\(\int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [859]

Optimal result

Integrand size = 33, antiderivative size = 148 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {A b^2 x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b^2 B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {A b^2 \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d}-\frac {b^2 B \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {17, 2827, 2715, 8, 2713} \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {A b^2 x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {A b^2 \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{2 d}-\frac {b^2 B \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 B \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \int \cos ^2(c+d x) (A+B \cos (c+d x)) \, dx}{\sqrt {\cos (c+d x)}} \\ & = \frac {\left (A b^2 \sqrt {b \cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{\sqrt {\cos (c+d x)}}+\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 {A b^2 \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d}+\frac {\left (A b^2 \sqrt {b \cos (c+d x)}\right ) \int 1 \, dx}{2 \sqrt {\cos (c+d x)}}-\frac {\left (b^2 B \sqrt {b \cos (c+d x)}\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d \sqrt {\cos (c+d x)}} \\ & = \frac {A b^2 x \sqrt {b \cos (c+d x)}}{2 \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 {A b^2 \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 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*}

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.47 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {(b \cos (c+d x))^{5/2} (6 A c+6 A d x+9 B \sin (c+d x)+3 A \sin (2 (c+d x))+B \sin (3 (c+d x)))}{12 d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

((b*Cos[c + d*x])^(5/2)*(6*A*c + 6*A*d*x + 9*B*Sin[c + d*x] + 3*A*Sin[2*(c + d*x)] + B*Sin[3*(c + d*x)]))/(12*
d*Cos[c + d*x]^(5/2))

Maple [A] (verified)

Time = 5.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.52

method	result	size
default	\(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (2 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+3 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 A \left (d x +c \right )+4 B \sin \left (d x +c \right )\right )}{6 d \sqrt {\cos \left (d x +c \right )}}\)	\(77\)
parts	\(\frac {A \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}+\frac {B \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}\, \sin \left (d x +c \right )}{3 d \sqrt {\cos \left (d x +c \right )}}\)	\(90\)
risch	\(\frac {A \,b^{2} x \sqrt {\cos \left (d x +c \right ) b}}{2 \sqrt {\cos \left (d x +c \right )}}+\frac {3 b^{2} B \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{4 d \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, B \sin \left (3 d x +3 c \right )}{12 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, A \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}\)	\(132\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.70 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\left [\frac {3 \, A \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (2 \, B b^{2} \cos \left (d x + c\right )^{2} + 3 \, A b^{2} \cos \left (d x + c\right ) + 4 \, B b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )}, \frac {3 \, A b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (2 \, B b^{2} \cos \left (d x + c\right )^{2} + 3 \, A b^{2} \cos \left (d x + c\right ) + 4 \, B b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )}\right ] \]

[In]

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

[Out]

[1/12*(3*A*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*B*b^2*cos(d*x + c)^2 + 3*A*b^2*cos(d*x + c) + 4*B*b^2)*sqrt(b*cos(d*x + c))*sqrt(c
os(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/6*(3*A*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*B*b^2*cos(d*x + c)^2 + 3*A*b^2*cos(d*x + c) + 4*B*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. \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.55 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {3 \, {\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt {b} + {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} B \sqrt {b}}{12 \, d} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.43 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (9\,B\,\sin \left (c+d\,x\right )+3\,A\,\sin \left (2\,c+2\,d\,x\right )+B\,\sin \left (3\,c+3\,d\,x\right )+6\,A\,d\,x\right )}{12\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]

[In]

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

[Out]

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