3.194 \(\int (a+a \cos (x))^{3/2} (A+B \sec (x)) \, dx\)

Optimal. Leaf size=72 \[ \frac{2 a^2 (4 A+3 B) \sin (x)}{3 \sqrt{a \cos (x)+a}}+2 a^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{a} \sin (x)}{\sqrt{a \cos (x)+a}}\right )+\frac{2}{3} a A \sin (x) \sqrt{a \cos (x)+a} \]

[Out]

2*a^(3/2)*B*ArcTanh[(Sqrt[a]*Sin[x])/Sqrt[a + a*Cos[x]]] + (2*a^2*(4*A + 3*B)*Sin[x])/(3*Sqrt[a + a*Cos[x]]) +
(2*a*A*Sqrt[a + a*Cos[x]]*Sin[x])/3

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Rubi [A]  time = 0.294085, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ \frac{2 a^2 (4 A+3 B) \sin (x)}{3 \sqrt{a \cos (x)+a}}+2 a^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{a} \sin (x)}{\sqrt{a \cos (x)+a}}\right )+\frac{2}{3} a A \sin (x) \sqrt{a \cos (x)+a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a^(3/2)*B*ArcTanh[(Sqrt[a]*Sin[x])/Sqrt[a + a*Cos[x]]] + (2*a^2*(4*A + 3*B)*Sin[x])/(3*Sqrt[a + a*Cos[x]]) +
(2*a*A*Sqrt[a + a*Cos[x]]*Sin[x])/3

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

Mathematica [A]  time = 0.104619, size = 62, normalized size = 0.86 \[ \frac{1}{3} a \sec \left (\frac{x}{2}\right ) \sqrt{a (\cos (x)+1)} \left (2 \sin \left (\frac{x}{2}\right ) (A \cos (x)+5 A+3 B)+3 \sqrt{2} B \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[x])]*Sec[x/2]*(3*Sqrt[2]*B*ArcTanh[Sqrt[2]*Sin[x/2]] + 2*(5*A + 3*B + A*Cos[x])*Sin[x/2]))/
3

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Maple [B]  time = 2.882, size = 199, normalized size = 2.8 \begin{align*}{\frac{1}{3}\sqrt{a}\cos \left ({\frac{x}{2}} \right ) \sqrt{a \left ( \sin \left ({\frac{x}{2}} \right ) \right ) ^{2}} \left ( -4\,A\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \sin \left ( x/2 \right ) \right ) ^{2}+12\,A\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}\sqrt{a}+6\,B\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}\sqrt{a}+3\,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+3\,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))^(3/2)*(A+B*sec(x)),x)

[Out]

1/3*a^(1/2)*cos(1/2*x)*(a*sin(1/2*x)^2)^(1/2)*(-4*A*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)*a^(1/2)*sin(1/2*x)^2+12*A*2
^(1/2)*(a*sin(1/2*x)^2)^(1/2)*a^(1/2)+6*B*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)*a^(1/2)+3*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+3*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 = 1.69244, size = 35, normalized size = 0.49 \begin{align*} \frac{1}{3} \,{\left (\sqrt{2} a \sin \left (\frac{3}{2} \, x\right ) + 9 \, \sqrt{2} a \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))^(3/2)*(A+B*sec(x)),x, algorithm="maxima")

[Out]

1/3*(sqrt(2)*a*sin(3/2*x) + 9*sqrt(2)*a*sin(1/2*x))*A*sqrt(a)

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Fricas [A]  time = 2.48128, size = 297, normalized size = 4.12 \begin{align*} \frac{3 \,{\left (B a \cos \left (x\right ) + B a\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 (A a \cos \left (x\right ) +{\left (5 \, A + 3 \, B\right )} a\right )} \sqrt{a \cos \left (x\right ) + a} \sin \left (x\right )}{6 \,{\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))^(3/2)*(A+B*sec(x)),x, algorithm="fricas")

[Out]

1/6*(3*(B*a*cos(x) + B*a)*sqrt(a)*log((a*cos(x)^3 - 7*a*cos(x)^2 - 4*sqrt(a*cos(x) + a)*sqrt(a)*(cos(x) - 2)*s
in(x) + 8*a)/(cos(x)^3 + cos(x)^2)) + 4*(A*a*cos(x) + (5*A + 3*B)*a)*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))**(3/2)*(A+B*sec(x)),x)

[Out]

Timed out

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Giac [B]  time = 3.23506, size = 209, normalized size = 2.9 \begin{align*} \frac{B a^{\frac{5}{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 (6 \, \sqrt{2} A a^{3} + 3 \, \sqrt{2} B a^{3} +{\left (4 \, \sqrt{2} A a^{3} + 3 \, \sqrt{2} B a^{3}\right )} \tan \left (\frac{1}{2} \, x\right )^{2}\right )} \tan \left (\frac{1}{2} \, x\right )}{3 \,{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + a\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*a^(5/2)*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) + 2/3*(6*sqrt(2)*A*a^3 + 3*sqrt(2)
*B*a^3 + (4*sqrt(2)*A*a^3 + 3*sqrt(2)*B*a^3)*tan(1/2*x)^2)*tan(1/2*x)/(a*tan(1/2*x)^2 + a)^(3/2)