3.305 \(\int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {13}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=198 \[ \frac {b (3 A+4 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {b (3 A+4 C) \sqrt {b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 d \sqrt {\cos (c+d x)}}+\frac {A b \sin (c+d x) \sqrt {b \cos (c+d x)}}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {b B \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {b B \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

1/4*A*b*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(9/2)+1/8*b*(3*A+4*C)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/c
os(d*x+c)^(5/2)+b*B*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(3/2)+1/3*b*B*sin(d*x+c)^3*(b*cos(d*x+c))^(1/
2)/d/cos(d*x+c)^(7/2)+1/8*b*(3*A+4*C)*arctanh(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {17, 3021, 2748, 3767, 3768, 3770} \[ \frac {b (3 A+4 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {b (3 A+4 C) \sqrt {b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 d \sqrt {\cos (c+d x)}}+\frac {A b \sin (c+d x) \sqrt {b \cos (c+d x)}}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {b B \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {b B \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(13/2),x]

[Out]

(b*(3*A + 4*C)*ArcTanh[Sin[c + d*x]]*Sqrt[b*Cos[c + d*x]])/(8*d*Sqrt[Cos[c + d*x]]) + (A*b*Sqrt[b*Cos[c + d*x]
]*Sin[c + d*x])/(4*d*Cos[c + d*x]^(9/2)) + (b*(3*A + 4*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(8*d*Cos[c + d*x]
^(5/2)) + (b*B*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)) + (b*B*Sqrt[b*Cos[c + d*x]]*Sin[c + d
*x]^3)/(3*d*Cos[c + d*x]^(7/2))

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m + 1/2)*b^(n - 1/2)*Sqrt[b*v])/Sqrt[a*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 2748

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 3021

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 3767

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 3768

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 3770

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 {(b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {13}{2}}(c+d x)} \, dx &=\frac {\left (b \sqrt {b \cos (c+d x)}\right ) \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {\left (b \sqrt {b \cos (c+d x)}\right ) \int (4 B+(3 A+4 C) \cos (c+d x)) \sec ^4(c+d x) \, dx}{4 \sqrt {\cos (c+d x)}}\\ &=\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {\left (b B \sqrt {b \cos (c+d x)}\right ) \int \sec ^4(c+d x) \, dx}{\sqrt {\cos (c+d x)}}+\frac {\left (b (3 A+4 C) \sqrt {b \cos (c+d x)}\right ) \int \sec ^3(c+d x) \, dx}{4 \sqrt {\cos (c+d x)}}\\ &=\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {b (3 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (b (3 A+4 C) \sqrt {b \cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{8 \sqrt {\cos (c+d x)}}-\frac {\left (b B \sqrt {b \cos (c+d x)}\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d \sqrt {\cos (c+d x)}}\\ &=\frac {b (3 A+4 C) \tanh ^{-1}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{8 d \sqrt {\cos (c+d x)}}+\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {b (3 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {b B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}+\frac {b B \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \cos ^{\frac {7}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 111, normalized size = 0.56 \[ \frac {b \sqrt {b \cos (c+d x)} \left (\sin (c+d x) \left (3 (3 A+4 C) \cos ^2(c+d x)+6 A+24 B \cos ^3(c+d x)+8 B \sin ^2(c+d x) \cos (c+d x)\right )+3 (3 A+4 C) \cos ^4(c+d x) \tanh ^{-1}(\sin (c+d x))\right )}{24 d \cos ^{\frac {9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(13/2),x]

[Out]

(b*Sqrt[b*Cos[c + d*x]]*(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]^(9/2))

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fricas [A]  time = 0.78, size = 308, normalized size = 1.56 \[ \left [\frac {3 \, {\left (3 \, A + 4 \, C\right )} b^{\frac {3}{2}} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (16 \, B b \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{2} + 8 \, B b \cos \left (d x + c\right ) + 6 \, A b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{5}}, -\frac {3 \, {\left (3 \, A + 4 \, C\right )} \sqrt {-b} b \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - {\left (16 \, B b \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{2} + 8 \, B b \cos \left (d x + c\right ) + 6 \, A b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x, algorithm="fricas")

[Out]

[1/48*(3*(3*A + 4*C)*b^(3/2)*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*b*cos(d*x + c)^3 + 3*(3*A + 4*C)*b*cos(d*x
+ c)^2 + 8*B*b*cos(d*x + c) + 6*A*b)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)
, -1/24*(3*(3*A + 4*C)*sqrt(-b)*b*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*co
s(d*x + c)^5 - (16*B*b*cos(d*x + c)^3 + 3*(3*A + 4*C)*b*cos(d*x + c)^2 + 8*B*b*cos(d*x + c) + 6*A*b)*sqrt(b*co
s(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {13}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(3/2)/cos(d*x + c)^(13/2), x)

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maple [A]  time = 0.30, size = 246, normalized size = 1.24 \[ -\frac {\left (9 A \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-9 A \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )+12 C \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-12 C \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-16 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-9 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-12 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )-8 B \cos \left (d x +c \right ) \sin \left (d x +c \right )-6 A \sin \left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{\frac {3}{2}}}{24 d \cos \left (d x +c \right )^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x)

[Out]

-1/24/d*(9*A*cos(d*x+c)^4*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-9*A*cos(d*x+c)^4*ln((1-cos(d*x+c)+sin(d*x
+c))/sin(d*x+c))+12*C*cos(d*x+c)^4*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-12*C*cos(d*x+c)^4*ln((1-cos(d*x+
c)+sin(d*x+c))/sin(d*x+c))-16*B*sin(d*x+c)*cos(d*x+c)^3-9*A*cos(d*x+c)^2*sin(d*x+c)-12*C*sin(d*x+c)*cos(d*x+c)
^2-8*B*cos(d*x+c)*sin(d*x+c)-6*A*sin(d*x+c))*(b*cos(d*x+c))^(3/2)/cos(d*x+c)^(11/2)

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maxima [B]  time = 0.88, size = 2732, normalized size = 13.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(13/2),x, algorithm="maxima")

[Out]

-1/48*(3*(12*(b*sin(8*d*x + 8*c) + 4*b*sin(6*d*x + 6*c) + 6*b*sin(4*d*x + 4*c) + 4*b*sin(2*d*x + 2*c))*cos(7/2
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 44*(b*sin(8*d*x + 8*c) + 4*b*sin(6*d*x + 6*c) + 6*b*sin(4*d*x
+ 4*c) + 4*b*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(b*sin(8*d*x + 8*c) +
4*b*sin(6*d*x + 6*c) + 6*b*sin(4*d*x + 4*c) + 4*b*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d
*x + 2*c))) - 12*(b*sin(8*d*x + 8*c) + 4*b*sin(6*d*x + 6*c) + 6*b*sin(4*d*x + 4*c) + 4*b*sin(2*d*x + 2*c))*cos
(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(b*cos(8*d*x + 8*c)^2 + 16*b*cos(6*d*x + 6*c)^2 + 36*b*c
os(4*d*x + 4*c)^2 + 16*b*cos(2*d*x + 2*c)^2 + b*sin(8*d*x + 8*c)^2 + 16*b*sin(6*d*x + 6*c)^2 + 36*b*sin(4*d*x
+ 4*c)^2 + 48*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*b*sin(2*d*x + 2*c)^2 + 2*(4*b*cos(6*d*x + 6*c) + 6*b*co
s(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*cos(8*d*x + 8*c) + 8*(6*b*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) +
b)*cos(6*d*x + 6*c) + 12*(4*b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 8*b*cos(2*d*x + 2*c) + 4*(2*b*sin(6*d*
x + 6*c) + 3*b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*(3*b*sin(4*d*x + 4*c) + 2*b*sin(
2*d*x + 2*c))*sin(6*d*x + 6*c) + b)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arcta
n2(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*(b
*cos(8*d*x + 8*c)^2 + 16*b*cos(6*d*x + 6*c)^2 + 36*b*cos(4*d*x + 4*c)^2 + 16*b*cos(2*d*x + 2*c)^2 + b*sin(8*d*
x + 8*c)^2 + 16*b*sin(6*d*x + 6*c)^2 + 36*b*sin(4*d*x + 4*c)^2 + 48*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*b
*sin(2*d*x + 2*c)^2 + 2*(4*b*cos(6*d*x + 6*c) + 6*b*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*cos(8*d*x + 8
*c) + 8*(6*b*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*cos(6*d*x + 6*c) + 12*(4*b*cos(2*d*x + 2*c) + b)*cos
(4*d*x + 4*c) + 8*b*cos(2*d*x + 2*c) + 4*(2*b*sin(6*d*x + 6*c) + 3*b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*
sin(8*d*x + 8*c) + 16*(3*b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + b)*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*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 12*(b*cos(8*d*x + 8*c) + 4*b*cos(6*d*x + 6*c) + 6*b*cos(4*d*
x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(b*cos(8*d*x +
8*c) + 4*b*cos(6*d*x + 6*c) + 6*b*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*sin(5/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c))) + 44*(b*cos(8*d*x + 8*c) + 4*b*cos(6*d*x + 6*c) + 6*b*cos(4*d*x + 4*c) + 4*b*cos(2*d*x +
2*c) + b)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 12*(b*cos(8*d*x + 8*c) + 4*b*cos(6*d*x + 6*c
) + 6*b*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*A*s
qrt(b)/(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*s
in(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) + 64*(3*b*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) + 9*b*cos
(4*d*x + 4*c)*sin(2*d*x + 2*c) - (3*b*cos(2*d*x + 2*c) + b)*sin(6*d*x + 6*c) - 3*(3*b*cos(2*d*x + 2*c) + b)*si
n(4*d*x + 4*c))*B*sqrt(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) + 12*(4*(b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x
+ 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(b*sin(4*d*x + 4*c) + 2*b*sin(2*d*x + 2*c))*c
os(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (b*cos(4*d*x + 4*c)^2 + 4*b*cos(2*d*x + 2*c)^2 + b*sin(4
*d*x + 4*c)^2 + 4*b*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b*sin(2*d*x + 2*c)^2 + 2*(2*b*cos(2*d*x + 2*c) + b)*
cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*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) + (b*cos(4*d*x + 4*c)^2 + 4*b*cos(2*d*x + 2*c)^2 + b*sin(4*d*x + 4*c)^2 + 4*b*sin(4*d*x + 4*c)*sin(2*d*x +
2*c) + 4*b*sin(2*d*x + 2*c)^2 + 2*(2*b*cos(2*d*x + 2*c) + b)*cos(4*d*x + 4*c) + 4*b*cos(2*d*x + 2*c) + b)*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*(b*cos(4*d*x + 4*c) + 2*b*cos(2*d*x + 2*c)
+ b)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(b*cos(4*d*x + 4*c) + 2*b*cos(2*d*x + 2*c) + b)
*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*C*sqrt(b)/(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))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{13/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(13/2),x)

[Out]

int(((b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(13/2), 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((b*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(13/2),x)

[Out]

Timed out

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