3.1274 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=163 \[ \frac{\sqrt{2} (A-B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(Sqrt[2]*(A - B + C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqr
t[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d) - (2*(A - 3*B)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a +
a*Sec[c + d*x]]) + (2*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.486442, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4265, 4086, 4013, 3808, 206} \[ \frac{\sqrt{2} (A-B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(A - B + C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqr
t[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d) - (2*(A - 3*B)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a +
a*Sec[c + d*x]]) + (2*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])

Rule 4265

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 4086

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

Rule 4013

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 3808

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b*d)
/(a*f), Subst[Int[1/(2*b - d*x^2), x], x, (b*Cot[e + f*x])/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])], x
] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^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 \frac{\cos ^{\frac{3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 A \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{2} a (A-3 B)+\frac{1}{2} a (2 A+3 C) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{3 a}\\ &=-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 A \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}+\left ((A-B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 A \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}-\frac{\left (2 (A-B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{\sqrt{a} d}-\frac{2 (A-3 B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 A \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.700131, size = 88, normalized size = 0.54 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left (3 (A-B+C) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) (A \cos (c+d x)-A+3 B)\right )}{3 d \sqrt{\cos (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*(3*(A - B + C)*ArcTanh[Sin[(c + d*x)/2]] + 2*(-A + 3*B + A*Cos[c + d*x])*Sin[(c + d*x)/2])
)/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [A]  time = 0.364, size = 220, normalized size = 1.4 \begin{align*} -{\frac{-2+2\,\cos \left ( dx+c \right ) }{3\,ad \left ( \sin \left ( dx+c \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-A\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+3\,B\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) +3\,A\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) -3\,B\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) +3\,C\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \right ) \sqrt{\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/3/d*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(A*cos(d*x+c)*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)-
A*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+3*B*(-2/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+3*A*arctan(1/2*sin(d*x+c)*(-2/
(cos(d*x+c)+1))^(1/2))-3*B*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))+3*C*arctan(1/2*sin(d*x+c)*(-2/(cos
(d*x+c)+1))^(1/2)))*cos(d*x+c)^(1/2)/a/sin(d*x+c)^2/(-2/(cos(d*x+c)+1))^(1/2)

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Maxima [B]  time = 2.41577, size = 764, normalized size = 4.69 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/6*((3*sqrt(2)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 3*sqrt(2)
*cos(3/2*d*x + 3/2*c)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 3*sqrt(2)*log(cos(1/3*arc
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*
c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 3*sqrt(2)*log(cos(1/3*arctan2(s
in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2
- 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 2*sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sq
rt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*A/sqrt(a) + 3*(sqrt(2)*log(cos(1/4*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - sqrt(2)*log(cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c)))^2 + sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d
*x + 2*c))) + 1) - 4*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*B/sqrt(a) - 3*(sqrt(2)*log(
cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - sqrt(2)*log(cos(1/2*d*x + 1/2*
c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*C/sqrt(a))/d

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Fricas [A]  time = 0.552341, size = 914, normalized size = 5.61 \begin{align*} \left [\frac{4 \,{\left (A \cos \left (d x + c\right ) - A + 3 \, B\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \frac{3 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) +{\left (A - B + C\right )} a\right )} \log \left (-\frac{\cos \left (d x + c\right )^{2} - \frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt{a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt{a}}}{6 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac{3 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) +{\left (A - B + C\right )} a\right )} \sqrt{-\frac{1}{a}} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \sqrt{\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - 2 \,{\left (A \cos \left (d x + c\right ) - A + 3 \, B\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/6*(4*(A*cos(d*x + c) - A + 3*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 3
*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*log(-(cos(d*x + c)^2 - 2*sqrt(2)*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c
) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*d), -1/3*(3*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*sqrt(-
1/a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))/sin(d*x + c)) - 2*(A
*cos(d*x + c) - A + 3*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x
+ c) + a*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)**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{\sqrt{a \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/sqrt(a*sec(d*x + c) + a), x)