3.103 \(\int \sqrt{a+a \cos (c+d x)} \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=138 \[ \frac{5 a \tan (c+d x)}{8 d \sqrt{a \cos (c+d x)+a}}+\frac{5 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 d}+\frac{a \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt{a \cos (c+d x)+a}}+\frac{5 a \tan (c+d x) \sec (c+d x)}{12 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

(5*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) + (5*a*Tan[c + d*x])/(8*d*Sqrt[a +
a*Cos[c + d*x]]) + (5*a*Sec[c + d*x]*Tan[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]]) + (a*Sec[c + d*x]^2*Tan[c +
d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.218053, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2772, 2773, 206} \[ \frac{5 a \tan (c+d x)}{8 d \sqrt{a \cos (c+d x)+a}}+\frac{5 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 d}+\frac{a \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt{a \cos (c+d x)+a}}+\frac{5 a \tan (c+d x) \sec (c+d x)}{12 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^4,x]

[Out]

(5*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) + (5*a*Tan[c + d*x])/(8*d*Sqrt[a +
a*Cos[c + d*x]]) + (5*a*Sec[c + d*x]*Tan[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]]) + (a*Sec[c + d*x]^2*Tan[c +
d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]])

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 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 \sqrt{a+a \cos (c+d x)} \sec ^4(c+d x) \, dx &=\frac{a \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}+\frac{5}{6} \int \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \, dx\\ &=\frac{5 a \sec (c+d x) \tan (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{a \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}+\frac{5}{8} \int \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac{5 a \tan (c+d x)}{8 d \sqrt{a+a \cos (c+d x)}}+\frac{5 a \sec (c+d x) \tan (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{a \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}+\frac{5}{16} \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac{5 a \tan (c+d x)}{8 d \sqrt{a+a \cos (c+d x)}}+\frac{5 a \sec (c+d x) \tan (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{a \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}-\frac{(5 a) \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 )}{8 d}\\ &=\frac{5 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{8 d}+\frac{5 a \tan (c+d x)}{8 d \sqrt{a+a \cos (c+d x)}}+\frac{5 a \sec (c+d x) \tan (c+d x)}{12 d \sqrt{a+a \cos (c+d x)}}+\frac{a \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.334332, size = 109, normalized size = 0.79 \[ \frac{\sec \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \sqrt{a (\cos (c+d x)+1)} \left (42 \sin \left (\frac{1}{2} (c+d x)\right )+5 \left (\sin \left (\frac{3}{2} (c+d x)\right )+3 \sin \left (\frac{5}{2} (c+d x)\right )\right )+30 \sqrt{2} \cos ^3(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^4,x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^3*(30*Sqrt[2]*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[
c + d*x]^3 + 42*Sin[(c + d*x)/2] + 5*(Sin[(3*(c + d*x))/2] + 3*Sin[(5*(c + d*x))/2])))/(96*d)

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Maple [B]  time = 2.813, size = 709, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^(1/2)*sec(d*x+c)^4,x)

[Out]

1/6*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-120*a*(ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*
2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))+ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*
(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a)))*sin(1/2*d*x+1/2*c)^6+60*(2
*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+3*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(
1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+3*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*
2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^4+(-160*a^(1/2
)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-90*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*
x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a-90*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/
2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^2+66*a^(1/2)*2^(1/2
)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+15*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)
^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+15*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*si
n(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)/a^(1/2)/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^3/(2*co
s(1/2*d*x+1/2*c)-2^(1/2))^3/sin(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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Maxima [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))^(1/2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.69106, size = 444, normalized size = 3.22 \begin{align*} \frac{15 \,{\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\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 \, \sqrt{a \cos \left (d x + c\right ) + a}{\left (15 \, \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{96 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/96*(15*(cos(d*x + c)^4 + cos(d*x + c)^3)*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*sqrt(a*cos(
d*x + c) + a)*(15*cos(d*x + c)^2 + 10*cos(d*x + c) + 8)*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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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(d*x+c))**(1/2)*sec(d*x+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 3.83765, size = 554, normalized size = 4.01 \begin{align*} \frac{\sqrt{2}{\left (\frac{15 \, \sqrt{2} a^{\frac{3}{2}} \log \left (\frac{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} + \frac{8 \,{\left (63 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{10} a^{\frac{3}{2}} - 369 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{8} a^{\frac{5}{2}} + 1638 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{6} a^{\frac{7}{2}} - 1074 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} a^{\frac{9}{2}} + 171 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a^{\frac{11}{2}} - 13 \, a^{\frac{13}{2}}\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )}^{3}}\right )}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/96*sqrt(2)*(15*sqrt(2)*a^(3/2)*log(abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))
^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*s
qrt(2)*abs(a) - 6*a))/abs(a) + 8*(63*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*a^
(3/2) - 369*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*a^(5/2) + 1638*(sqrt(a)*tan(
1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*a^(7/2) - 1074*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a
*tan(1/2*d*x + 1/2*c)^2 + a))^4*a^(9/2) + 171*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 +
a))^2*a^(11/2) - 13*a^(13/2))/((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt
(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^3)/d