3.4.37 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx\) [337]

Optimal. Leaf size=164 \[ \frac {B \tanh ^{-1}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 b^2 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{3 b^2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{2 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(2 A+3 C) \sin (c+d x)}{3 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]

[Out]

1/3*A*sin(d*x+c)/b^2/d/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(1/2)+1/2*B*sin(d*x+c)/b^2/d/cos(d*x+c)^(3/2)/(b*cos(d*
x+c))^(1/2)+1/3*(2*A+3*C)*sin(d*x+c)/b^2/d/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)+1/2*B*arctanh(sin(d*x+c))*cos
(d*x+c)^(1/2)/b^2/d/(b*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {18, 3100, 2827, 3853, 3855, 3852, 8} \begin {gather*} \frac {(2 A+3 C) \sin (c+d x)}{3 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{3 b^2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{2 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 b^2 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^(5/2)),x]

[Out]

(B*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(2*b^2*d*Sqrt[b*Cos[c + d*x]]) + (A*Sin[c + d*x])/(3*b^2*d*Cos[c
+ d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]) + (B*Sin[c + d*x])/(2*b^2*d*Cos[c + d*x]^(3/2)*Sqrt[b*Cos[c + d*x]]) + ((2*
A + 3*C)*Sin[c + d*x])/(3*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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*} \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {A \sin (c+d x)}{3 b^2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\sqrt {\cos (c+d x)} \int (3 B+(2 A+3 C) \cos (c+d x)) \sec ^3(c+d x) \, dx}{3 b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {A \sin (c+d x)}{3 b^2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\left (B \sqrt {\cos (c+d x)}\right ) \int \sec ^3(c+d x) \, dx}{b^2 \sqrt {b \cos (c+d x)}}+\frac {\left ((2 A+3 C) \sqrt {\cos (c+d x)}\right ) \int \sec ^2(c+d x) \, dx}{3 b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {A \sin (c+d x)}{3 b^2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{2 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\left (B \sqrt {\cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{2 b^2 \sqrt {b \cos (c+d x)}}-\frac {\left ((2 A+3 C) \sqrt {\cos (c+d x)}\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 b^2 d \sqrt {b \cos (c+d x)}}\\ &=\frac {B \tanh ^{-1}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 b^2 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{3 b^2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x)}{2 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {(2 A+3 C) \sin (c+d x)}{3 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 87, normalized size = 0.53 \begin {gather*} \frac {\sqrt {\cos (c+d x)} \left (3 B \tanh ^{-1}(\sin (c+d x)) \cos ^2(c+d x)+(4 A+3 C+3 B \cos (c+d x)+(2 A+3 C) \cos (2 (c+d x))) \tan (c+d x)\right )}{6 d (b \cos (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^(5/2)),x]

[Out]

(Sqrt[Cos[c + d*x]]*(3*B*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + (4*A + 3*C + 3*B*Cos[c + d*x] + (2*A + 3*C)*Co
s[2*(c + d*x)])*Tan[c + d*x]))/(6*d*(b*Cos[c + d*x])^(5/2))

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Maple [A]
time = 0.23, size = 157, normalized size = 0.96

method	result	size
default	\(\frac {-3 B \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 B \ln \left (-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right )+4 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+3 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 A \sin \left (d x +c \right )}{6 d \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {\cos \left (d x +c \right )}}\)	\(157\)
risch	\(-\frac {i \left (3 B \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,{\mathrm e}^{3 i \left (d x +c \right )}-3 B +\left (-16 A -18 C \right ) \cos \left (d x +c \right )+i \left (-8 A -6 C \right ) \sin \left (d x +c \right )\right )}{6 b^{2} \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 b^{2} \sqrt {b \cos \left (d x +c \right )}\, d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 b^{2} \sqrt {b \cos \left (d x +c \right )}\, d}\)	\(180\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/6/d*(-3*B*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))*cos(d*x+c)^3+3*B*ln(-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x
+c))*cos(d*x+c)^3+4*A*cos(d*x+c)^2*sin(d*x+c)+6*C*sin(d*x+c)*cos(d*x+c)^2+3*B*cos(d*x+c)*sin(d*x+c)+2*A*sin(d*
x+c))/(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1098 vs. \(2 (140) = 280\).
time = 0.71, size = 1098, normalized size = 6.70 \begin {gather*} \frac {\frac {24 \, C \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{3} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{3} \cos \left (2 \, d x + 2 \, c\right ) + b^{3}} + \frac {16 \, {\left ({\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (6 \, d x + 6 \, c\right ) + 3 \, {\left (3 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 3 \, \cos \left (6 \, d x + 6 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) - 9 \, \cos \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right )\right )} A}{{\left (b^{2} \cos \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b^{2} \cos \left (4 \, d x + 4 \, c\right )^{2} + 9 \, b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (6 \, d x + 6 \, c\right )^{2} + 9 \, b^{2} \sin \left (4 \, d x + 4 \, c\right )^{2} + 18 \, b^{2} \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 9 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 6 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2} + 2 \, {\left (3 \, b^{2} \cos \left (4 \, d x + 4 \, c\right ) + 3 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \cos \left (6 \, d x + 6 \, c\right ) + 6 \, {\left (3 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right ) + 6 \, {\left (b^{2} \sin \left (4 \, d x + 4 \, c\right ) + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \sin \left (6 \, d x + 6 \, c\right )\right )} \sqrt {b}} - \frac {3 \, {\left (4 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + 2 \, \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - 4 \, {\left (\sin \left (4 \, d x + 4 \, c\right ) + 2 \, \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - {\left (2 \, {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) + {\left (2 \, {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) - 4 \, {\left (\cos \left (4 \, d x + 4 \, c\right ) + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 4 \, {\left (\cos \left (4 \, d x + 4 \, c\right ) + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )\right )} B}{{\left (b^{2} \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, b^{2} \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2} + 2 \, {\left (2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )\right )} \sqrt {b}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/12*(24*C*sqrt(b)*sin(2*d*x + 2*c)/(b^3*cos(2*d*x + 2*c)^2 + b^3*sin(2*d*x + 2*c)^2 + 2*b^3*cos(2*d*x + 2*c)
+ b^3) + 16*((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))*A/((b^2*cos(6*d*x + 6*c)^2 + 9*b^2*cos(4*d
*x + 4*c)^2 + 9*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(6*d*x + 6*c)^2 + 9*b^2*sin(4*d*x + 4*c)^2 + 18*b^2*sin(4*d*x
+ 4*c)*sin(2*d*x + 2*c) + 9*b^2*sin(2*d*x + 2*c)^2 + 6*b^2*cos(2*d*x + 2*c) + b^2 + 2*(3*b^2*cos(4*d*x + 4*c)
+ 3*b^2*cos(2*d*x + 2*c) + b^2)*cos(6*d*x + 6*c) + 6*(3*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c) + 6*(b^2*
sin(4*d*x + 4*c) + b^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*sqrt(b)) - 3*(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*arctan2(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), c
os(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))))*B/((b^2*cos(4*d*x + 4*c)^2 + 4*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(4*d*x + 4*c)^2 + 4*b^2*sin(4*d*x +
4*c)*sin(2*d*x + 2*c) + 4*b^2*sin(2*d*x + 2*c)^2 + 4*b^2*cos(2*d*x + 2*c) + b^2 + 2*(2*b^2*cos(2*d*x + 2*c) +
b^2)*cos(4*d*x + 4*c))*sqrt(b)))/d

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Fricas [A]
time = 0.43, size = 271, normalized size = 1.65 \begin {gather*} \left [\frac {3 \, B \sqrt {b} \cos \left (d x + c\right )^{4} \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 (2 \, {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, b^{3} d \cos \left (d x + c\right )^{4}}, -\frac {3 \, B \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 )^{4} - {\left (2 \, {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, b^{3} d \cos \left (d x + c\right )^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*B*sqrt(b)*cos(d*x + c)^4*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*s
in(d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*(2*(2*A + 3*C)*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 2*A)*sq
rt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^4), -1/6*(3*B*sqrt(-b)*arctan(sqrt(b*c
os(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d*x + c)^4 - (2*(2*A + 3*C)*cos(d*x + c)^2 + 3*
B*cos(d*x + c) + 2*A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(b*cos(d*x+c))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6190 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((b*cos(d*x + c))^(5/2)*cos(d*x + c)^(3/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (b\,\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((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(b*cos(c + d*x))^(5/2)),x)

[Out]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(b*cos(c + d*x))^(5/2)), x)

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