3.2.80 \(\int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) [180]

Optimal. Leaf size=181 \[ \frac {2 a^2 (8 A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (52 A+63 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {4 a^2 (52 A+63 B) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[Out]

2/35*a^2*(8*A+7*B)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2)+2/105*a^2*(52*A+63*B)*sin(d*x+c)/d/cos
(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+4/105*a^2*(52*A+63*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2
)+2/7*a*A*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(7/2)

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Rubi [A]
time = 0.28, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3054, 3059, 2851, 2850} \begin {gather*} \frac {2 a^2 (52 A+63 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a^2 (52 A+63 B) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(9/2),x]

[Out]

(2*a^2*(8*A + 7*B)*Sin[c + d*x])/(35*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(52*A + 63*B)*Sin
[c + d*x])/(105*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (4*a^2*(52*A + 63*B)*Sin[c + d*x])/(105*d*Sqr
t[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7
/2))

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 3054

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
*Sin[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 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*} \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx &=\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{2} a (8 A+7 B)+\frac {1}{2} a (4 A+7 B) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{35} (a (52 A+63 B)) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (52 A+63 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {1}{105} (2 a (52 A+63 B)) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (52 A+63 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {4 a^2 (52 A+63 B) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 0.57, size = 102, normalized size = 0.56 \begin {gather*} \frac {a \sqrt {a (1+\cos (c+d x))} (82 A+63 B+3 (78 A+77 B) \cos (c+d x)+(52 A+63 B) \cos (2 (c+d x))+52 A \cos (3 (c+d x))+63 B \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{105 d \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(9/2),x]

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(82*A + 63*B + 3*(78*A + 77*B)*Cos[c + d*x] + (52*A + 63*B)*Cos[2*(c + d*x)] + 5
2*A*Cos[3*(c + d*x)] + 63*B*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(105*d*Cos[c + d*x]^(7/2))

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Maple [A]
time = 0.27, size = 109, normalized size = 0.60

method	result	size
default	\(-\frac {2 a \left (-1+\cos \left (d x +c \right )\right ) \left (104 A \left (\cos ^{3}\left (d x +c \right )\right )+126 B \left (\cos ^{3}\left (d x +c \right )\right )+52 A \left (\cos ^{2}\left (d x +c \right )\right )+63 B \left (\cos ^{2}\left (d x +c \right )\right )+39 A \cos \left (d x +c \right )+21 B \cos \left (d x +c \right )+15 A \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{105 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{\frac {7}{2}}}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-2/105/d*a*(-1+cos(d*x+c))*(104*A*cos(d*x+c)^3+126*B*cos(d*x+c)^3+52*A*cos(d*x+c)^2+63*B*cos(d*x+c)^2+39*A*cos
(d*x+c)+21*B*cos(d*x+c)+15*A)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/cos(d*x+c)^(7/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (157) = 314\).
time = 0.57, size = 481, normalized size = 2.66 \begin {gather*} \frac {4 \, {\left (\frac {21 \, {\left (\frac {5 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {7 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} B {\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 {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{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 {{\left (\frac {105 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {245 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {273 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {171 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {38 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )} A {\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 {9}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{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 )}}{105 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

4/105*(21*(5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 7*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 2*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x +
c) + 1)^7)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(-sin(d*
x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 + 1)) + (105*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 245*sqrt(2)*a^(3/2)*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 273*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 171*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(c
os(d*x + c) + 1)^7 + 38*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*A*(sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(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

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Fricas [A]
time = 0.36, size = 107, normalized size = 0.59 \begin {gather*} \frac {2 \, {\left (2 \, {\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right )^{3} + {\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (13 \, A + 7 \, B\right )} a \cos \left (d x + c\right ) + 15 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

2/105*(2*(52*A + 63*B)*a*cos(d*x + c)^3 + (52*A + 63*B)*a*cos(d*x + c)^2 + 3*(13*A + 7*B)*a*cos(d*x + c) + 15*
A*a)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(9/2),x)

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [B]
time = 6.72, size = 236, normalized size = 1.30 \begin {gather*} \frac {\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (-\frac {8\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A+3\,B\right )}{3\,d}+\frac {16\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (13\,A+12\,B\right )}{15\,d}+\frac {8\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (52\,A+63\,B\right )}{105\,d}\right )}{6\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )+2\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )+2\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(3/2))/cos(c + d*x)^(9/2),x)

[Out]

((a + a*cos(c + d*x))^(1/2)*((16*a*exp((c*7i)/2 + (d*x*7i)/2)*sin((3*c)/2 + (3*d*x)/2)*(13*A + 12*B))/(15*d) -
(8*a*exp((c*7i)/2 + (d*x*7i)/2)*sin(c/2 + (d*x)/2)*(2*A + 3*B))/(3*d) + (8*a*exp((c*7i)/2 + (d*x*7i)/2)*sin((
7*c)/2 + (7*d*x)/2)*(52*A + 63*B))/(105*d)))/(6*cos(c + d*x)^(1/2)*exp((c*7i)/2 + (d*x*7i)/2)*cos(c/2 + (d*x)/
2) + 6*cos(c + d*x)^(1/2)*exp((c*7i)/2 + (d*x*7i)/2)*cos((3*c)/2 + (3*d*x)/2) + 2*cos(c + d*x)^(1/2)*exp((c*7i
)/2 + (d*x*7i)/2)*cos((5*c)/2 + (5*d*x)/2) + 2*cos(c + d*x)^(1/2)*exp((c*7i)/2 + (d*x*7i)/2)*cos((7*c)/2 + (7*
d*x)/2))

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