3.1131 \(\int \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=135 \[ \frac{a (2 A-C) \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{C \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{d \sqrt{\cos (c+d x)}}+\frac{\sqrt{a} C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d} \]

[Out]

(Sqrt[a]*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d +
(a*(2*A - C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (C*Sqrt[a + a*Sec[c + d*x]]*Sin[
c + d*x])/(d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.409491, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {4265, 4089, 4015, 3801, 215} \[ \frac{a (2 A-C) \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{C \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{d \sqrt{\cos (c+d x)}}+\frac{\sqrt{a} C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d +
(a*(2*A - C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (C*Sqrt[a + a*Sec[c + d*x]]*Sin[
c + d*x])/(d*Sqrt[Cos[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 4089

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

Rule 4015

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

Rule 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

Rule 215

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

Rubi steps

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

Mathematica [A]  time = 0.631644, size = 90, normalized size = 0.67 \[ \frac{\sqrt{\cos (c+d x)} \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} \left (2 \sin \left (\frac{1}{2} (c+d x)\right ) (2 A+C \sec (c+d x))+\sqrt{2} C \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.358, size = 210, normalized size = 1.6 \begin{align*} -{\frac{-1+\cos \left ( dx+c \right ) }{2\,d \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 ( 4\,A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+C\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \cos \left ( dx+c \right ) -C\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \cos \left ( dx+c \right ) +2\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}}}{\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(4*A*cos(d*x+c)*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2
)+C*2^(1/2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)-C*2^(1/2)*arcta
n(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)+2*C*(-2/(cos(d*x+c)+1))^(1/2)*si
n(d*x+c))/sin(d*x+c)^2/(-2/(cos(d*x+c)+1))^(1/2)/cos(d*x+c)^(1/2)

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Maxima [B]  time = 2.22962, size = 987, normalized size = 7.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) - (4*sqrt(2)*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) + 4*sqrt(2)*c
os(3/2*d*x + 3/2*c)*sin(2*d*x + 2*c) - 4*sqrt(2)*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) - (cos(2*d*x + 2*c)^2 +
sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2
*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*s
in(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*
c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^
2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x +
c))) + 2) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x
+ c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x
+ c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + (cos(2*d*x + 2*c)^2 + sin
(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arc
tan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1
/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 4*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(5/2*d*x + 5/2*c) + 4
*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(1/2*arct
an2(sin(d*x + c), cos(d*x + c))) - 4*sqrt(2)*sin(3/2*d*x + 3/2*c))*C*sqrt(a)/(cos(2*d*x + 2*c)^2 + sin(2*d*x +
2*c)^2 + 2*cos(2*d*x + 2*c) + 1))/d

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Fricas [A]  time = 0.57613, size = 942, normalized size = 6.98 \begin{align*} \left [\frac{4 \,{\left (2 \, A \cos \left (d x + c\right ) + C\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 ) +{\left (C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 4 \, \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}{\left (\cos \left (d x + c\right ) - 2\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{4 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac{2 \,{\left (2 \, A \cos \left (d x + c\right ) + C\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 ) +{\left (C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \sqrt{-a} \arctan \left (\frac{2 \, \sqrt{-a} \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 )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{2 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(2*A*cos(d*x + c) + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + (C*co
s(d*x + c)^2 + C*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c
))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x +
c)^2)))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), 1/2*(2*(2*A*cos(d*x + c) + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sqrt(cos(d*x + c))*sin(d*x + c) + (C*cos(d*x + c)^2 + C*cos(d*x + c))*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*
cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)))/(d
*cos(d*x + c)^2 + d*cos(d*x + c))]

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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+C*sec(d*x+c)**2)*cos(d*x+c)**(1/2)*(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{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{a \sec \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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