\(\int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [308]

Optimal result

Integrand size = 43, antiderivative size = 155 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {b^2 B x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b^2 (3 A+2 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d}+\frac {b^2 C \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {17, 3102, 2813} \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {b^2 (3 A+2 C) \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 B x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b^2 B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{2 d}+\frac {b^2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{3 d} \]

[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]^(3/2),x]

[Out]

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

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 2813

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*a*c +
b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Cos[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

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*} \text {integral}& = \frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \int \cos (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 {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \int \cos (c+d x) (3 A+2 C+3 B \cos (c+d x)) \, dx}{3 \sqrt {\cos (c+d x)}} \\ & = \frac {b^2 B x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b^2 (3 A+2 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d}+\frac {b^2 C \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.48 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} (6 B c+6 B d x+3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+C \sin (3 (c+d x)))}{12 d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(6*B*c + 6*B*d*x + 3*(4*A + 3*C)*Sin[c + d*x] + 3*B*Sin[2*(c + d*x)] + C*Sin[3*(c + d*
x)]))/(12*d*Cos[c + d*x]^(5/2))

Maple [A] (verified)

Time = 9.72 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.55

method	result	size
default	\(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (2 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+6 A \sin \left (d x +c \right )+3 B \left (d x +c \right )+4 \sin \left (d x +c \right ) C \right )}{6 d \sqrt {\cos \left (d x +c \right )}}\)	\(86\)
parts	\(\frac {A \,b^{2} \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {B \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}+\frac {C \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{3 d \sqrt {\cos \left (d x +c \right )}}\)	\(122\)
risch	\(\frac {b^{2} B x \sqrt {\cos \left (d x +c \right ) b}}{2 \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (4 A +3 C \right ) \sin \left (d x +c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, C \sin \left (3 d x +3 c \right )}{12 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, B \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}\)	\(138\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.70 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {3 \, B \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + 3 \, B b^{2} \cos \left (d x + c\right ) + 2 \, {\left (3 \, A + 2 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )}, \frac {3 \, B b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + 3 \, B b^{2} \cos \left (d x + c\right ) + 2 \, {\left (3 \, A + 2 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )}\right ] \]

[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)^(3/2),x, algorithm="fricas")

[Out]

[1/12*(3*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*(2*C*b^2*cos(d*x + c)^2 + 3*B*b^2*cos(d*x + c) + 2*(3*A + 2*C)*b^2)*sqrt(b*cos(d*x +
c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/6*(3*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) + (2*C*b^2*cos(d*x + c)^2 + 3*B*b^2*cos(d*x + c) + 2*(3*A + 2*C)
*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[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)**(3/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.61 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {12 \, A b^{\frac {5}{2}} \sin \left (d x + c\right ) + 3 \, {\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} B \sqrt {b} + {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} C \sqrt {b}}{12 \, d} \]

[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)^(3/2),x, algorithm="maxima")

[Out]

1/12*(12*A*b^(5/2)*sin(d*x + c) + 3*(2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*B*sqrt(b) + (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))))*C*sqrt(b))/d

Giac [F]

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

[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)^(3/2),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)

Mupad [B] (verification not implemented)

Time = 0.79 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.47 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (12\,A\,\sin \left (c+d\,x\right )+9\,C\,\sin \left (c+d\,x\right )+3\,B\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (3\,c+3\,d\,x\right )+6\,B\,d\,x\right )}{12\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]

[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)^(3/2),x)

[Out]

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