\(\int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) [842]

Optimal result

Integrand size = 33, antiderivative size = 172 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {3 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {A \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {3 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {A \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {17, 2827, 2713, 2715, 8} \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=-\frac {A \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {A \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {3 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d}+\frac {3 B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.47 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {b \cos (c+d x)} (36 B c+36 B d x+72 A \sin (c+d x)+24 B \sin (2 (c+d x))+8 A \sin (3 (c+d x))+3 B \sin (4 (c+d x)))}{96 d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 5.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.53

method	result	size
default	\(\frac {\sqrt {\cos \left (d x +c \right ) b}\, \left (6 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+8 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+9 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+16 A \sin \left (d x +c \right )+9 B \left (d x +c \right )\right )}{24 d \sqrt {\cos \left (d x +c \right )}}\)	\(91\)
parts	\(\frac {A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{3 d \sqrt {\cos \left (d x +c \right )}}+\frac {B \sqrt {\cos \left (d x +c \right ) b}\, \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d \sqrt {\cos \left (d x +c \right )}}\)	\(104\)
risch	\(\frac {3 \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{i \left (d x +c \right )} x B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{5 i \left (d x +c \right )} B}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{4 i \left (d x +c \right )} A}{12 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {3 i \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{2 i \left (d x +c \right )} A}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {3 i \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) A}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-i \left (d x +c \right )} B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-2 i \left (d x +c \right )} A}{12 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {7 i \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) B \cos \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {9 \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) B \sin \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\)	\(412\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.47 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\left [\frac {9 \, B \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (6 \, B \cos \left (d x + c\right )^{3} + 8 \, A \cos \left (d x + c\right )^{2} + 9 \, B \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )}, \frac {9 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (6 \, B \cos \left (d x + c\right )^{3} + 8 \, A \cos \left (d x + c\right )^{2} + 9 \, B \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )}\right ] \]

[In]

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

[Out]

[1/48*(9*B*sqrt(-b)*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))*s
in(d*x + c) - b) + 2*(6*B*cos(d*x + c)^3 + 8*A*cos(d*x + c)^2 + 9*B*cos(d*x + c) + 16*A)*sqrt(b*cos(d*x + c))*
sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/24*(9*B*sqrt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/
(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + (6*B*cos(d*x + c)^3 + 8*A*cos(d*x + c)^2 + 9*B*cos(d*x + c) + 16*
A)*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. \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.54 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} B \sqrt {b} + 8 \, A \sqrt {b} {\left (\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )}}{96 \, d} \]

[In]

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

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))*B*sqrt(b)
+ 8*A*sqrt(b)*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))))/d

Giac [A] (verification not implemented)

none

Time = 3.40 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.62 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {9 \, B \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 36 \, B \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 48 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 54 \, B \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 80 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, B \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 80 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B \sqrt {b} d x + 48 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 30 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{24 \, {\left (d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 4 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 6 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + d\right )}} \]

[In]

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

[Out]

1/24*(9*B*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^8 + 36*B*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^6 + 48*A*sqrt(b)*tan(1/2*
d*x + 1/2*c)^7 - 30*B*sqrt(b)*tan(1/2*d*x + 1/2*c)^7 + 54*B*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^4 + 80*A*sqrt(b)*
tan(1/2*d*x + 1/2*c)^5 + 18*B*sqrt(b)*tan(1/2*d*x + 1/2*c)^5 + 36*B*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^2 + 80*A*
sqrt(b)*tan(1/2*d*x + 1/2*c)^3 - 18*B*sqrt(b)*tan(1/2*d*x + 1/2*c)^3 + 9*B*sqrt(b)*d*x + 48*A*sqrt(b)*tan(1/2*
d*x + 1/2*c) + 30*B*sqrt(b)*tan(1/2*d*x + 1/2*c))/(d*tan(1/2*d*x + 1/2*c)^8 + 4*d*tan(1/2*d*x + 1/2*c)^6 + 6*d
*tan(1/2*d*x + 1/2*c)^4 + 4*d*tan(1/2*d*x + 1/2*c)^2 + d)

Mupad [B] (verification not implemented)

Time = 16.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.61 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (24\,B\,\sin \left (c+d\,x\right )+80\,A\,\sin \left (2\,c+2\,d\,x\right )+8\,A\,\sin \left (4\,c+4\,d\,x\right )+27\,B\,\sin \left (3\,c+3\,d\,x\right )+3\,B\,\sin \left (5\,c+5\,d\,x\right )+72\,B\,d\,x\,\cos \left (c+d\,x\right )\right )}{96\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

[In]

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

[Out]

(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(24*B*sin(c + d*x) + 80*A*sin(2*c + 2*d*x) + 8*A*sin(4*c + 4*d*x) +
27*B*sin(3*c + 3*d*x) + 3*B*sin(5*c + 5*d*x) + 72*B*d*x*cos(c + d*x)))/(96*d*(cos(2*c + 2*d*x) + 1))