\(\int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [172]

Optimal result

Integrand size = 35, antiderivative size = 130 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a (4 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {4 a (4 A+5 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]

[Out]

2/5*a*A*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2)+2/15*a*(4*A+5*B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a
+a*cos(d*x+c))^(1/2)+4/15*a*(4*A+5*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3059, 2851, 2850} \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a (4 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a (4 A+5 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}} \]

[In]

Int[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(7/2),x]

[Out]

(2*a*A*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(4*A + 5*B)*Sin[c + d*x])/(15*d*
Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a*(4*A + 5*B)*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a
+ a*Cos[c + d*x]])

Rule 2850

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2851

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x]
+ Dist[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 3059

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {1}{5} (4 A+5 B) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a (4 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {1}{15} (2 (4 A+5 B)) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a (4 A+5 B) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {4 a (4 A+5 B) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a (1+\cos (c+d x))} (7 A+5 B+(4 A+5 B) \cos (c+d x)+(4 A+5 B) \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{15 d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

Integrate[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(7/2),x]

[Out]

(2*Sqrt[a*(1 + Cos[c + d*x])]*(7*A + 5*B + (4*A + 5*B)*Cos[c + d*x] + (4*A + 5*B)*Cos[2*(c + d*x)])*Tan[(c + d
*x)/2])/(15*d*Cos[c + d*x]^(5/2))

Maple [A] (verified)

Time = 7.38 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {2 \sin \left (d x +c \right ) \left (8 A \left (\cos ^{2}\left (d x +c \right )\right )+10 B \left (\cos ^{2}\left (d x +c \right )\right )+4 A \cos \left (d x +c \right )+5 B \cos \left (d x +c \right )+3 A \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{15 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {5}{2}}}\)	\(86\)
parts	\(\frac {2 A \sin \left (d x +c \right ) \left (8 \left (\cos ^{2}\left (d x +c \right )\right )+4 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{15 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {5}{2}}}+\frac {2 B \sin \left (d x +c \right ) \left (2 \cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{3 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {3}{2}}}\)	\(116\)

[In]

int((a+cos(d*x+c)*a)^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/15/d*sin(d*x+c)*(8*A*cos(d*x+c)^2+10*B*cos(d*x+c)^2+4*A*cos(d*x+c)+5*B*cos(d*x+c)+3*A)*(a*(1+cos(d*x+c)))^(1
/2)/(1+cos(d*x+c))/cos(d*x+c)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (2 \, {\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 3 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]

[In]

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/15*(2*(4*A + 5*B)*cos(d*x + c)^2 + (4*A + 5*B)*cos(d*x + c) + 3*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c
))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(7/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (112) = 224\).

Time = 0.34 (sec) , antiderivative size = 428, normalized size of antiderivative = 3.29 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 \, {\left (\frac {5 \, B {\left (\frac {3 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}} + \frac {A {\left (\frac {15 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {25 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {17 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {7 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}}\right )}}{15 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

2/15*(5*B*(3*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 4*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((si
n(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(2*sin(d*x + c)^2/(cos(d
*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + A*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c)
+ 1) - 25*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 17*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c
) + 1)^5 - 7*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/
((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(3*sin(d*x + c)^2/(c
os(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)))/d

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 3.44 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\left (14\,A\,\sin \left (c+d\,x\right )+10\,B\,\sin \left (c+d\,x\right )+8\,A\,\sin \left (2\,c+2\,d\,x\right )+18\,A\,\sin \left (3\,c+3\,d\,x\right )+4\,A\,\sin \left (4\,c+4\,d\,x\right )+4\,A\,\sin \left (5\,c+5\,d\,x\right )+10\,B\,\sin \left (2\,c+2\,d\,x\right )+15\,B\,\sin \left (3\,c+3\,d\,x\right )+5\,B\,\sin \left (4\,c+4\,d\,x\right )+5\,B\,\sin \left (5\,c+5\,d\,x\right )\right )}{15\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (10\,\cos \left (c+d\,x\right )+8\,\cos \left (2\,c+2\,d\,x\right )+5\,\cos \left (3\,c+3\,d\,x\right )+2\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (5\,c+5\,d\,x\right )+6\right )} \]

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(1/2))/cos(c + d*x)^(7/2),x)

[Out]

(4*(a*(cos(c + d*x) + 1))^(1/2)*(14*A*sin(c + d*x) + 10*B*sin(c + d*x) + 8*A*sin(2*c + 2*d*x) + 18*A*sin(3*c +
3*d*x) + 4*A*sin(4*c + 4*d*x) + 4*A*sin(5*c + 5*d*x) + 10*B*sin(2*c + 2*d*x) + 15*B*sin(3*c + 3*d*x) + 5*B*si
n(4*c + 4*d*x) + 5*B*sin(5*c + 5*d*x)))/(15*d*cos(c + d*x)^(1/2)*(10*cos(c + d*x) + 8*cos(2*c + 2*d*x) + 5*cos
(3*c + 3*d*x) + 2*cos(4*c + 4*d*x) + cos(5*c + 5*d*x) + 6))