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

Optimal. Leaf size=98 \[ \frac{2 a^3 (32 A+35 B) \sin (x)}{15 \sqrt{a \cos (x)+a}}+\frac{2}{15} a^2 (8 A+5 B) \sin (x) \sqrt{a \cos (x)+a}+2 a^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{a} \sin (x)}{\sqrt{a \cos (x)+a}}\right )+\frac{2}{5} a A \sin (x) (a \cos (x)+a)^{3/2} \]

[Out]

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

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Rubi [A]  time = 0.431637, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2828, 2976, 2981, 2773, 206} \[ \frac{2 a^3 (32 A+35 B) \sin (x)}{15 \sqrt{a \cos (x)+a}}+\frac{2}{15} a^2 (8 A+5 B) \sin (x) \sqrt{a \cos (x)+a}+2 a^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{a} \sin (x)}{\sqrt{a \cos (x)+a}}\right )+\frac{2}{5} a A \sin (x) (a \cos (x)+a)^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[((a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n)/Sin[e + f*x]^n, x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[n]

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]

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 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])

Rubi steps

\begin{align*} \int (a+a \cos (x))^{5/2} (A+B \sec (x)) \, dx &=\int (a+a \cos (x))^{5/2} (B+A \cos (x)) \sec (x) \, dx\\ &=\frac{2}{5} a A (a+a \cos (x))^{3/2} \sin (x)+\frac{2}{5} \int (a+a \cos (x))^{3/2} \left (\frac{5 a B}{2}+\frac{1}{2} a (8 A+5 B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac{2}{15} a^2 (8 A+5 B) \sqrt{a+a \cos (x)} \sin (x)+\frac{2}{5} a A (a+a \cos (x))^{3/2} \sin (x)+\frac{4}{15} \int \sqrt{a+a \cos (x)} \left (\frac{15 a^2 B}{4}+\frac{1}{4} a^2 (32 A+35 B) \cos (x)\right ) \sec (x) \, dx\\ &=\frac{2 a^3 (32 A+35 B) \sin (x)}{15 \sqrt{a+a \cos (x)}}+\frac{2}{15} a^2 (8 A+5 B) \sqrt{a+a \cos (x)} \sin (x)+\frac{2}{5} a A (a+a \cos (x))^{3/2} \sin (x)+\left (a^2 B\right ) \int \sqrt{a+a \cos (x)} \sec (x) \, dx\\ &=\frac{2 a^3 (32 A+35 B) \sin (x)}{15 \sqrt{a+a \cos (x)}}+\frac{2}{15} a^2 (8 A+5 B) \sqrt{a+a \cos (x)} \sin (x)+\frac{2}{5} a A (a+a \cos (x))^{3/2} \sin (x)-\left (2 a^3 B\right ) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (x)}{\sqrt{a+a \cos (x)}}\right )\\ &=2 a^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{a} \sin (x)}{\sqrt{a+a \cos (x)}}\right )+\frac{2 a^3 (32 A+35 B) \sin (x)}{15 \sqrt{a+a \cos (x)}}+\frac{2}{15} a^2 (8 A+5 B) \sqrt{a+a \cos (x)} \sin (x)+\frac{2}{5} a A (a+a \cos (x))^{3/2} \sin (x)\\ \end{align*}

Mathematica [A]  time = 0.164598, size = 78, normalized size = 0.8 \[ \frac{1}{30} a^2 \sec \left (\frac{x}{2}\right ) \sqrt{a (\cos (x)+1)} \left (2 \sin \left (\frac{x}{2}\right ) (2 (14 A+5 B) \cos (x)+3 A \cos (2 x)+89 A+80 B)+30 \sqrt{2} B \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[x])]*Sec[x/2]*(30*Sqrt[2]*B*ArcTanh[Sqrt[2]*Sin[x/2]] + 2*(89*A + 80*B + 2*(14*A + 5*B)*C
os[x] + 3*A*Cos[2*x])*Sin[x/2]))/30

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Maple [B]  time = 2.795, size = 228, normalized size = 2.3 \begin{align*}{\frac{1}{15}{a}^{{\frac{3}{2}}}\cos \left ({\frac{x}{2}} \right ) \sqrt{a \left ( \sin \left ({\frac{x}{2}} \right ) \right ) ^{2}} \left ( 24\,A\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \sin \left ( x/2 \right ) \right ) ^{4}-20\,\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}\sqrt{a} \left ( 4\,A+B \right ) \left ( \sin \left ( x/2 \right ) \right ) ^{2}+120\,A\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}\sqrt{a}+90\,B\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}\sqrt{a}+15\,B\ln \left ( -4\,{\frac{\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}-a\sqrt{2}\cos \left ( x/2 \right ) +2\,a}{-2\,\cos \left ( x/2 \right ) +\sqrt{2}}} \right ) a+15\,B\ln \left ( 4\,{\frac{a\sqrt{2}\cos \left ( x/2 \right ) +\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}+2\,a}{2\,\cos \left ( x/2 \right ) +\sqrt{2}}} \right ) a \right ) \left ( \sin \left ({\frac{x}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{x}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.73421, size = 58, normalized size = 0.59 \begin{align*} \frac{1}{30} \,{\left (3 \, \sqrt{2} a^{2} \sin \left (\frac{5}{2} \, x\right ) + 25 \, \sqrt{2} a^{2} \sin \left (\frac{3}{2} \, x\right ) + 150 \, \sqrt{2} a^{2} \sin \left (\frac{1}{2} \, x\right )\right )} A \sqrt{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.48419, size = 354, normalized size = 3.61 \begin{align*} \frac{15 \,{\left (B a^{2} \cos \left (x\right ) + B a^{2}\right )} \sqrt{a} \log \left (\frac{a \cos \left (x\right )^{3} - 7 \, a \cos \left (x\right )^{2} - 4 \, \sqrt{a \cos \left (x\right ) + a} \sqrt{a}{\left (\cos \left (x\right ) - 2\right )} \sin \left (x\right ) + 8 \, a}{\cos \left (x\right )^{3} + \cos \left (x\right )^{2}}\right ) + 4 \,{\left (3 \, A a^{2} \cos \left (x\right )^{2} +{\left (14 \, A + 5 \, B\right )} a^{2} \cos \left (x\right ) +{\left (43 \, A + 40 \, B\right )} a^{2}\right )} \sqrt{a \cos \left (x\right ) + a} \sin \left (x\right )}{30 \,{\left (\cos \left (x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(x))**(5/2)*(A+B*sec(x)),x)

[Out]

Timed out

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Giac [B]  time = 3.30931, size = 244, normalized size = 2.49 \begin{align*} \frac{B a^{\frac{7}{2}} \log \left (\frac{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, x\right ) - \sqrt{a \tan \left (\frac{1}{2} \, x\right )^{2} + a}\right )}^{2} - 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, x\right ) - \sqrt{a \tan \left (\frac{1}{2} \, x\right )^{2} + a}\right )}^{2} + 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} + \frac{2 \,{\left (60 \, \sqrt{2} A a^{5} + 45 \, \sqrt{2} B a^{5} +{\left (80 \, \sqrt{2} A a^{5} + 80 \, \sqrt{2} B a^{5} +{\left (32 \, \sqrt{2} A a^{5} + 35 \, \sqrt{2} B a^{5}\right )} \tan \left (\frac{1}{2} \, x\right )^{2}\right )} \tan \left (\frac{1}{2} \, x\right )^{2}\right )} \tan \left (\frac{1}{2} \, x\right )}{15 \,{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + a\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*a^(7/2)*log(abs(2*(sqrt(a)*tan(1/2*x) - sqrt(a*tan(1/2*x)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(a)
*tan(1/2*x) - sqrt(a*tan(1/2*x)^2 + a))^2 + 4*sqrt(2)*abs(a) - 6*a))/abs(a) + 2/15*(60*sqrt(2)*A*a^5 + 45*sqrt
(2)*B*a^5 + (80*sqrt(2)*A*a^5 + 80*sqrt(2)*B*a^5 + (32*sqrt(2)*A*a^5 + 35*sqrt(2)*B*a^5)*tan(1/2*x)^2)*tan(1/2
*x)^2)*tan(1/2*x)/(a*tan(1/2*x)^2 + a)^(5/2)