3.189 \(\int \frac{(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=228 \[ \frac{2 a^3 (292 A+345 B) \sin (c+d x)}{315 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (4 A+3 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 a^3 (292 A+345 B) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac{9}{2}}(c+d x)} \]

[Out]

(2*a^3*(124*A + 135*B)*Sin[c + d*x])/(315*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(292*A + 345
*B)*Sin[c + d*x])/(315*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a^3*(292*A + 345*B)*Sin[c + d*x])/(
315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(4*A + 3*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])
/(21*d*Cos[c + d*x]^(7/2)) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.7035, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {2975, 2980, 2772, 2771} \[ \frac{2 a^3 (292 A+345 B) \sin (c+d x)}{315 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (4 A+3 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{4 a^3 (292 A+345 B) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 a A \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{9 d \cos ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(11/2),x]

[Out]

(2*a^3*(124*A + 135*B)*Sin[c + d*x])/(315*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a^3*(292*A + 345
*B)*Sin[c + d*x])/(315*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a^3*(292*A + 345*B)*Sin[c + d*x])/(
315*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(4*A + 3*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])
/(21*d*Cos[c + d*x]^(7/2)) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2))

Rule 2975

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((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]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], 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] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 2980

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]

Rule 2772

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 2771

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]

Rubi steps

\begin{align*} \int \frac{(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac{11}{2}}(c+d x)} \, dx &=\frac{2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{2}{9} \int \frac{(a+a \cos (c+d x))^{3/2} \left (\frac{3}{2} a (4 A+3 B)+\frac{1}{2} a (4 A+9 B) \cos (c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (4 A+3 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{4}{63} \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{4} a^2 (124 A+135 B)+\frac{1}{4} a^2 (76 A+99 B) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (4 A+3 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{1}{105} \left (a^2 (292 A+345 B)\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (292 A+345 B) \sin (c+d x)}{315 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (4 A+3 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}+\frac{1}{315} \left (2 a^2 (292 A+345 B)\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a^3 (292 A+345 B) \sin (c+d x)}{315 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{4 a^3 (292 A+345 B) \sin (c+d x)}{315 d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (4 A+3 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{21 d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 a A (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{9 d \cos ^{\frac{9}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.829134, size = 126, normalized size = 0.55 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} ((1396 A+1215 B) \cos (c+d x)+2 (803 A+870 B) \cos (2 (c+d x))+292 A \cos (3 (c+d x))+292 A \cos (4 (c+d x))+1454 A+345 B \cos (3 (c+d x))+345 B \cos (4 (c+d x))+1395 B)}{630 d \cos ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(11/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*(1454*A + 1395*B + (1396*A + 1215*B)*Cos[c + d*x] + 2*(803*A + 870*B)*Cos[2*(c
+ d*x)] + 292*A*Cos[3*(c + d*x)] + 345*B*Cos[3*(c + d*x)] + 292*A*Cos[4*(c + d*x)] + 345*B*Cos[4*(c + d*x)])*
Tan[(c + d*x)/2])/(630*d*Cos[c + d*x]^(9/2))

________________________________________________________________________________________

Maple [A]  time = 0.621, size = 133, normalized size = 0.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 584\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+690\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+292\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+345\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+219\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+180\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+130\,A\cos \left ( dx+c \right ) +45\,B\cos \left ( dx+c \right ) +35\,A \right ) }{315\,d\sin \left ( dx+c \right ) }\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(11/2),x)

[Out]

-2/315/d*a^2*(-1+cos(d*x+c))*(584*A*cos(d*x+c)^4+690*B*cos(d*x+c)^4+292*A*cos(d*x+c)^3+345*B*cos(d*x+c)^3+219*
A*cos(d*x+c)^2+180*B*cos(d*x+c)^2+130*A*cos(d*x+c)+45*B*cos(d*x+c)+35*A)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/c
os(d*x+c)^(9/2)

________________________________________________________________________________________

Maxima [B]  time = 2.45213, size = 720, normalized size = 3.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

8/315*(15*(21*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 56*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c
) + 1)^3 + 63*sqrt(2)*a^(5/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 36*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x
+ c) + 1)^7 + 8*sqrt(2)*a^(5/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
1)^2/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(2*sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + (315*sqrt(2)*a^(5/2)*sin(d*x + c)/(cos(
d*x + c) + 1) - 945*sqrt(2)*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 1449*sqrt(2)*a^(5/2)*sin(d*x + c)^5/
(cos(d*x + c) + 1)^5 - 1287*sqrt(2)*a^(5/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 572*sqrt(2)*a^(5/2)*sin(d*x
+ c)^9/(cos(d*x + c) + 1)^9 - 104*sqrt(2)*a^(5/2)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)*A*(sin(d*x + c)^2/(co
s(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^
(11/2)*(3*sin(d*x + c)^2/(cos(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

________________________________________________________________________________________

Fricas [A]  time = 1.19209, size = 351, normalized size = 1.54 \begin{align*} \frac{2 \,{\left (2 \,{\left (292 \, A + 345 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} +{\left (292 \, A + 345 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (73 \, A + 60 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \,{\left (26 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right ) + 35 \, A a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \,{\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

2/315*(2*(292*A + 345*B)*a^2*cos(d*x + c)^4 + (292*A + 345*B)*a^2*cos(d*x + c)^3 + 3*(73*A + 60*B)*a^2*cos(d*x
+ c)^2 + 5*(26*A + 9*B)*a^2*cos(d*x + c) + 35*A*a^2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)
/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(11/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\cos \left (d x + c\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(11/2), x)