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*Sqrt[b*Cos[c + d*x]]),x]
[Out]
(3*(3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^4 + Sin[c + d*x]*(6*A + 3*(3*A + 4*C)*Cos[c + d*x]^2 + 24*B*
Cos[c + d*x]^3 + 8*B*Cos[c + d*x]*Sin[c + d*x]^2))/(24*d*Cos[c + d*x]^(7/2)*Sqrt[b*Cos[c + d*x]])
Maple [A] (verified)
Time = 10.16 (sec) , antiderivative size = 212, normalized size of antiderivative =
1.10
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method | result | size |
| | |
default |
\(\frac {-9 A \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+9 A \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-12 C \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+12 C \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+16 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+9 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+12 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+8 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+6 A \sin \left (d x +c \right )}{24 d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {7}{2}}}\) |
\(212\) |
parts |
\(-\frac {A \left (3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )-3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2 \sin \left (d x +c \right )\right )}{8 d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {7}{2}}}+\frac {B \left (2 \left (\cos ^{2}\left (d x +c \right )\right )+1\right ) \sin \left (d x +c \right )}{3 d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {5}{2}}}+\frac {C \left (-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\sin \left (d x +c \right )\right )}{2 d \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {3}{2}}}\) |
\(231\) |
risch |
\(-\frac {i \left (9 A \,{\mathrm e}^{6 i \left (d x +c \right )}+12 C \,{\mathrm e}^{6 i \left (d x +c \right )}+33 A \,{\mathrm e}^{4 i \left (d x +c \right )}+12 C \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B \,{\mathrm e}^{3 i \left (d x +c \right )}-33 A \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-9 A -12 C -80 B \cos \left (d x +c \right )-48 i B \sin \left (d x +c \right )\right )}{24 \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (3 A +4 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 \sqrt {\cos \left (d x +c \right ) b}\, d}\) |
\(235\) |
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[In]
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/24/d*(-9*A*cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)-1)+9*A*cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)+1)-12*C*cos(
d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)-1)+12*C*cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)+1)+16*B*sin(d*x+c)*cos(d*x+c
)^3+9*A*sin(d*x+c)*cos(d*x+c)^2+12*C*cos(d*x+c)^2*sin(d*x+c)+8*B*sin(d*x+c)*cos(d*x+c)+6*A*sin(d*x+c))/(cos(d*
x+c)*b)^(1/2)/cos(d*x+c)^(7/2)
Fricas [A] (verification not implemented)
none
Time = 0.32 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.58
\[
\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\left [\frac {3 \, {\left (3 \, A + 4 \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (16 \, B \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, b d \cos \left (d x + c\right )^{5}}, -\frac {3 \, {\left (3 \, A + 4 \, C\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - {\left (16 \, B \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, b d \cos \left (d x + c\right )^{5}}\right ]
\]
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
[1/48*(3*(3*A + 4*C)*sqrt(b)*cos(d*x + c)^5*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d
*x + c))*sin(d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*(16*B*cos(d*x + c)^3 + 3*(3*A + 4*C)*cos(d*x + c
)^2 + 8*B*cos(d*x + c) + 6*A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c)^5), -1/2
4*(3*(3*A + 4*C)*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d*x +
c)^5 - (16*B*cos(d*x + c)^3 + 3*(3*A + 4*C)*cos(d*x + c)^2 + 8*B*cos(d*x + c) + 6*A)*sqrt(b*cos(d*x + c))*sqrt
(cos(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c)^5)]
Sympy [F(-1)]
Timed out. \[
\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Timed out}
\]
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2)/(b*cos(d*x+c))**(1/2),x)
[Out]
Timed out
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 2611 vs. \(2 (165) = 330\).
Time = 0.57 (sec) , antiderivative size = 2611, normalized size of antiderivative = 13.53
\[
\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\text {Too large to display}
\]
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
-1/48*(3*(12*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(7/2*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 44*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin
(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c
) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 12*(sin(8*
d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), c
os(2*d*x + 2*c))) - 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) +
cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + 16*cos(6*d*x + 6*c)^2
+ 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d
*x + 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 8*c)^2 + 16*(3*sin(4*d*x +
4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16*sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x +
4*c)*sin(2*d*x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) + 1) + 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*
d*x + 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + 16*cos(6*
d*x + 6*c)^2 + 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 +
4*(2*sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 8*c)^2 + 16*(3
*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16*sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*
sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sin(1/2*arctan2(
sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 12*(cos(8*d*x + 8*c) + 4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4
*cos(2*d*x + 2*c) + 1)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(cos(8*d*x + 8*c) + 4*cos(6*d
*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))
+ 44*(cos(8*d*x + 8*c) + 4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*sin(3/2*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c))) + 12*(cos(8*d*x + 8*c) + 4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d
*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*A/((2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x +
4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x +
2*c) + 1)*cos(6*d*x + 6*c) + 16*cos(6*d*x + 6*c)^2 + 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*cos(4*d
*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d
*x + 8*c) + sin(8*d*x + 8*c)^2 + 16*(3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16*sin(6*d*x
+ 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2*d*x
+ 2*c) + 1)*sqrt(b)) - 64*((3*cos(2*d*x + 2*c) + 1)*sin(6*d*x + 6*c) + 3*(3*cos(2*d*x + 2*c) + 1)*sin(4*d*x +
4*c) - 3*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 9*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*B/((2*(3*cos(4*d*x + 4*c) +
3*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) +
9*cos(4*d*x + 4*c)^2 + 9*cos(2*d*x + 2*c)^2 + 6*(sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6
*d*x + 6*c)^2 + 9*sin(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sin(2*d*x + 2*c)^2 + 6*cos(2*d
*x + 2*c) + 1)*sqrt(b)) + 12*(4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c))) - 4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))
) - (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c
)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x +
4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^
2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(cos(4*d*x
+ 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) +
2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C/((2*(2*cos(2*d*x + 2*c) + 1)*
cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d
*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*sqrt(b)))/d
Giac [F]
\[
\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\sqrt {b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{\frac {9}{2}}} \,d x }
\]
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/(sqrt(b*cos(d*x + c))*cos(d*x + c)^(9/2)), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{9/2}\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x
\]
[In]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(9/2)*(b*cos(c + d*x))^(1/2)),x)
[Out]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(9/2)*(b*cos(c + d*x))^(1/2)), x)