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

Optimal. Leaf size=142 \[ \frac {2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^3 (35 A+32 B) \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (5 A+8 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 d}+\frac {2 a B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]

[Out]

2*a^(5/2)*A*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+2/5*a*B*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2
/15*a^3*(35*A+32*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/15*a^2*(5*A+8*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.41, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2976, 2981, 2773, 206} \[ \frac {2 a^3 (35 A+32 B) \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (5 A+8 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 d}+\frac {2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*A*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a^3*(35*A + 32*B)*Sin[c + d*x])/
(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(5*A + 8*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a*B*(a
+ a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*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 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx &=\frac {2 a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {2}{5} \int (a+a \cos (c+d x))^{3/2} \left (\frac {5 a A}{2}+\frac {1}{2} a (5 A+8 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {2 a^2 (5 A+8 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {4}{15} \int \sqrt {a+a \cos (c+d x)} \left (\frac {15 a^2 A}{4}+\frac {1}{4} a^2 (35 A+32 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {2 a^3 (35 A+32 B) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (5 A+8 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\left (a^2 A\right ) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac {2 a^3 (35 A+32 B) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (5 A+8 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac {\left (2 a^3 A\right ) \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}\\ &=\frac {2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {2 a^3 (35 A+32 B) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (5 A+8 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 104, normalized size = 0.73 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} \left (\sin \left (\frac {1}{2} (c+d x)\right ) (2 (5 A+14 B) \cos (c+d x)+80 A+3 B \cos (2 (c+d x))+89 B)+15 \sqrt {2} A \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(15*Sqrt[2]*A*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + (80*A + 89*
B + 2*(5*A + 14*B)*Cos[c + d*x] + 3*B*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(15*d)

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fricas [A]  time = 0.44, size = 177, normalized size = 1.25 \[ \frac {15 \, {\left (A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, B a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 14 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (40 \, A + 43 \, B\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{30 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

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))*sec(d*x+c),x, algorithm="fricas")

[Out]

1/30*(15*(A*a^2*cos(d*x + c) + A*a^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x +
c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3*B*a^2*cos(d*x
+ c)^2 + (5*A + 14*B)*a^2*cos(d*x + c) + (40*A + 43*B)*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x
+ c) + d)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))*sec(d*x+c),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.28, size = 311, normalized size = 2.19 \[ \frac {a^{\frac {3}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (24 B \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \left (A +4 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+90 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+15 A \ln \left (\frac {4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +15 A \ln \left (-\frac {4 \left (\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 a \right )}{-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +120 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^
(1/2)*sin(1/2*d*x+1/2*c)^4-20*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*2^(1/2)*(A+4*B)*sin(1/2*d*x+1/2*c)^2+90*A
*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+15*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d
*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+15*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^
(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+120*B*2^(1/2)*(a*sin(1/2*d*x
+1/2*c)^2)^(1/2)*a^(1/2))/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [A]  time = 0.87, size = 61, normalized size = 0.43 \[ \frac {{\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 25 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 150 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{30 \, d} \]

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))*sec(d*x+c),x, algorithm="maxima")

[Out]

1/30*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x +
1/2*c))*B*sqrt(a)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))*sec(d*x+c),x)

[Out]

Timed out

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