3.4.6 \(\int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [306]

Optimal. Leaf size=241 \[ \frac {3 b^2 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 (5 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {3 b^2 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b^2 B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {b^2 C \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d}-\frac {b^2 (5 A+4 C) \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{15 d \sqrt {\cos (c+d x)}} \]

[Out]

1/4*b^2*B*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+1/5*b^2*C*cos(d*x+c)^(7/2)*sin(d*x+c)*(b*cos(d*x+
c))^(1/2)/d+3/8*b^2*B*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+1/5*b^2*(5*A+4*C)*sin(d*x+c)*(b*cos(d*x+c))^(1/2
)/d/cos(d*x+c)^(1/2)-1/15*b^2*(5*A+4*C)*sin(d*x+c)^3*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)+3/8*b^2*B*sin(d*x
+c)*cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.09, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {17, 3102, 2827, 2713, 2715, 8} \begin {gather*} -\frac {b^2 (5 A+4 C) \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{15 d \sqrt {\cos (c+d x)}}+\frac {b^2 (5 A+4 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {3 b^2 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d}+\frac {3 b^2 B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d}+\frac {b^2 C \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*B*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + (b^2*(5*A + 4*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/
(5*d*Sqrt[Cos[c + d*x]]) + (3*b^2*B*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(8*d) + (b^2*B*Cos[c
+ d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(4*d) + (b^2*C*Cos[c + d*x]^(7/2)*Sqrt[b*Cos[c + d*x]]*Sin[c
+ d*x])/(5*d) - (b^2*(5*A + 4*C)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^3)/(15*d*Sqrt[Cos[c + d*x]])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*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 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \int \cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {b^2 C \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d}+\frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \int \cos ^3(c+d x) (5 A+4 C+5 B \cos (c+d x)) \, dx}{5 \sqrt {\cos (c+d x)}}\\ &=\frac {b^2 C \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d}+\frac {\left (b^2 B \sqrt {b \cos (c+d x)}\right ) \int \cos ^4(c+d x) \, dx}{\sqrt {\cos (c+d x)}}+\frac {\left (b^2 (5 A+4 C) \sqrt {b \cos (c+d x)}\right ) \int \cos ^3(c+d x) \, dx}{5 \sqrt {\cos (c+d x)}}\\ &=\frac {b^2 B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {b^2 C \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d}+\frac {\left (3 b^2 B \sqrt {b \cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 \sqrt {\cos (c+d x)}}-\frac {\left (b^2 (5 A+4 C) \sqrt {b \cos (c+d x)}\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d \sqrt {\cos (c+d x)}}\\ &=\frac {b^2 (5 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {3 b^2 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b^2 B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {b^2 C \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d}-\frac {b^2 (5 A+4 C) \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {\left (3 b^2 B \sqrt {b \cos (c+d x)}\right ) \int 1 \, dx}{8 \sqrt {\cos (c+d x)}}\\ &=\frac {3 b^2 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 (5 A+4 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {3 b^2 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b^2 B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {b^2 C \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{5 d}-\frac {b^2 (5 A+4 C) \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{15 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.30, size = 109, normalized size = 0.45 \begin {gather*} \frac {(b \cos (c+d x))^{5/2} (180 B c+180 B d x+60 (6 A+5 C) \sin (c+d x)+120 B \sin (2 (c+d x))+40 A \sin (3 (c+d x))+50 C \sin (3 (c+d x))+15 B \sin (4 (c+d x))+6 C \sin (5 (c+d x)))}{480 d \cos ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(180*B*c + 180*B*d*x + 60*(6*A + 5*C)*Sin[c + d*x] + 120*B*Sin[2*(c + d*x)] + 40*A*Sin
[3*(c + d*x)] + 50*C*Sin[3*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 6*C*Sin[5*(c + d*x)]))/(480*d*Cos[c + d*x]^(5/
2))

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Maple [A]
time = 0.20, size = 134, normalized size = 0.56

method	result	size
default	\(\frac {\left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}} \left (24 C \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+30 B \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+40 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+32 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+45 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+80 A \sin \left (d x +c \right )+45 B \left (d x +c \right )+64 C \sin \left (d x +c \right )\right )}{120 d \cos \left (d x +c \right )^{\frac {5}{2}}}\)	\(134\)
risch	\(\frac {3 b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{i \left (d x +c \right )} B x}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{6 i \left (d x +c \right )} C}{80 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{5 i \left (d x +c \right )} B}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{2 i \left (d x +c \right )} \left (6 A +5 C \right )}{8 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (6 A +5 C \right )}{8 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-i \left (d x +c \right )} B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-2 i \left (d x +c \right )} \left (4 A +5 C \right )}{48 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (10 A +11 C \right ) \cos \left (4 d x +4 c \right )}{120 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (5 A +7 C \right ) \sin \left (4 d x +4 c \right )}{60 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {7 i b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) B \cos \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {9 b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) B \sin \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\)	\(568\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/120/d*(b*cos(d*x+c))^(5/2)*(24*C*cos(d*x+c)^4*sin(d*x+c)+30*B*cos(d*x+c)^3*sin(d*x+c)+40*A*cos(d*x+c)^2*sin(
d*x+c)+32*C*sin(d*x+c)*cos(d*x+c)^2+45*B*cos(d*x+c)*sin(d*x+c)+80*A*sin(d*x+c)+45*B*(d*x+c)+64*C*sin(d*x+c))/c
os(d*x+c)^(5/2)

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Maxima [A]
time = 0.71, size = 185, normalized size = 0.77 \begin {gather*} \frac {40 \, {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} A \sqrt {b} + 15 \, {\left (12 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b^{2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} B \sqrt {b} + 2 \, {\left (3 \, b^{2} \sin \left (5 \, d x + 5 \, c\right ) + 25 \, b^{2} \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right ) + 150 \, b^{2} \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, d x + 5 \, c\right ), \cos \left (5 \, d x + 5 \, c\right )\right )\right )\right )} C \sqrt {b}}{480 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/480*(40*(b^2*sin(3*d*x + 3*c) + 9*b^2*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*A*sqrt(b) + 15*(
12*(d*x + c)*b^2 + b^2*sin(4*d*x + 4*c) + 8*b^2*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))*B*sqrt(b
) + 2*(3*b^2*sin(5*d*x + 5*c) + 25*b^2*sin(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 150*b^2*sin(1/5*
arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))))*C*sqrt(b))/d

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Fricas [A]
time = 0.44, size = 331, normalized size = 1.37 \begin {gather*} \left [\frac {45 \, B \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (24 \, C b^{2} \cos \left (d x + c\right )^{4} + 30 \, B b^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 45 \, B b^{2} \cos \left (d x + c\right ) + 16 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )}, \frac {45 \, B b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (24 \, C b^{2} \cos \left (d x + c\right )^{4} + 30 \, B b^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} b^{2} \cos \left (d x + c\right )^{2} + 45 \, B b^{2} \cos \left (d x + c\right ) + 16 \, {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/240*(45*B*sqrt(-b)*b^2*cos(d*x + c)*log(2*b*cos(d*x + c)^2 - 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x +
c))*sin(d*x + c) - b) + 2*(24*C*b^2*cos(d*x + c)^4 + 30*B*b^2*cos(d*x + c)^3 + 8*(5*A + 4*C)*b^2*cos(d*x + c)
^2 + 45*B*b^2*cos(d*x + c) + 16*(5*A + 4*C)*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(
d*x + c)), 1/120*(45*B*b^(5/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x
+ c) + (24*C*b^2*cos(d*x + c)^4 + 30*B*b^2*cos(d*x + c)^3 + 8*(5*A + 4*C)*b^2*cos(d*x + c)^2 + 45*B*b^2*cos(d*
x + c) + 16*(5*A + 4*C)*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*cos(d*x+c)**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Simplification assuming sageVARc near 0S
implification assuming sageVARc near 0Simplification assuming sageVARc near 0Simplification assuming sageVARc
near 0Simplification

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Mupad [B]
time = 3.04, size = 144, normalized size = 0.60 \begin {gather*} \frac {b^2\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (120\,B\,\sin \left (c+d\,x\right )+400\,A\,\sin \left (2\,c+2\,d\,x\right )+40\,A\,\sin \left (4\,c+4\,d\,x\right )+135\,B\,\sin \left (3\,c+3\,d\,x\right )+15\,B\,\sin \left (5\,c+5\,d\,x\right )+350\,C\,\sin \left (2\,c+2\,d\,x\right )+56\,C\,\sin \left (4\,c+4\,d\,x\right )+6\,C\,\sin \left (6\,c+6\,d\,x\right )+360\,B\,d\,x\,\cos \left (c+d\,x\right )\right )}{480\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(120*B*sin(c + d*x) + 400*A*sin(2*c + 2*d*x) + 40*A*sin(4*c + 4
*d*x) + 135*B*sin(3*c + 3*d*x) + 15*B*sin(5*c + 5*d*x) + 350*C*sin(2*c + 2*d*x) + 56*C*sin(4*c + 4*d*x) + 6*C*
sin(6*c + 6*d*x) + 360*B*d*x*cos(c + d*x)))/(480*d*(cos(2*c + 2*d*x) + 1))

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