3.105 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=106 \[ \frac{7 a^2 \sin (c+d x)}{4 d \sqrt{a \sec (c+d x)+a}}+\frac{7 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(7*a^(3/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (7*a^2*Sin[c + d*x])/(4*d*Sqrt[a +
a*Sec[c + d*x]]) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.126602, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3813, 21, 3805, 3774, 203} \[ \frac{7 a^2 \sin (c+d x)}{4 d \sqrt{a \sec (c+d x)+a}}+\frac{7 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^(3/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (7*a^2*Sin[c + d*x])/(4*d*Sqrt[a +
a*Sec[c + d*x]]) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 3813

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[a/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d,
e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2
*m]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3805

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(a*Cot[
e + f*x]*(d*Csc[e + f*x])^n)/(f*n*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(a*(2*n + 1))/(2*b*d*n), Int[Sqrt[a + b
*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2
^(-1)] && IntegerQ[2*n]

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 203

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

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^{3/2} \, dx &=\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{2} a \int \frac{\cos (c+d x) \left (\frac{7 a}{2}+\frac{7}{2} a \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{4} (7 a) \int \cos (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{7 a^2 \sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt{a+a \sec (c+d x)}}+\frac{1}{8} (7 a) \int \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{7 a^2 \sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt{a+a \sec (c+d x)}}-\frac{\left (7 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}\\ &=\frac{7 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}+\frac{7 a^2 \sin (c+d x)}{4 d \sqrt{a+a \sec (c+d x)}}+\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.362199, size = 108, normalized size = 1.02 \[ \frac{a \cos (c+d x) \sqrt{a (\sec (c+d x)+1)} \left ((7 \sin (c+d x)+\sin (2 (c+d x))) \sqrt{1-\sec (c+d x)}+7 \tan (c+d x) \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{4 d (\cos (c+d x)+1) \sqrt{1-\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(Sqrt[1 - Sec[c + d*x]]*(7*Sin[c + d*x] + Sin[2*(c + d*x)]) + 7*Arc
Tanh[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x]))/(4*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])

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Maple [B]  time = 0.198, size = 222, normalized size = 2.1 \begin{align*}{\frac{a}{16\,d\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) } \left ( 7\,\sqrt{2} \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +7\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) -8\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}-20\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+28\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/16/d*a*(7*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)+7*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)-8*cos(d*x+c)^4-20*cos(d*x+c)^3+28
*cos(d*x+c)^2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)/sin(d*x+c)

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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(cos(d*x+c)^2*(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.88736, size = 736, normalized size = 6.94 \begin{align*} \left [\frac{7 \,{\left (a \cos \left (d x + c\right ) + a\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \,{\left (2 \, a \cos \left (d x + c\right )^{2} + 7 \, a \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac{7 \,{\left (a \cos \left (d x + c\right ) + a\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) -{\left (2 \, a \cos \left (d x + c\right )^{2} + 7 \, a \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \,{\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(7*(a*cos(d*x + c) + a)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(2*a*cos(d*x + c)^2 + 7*a*cos(d*x
+ c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/4*(7*(a*cos(d*x + c) + a
)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (2*a*cos(d*x +
c)^2 + 7*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]

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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(cos(d*x+c)**2*(a+a*sec(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 7.48015, size = 554, normalized size = 5.23 \begin{align*} -\frac{7 \, \sqrt{-a} a \log \left ({\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 )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right ) \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 7 \, \sqrt{-a} a \log \left ({\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 )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right ) \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + \frac{4 \, \sqrt{2}{\left (7 \,{\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} \sqrt{-a} a^{2} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 95 \,{\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} \sqrt{-a} a^{3} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) + 53 \,{\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} \sqrt{-a} a^{4} \mathrm{sgn}\left (\cos \left (d x + c\right )\right ) - 5 \, \sqrt{-a} a^{5} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )\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 )}^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(7*sqrt(-a)*a*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt
(2) + 3)))*sgn(cos(d*x + c)) - 7*sqrt(-a)*a*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2
*c)^2 + a))^2 + a*(2*sqrt(2) - 3)))*sgn(cos(d*x + c)) + 4*sqrt(2)*(7*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*
tan(1/2*d*x + 1/2*c)^2 + a))^6*sqrt(-a)*a^2*sgn(cos(d*x + c)) - 95*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*ta
n(1/2*d*x + 1/2*c)^2 + a))^4*sqrt(-a)*a^3*sgn(cos(d*x + c)) + 53*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(
1/2*d*x + 1/2*c)^2 + a))^2*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 5*sqrt(-a)*a^5*sgn(cos(d*x + c)))/((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)^2)/d