3.307 \(\int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\)

Optimal. Leaf size=199 \[ \frac {b^2 x (4 A+3 C) \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 (4 A+3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d}-\frac {b^2 B \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 B \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {b^2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d} \]

[Out]

1/4*b^2*C*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+1/8*b^2*(4*A+3*C)*x*(b*cos(d*x+c))^(1/2)/cos(d*x+
c)^(1/2)+b^2*B*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)-1/3*b^2*B*sin(d*x+c)^3*(b*cos(d*x+c))^(1/2)/
d/cos(d*x+c)^(1/2)+1/8*b^2*(4*A+3*C)*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 = 0.11, antiderivative size = 199, 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, 3023, 2748, 2635, 8, 2633} \[ \frac {b^2 x (4 A+3 C) \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 (4 A+3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d}-\frac {b^2 B \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 B \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {b^2 C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(4*A + 3*C)*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + (b^2*B*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d
*Sqrt[Cos[c + d*x]]) + (b^2*(4*A + 3*C)*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(8*d) + (b^2*C*C
os[c + d*x]^(5/2)*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(4*d) - (b^2*B*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^3)/(3*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 2633

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 2635

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] && Integer
Q[2*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 3023

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

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Mathematica [A]  time = 0.30, size = 92, normalized size = 0.46 \[ \frac {(b \cos (c+d x))^{5/2} (24 (A+C) \sin (2 (c+d x))+48 A c+48 A d x+72 B \sin (c+d x)+8 B \sin (3 (c+d x))+3 C \sin (4 (c+d x))+36 c C+36 C d x)}{96 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(48*A*c + 36*c*C + 48*A*d*x + 36*C*d*x + 72*B*Sin[c + d*x] + 24*(A + C)*Sin[2*(c + d*x
)] + 8*B*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(96*d*Cos[c + d*x]^(5/2))

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fricas [A]  time = 1.84, size = 303, normalized size = 1.52 \[ \left [\frac {3 \, {\left (4 \, A + 3 \, C\right )} \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (6 \, C b^{2} \cos \left (d x + c\right )^{3} + 8 \, B b^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} b^{2} \cos \left (d x + c\right ) + 16 \, B b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )}, \frac {3 \, {\left (4 \, A + 3 \, C\right )} b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (6 \, C b^{2} \cos \left (d x + c\right )^{3} + 8 \, B b^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} b^{2} \cos \left (d x + c\right ) + 16 \, B b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )}\right ] \]

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]

[1/48*(3*(4*A + 3*C)*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(c
os(d*x + c))*sin(d*x + c) - b) + 2*(6*C*b^2*cos(d*x + c)^3 + 8*B*b^2*cos(d*x + c)^2 + 3*(4*A + 3*C)*b^2*cos(d*
x + c) + 16*B*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/24*(3*(4*A + 3*C)
*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) + (6*C*b^2*cos(d*
x + c)^3 + 8*B*b^2*cos(d*x + c)^2 + 3*(4*A + 3*C)*b^2*cos(d*x + c) + 16*B*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{\sqrt {\cos \left (d x + c\right )}}\,{d x} \]

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]

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

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maple [A]  time = 0.50, size = 114, normalized size = 0.57 \[ \frac {\left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}} \left (6 C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+8 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+12 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+9 C \sin \left (d x +c \right ) \cos \left (d x +c \right )+12 A \left (d x +c \right )+16 B \sin \left (d x +c \right )+9 C \left (d x +c \right )\right )}{24 d \cos \left (d x +c \right )^{\frac {5}{2}}} \]

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)

[Out]

1/24/d*(b*cos(d*x+c))^(5/2)*(6*C*sin(d*x+c)*cos(d*x+c)^3+8*B*sin(d*x+c)*cos(d*x+c)^2+12*A*cos(d*x+c)*sin(d*x+c
)+9*C*sin(d*x+c)*cos(d*x+c)+12*A*(d*x+c)+16*B*sin(d*x+c)+9*C*(d*x+c))/cos(d*x+c)^(5/2)

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maxima [A]  time = 0.72, size = 140, normalized size = 0.70 \[ \frac {24 \, {\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt {b} + 8 \, {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} B \sqrt {b} + 3 \, {\left (12 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b^{2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} C \sqrt {b}}{96 \, d} \]

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]

1/96*(24*(2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*A*sqrt(b) + 8*(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))))*B*sqrt(b) + 3*(12*(d*x + c)*b^2 + b^2*sin(4*d*x + 4*c) + 8*b^2*sin(1/2*a
rctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))*C*sqrt(b))/d

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mupad [B]  time = 1.07, size = 94, normalized size = 0.47 \[ \frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (72\,B\,\sin \left (c+d\,x\right )+24\,A\,\sin \left (2\,c+2\,d\,x\right )+8\,B\,\sin \left (3\,c+3\,d\,x\right )+24\,C\,\sin \left (2\,c+2\,d\,x\right )+3\,C\,\sin \left (4\,c+4\,d\,x\right )+48\,A\,d\,x+36\,C\,d\,x\right )}{96\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*(b*cos(c + d*x))^(1/2)*(72*B*sin(c + d*x) + 24*A*sin(2*c + 2*d*x) + 8*B*sin(3*c + 3*d*x) + 24*C*sin(2*c +
2*d*x) + 3*C*sin(4*c + 4*d*x) + 48*A*d*x + 36*C*d*x))/(96*d*cos(c + d*x)^(1/2))

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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))**(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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