3.291 \(\int \frac{\sqrt{\sec (c+d x)} (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=154 \[ \frac{(19 A+3 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(9 A-7 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

((19*A + 3*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[
2]*a^(5/2)*d) - ((A + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((9*A - 7*C)*Sec[
c + d*x]^(3/2)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))

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Rubi [A]  time = 0.369346, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {4085, 4012, 3808, 206} \[ \frac{(19 A+3 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(9 A-7 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((19*A + 3*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[
2]*a^(5/2)*d) - ((A + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((9*A - 7*C)*Sec[
c + d*x]^(3/2)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))

Rule 4085

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

Rule 4012

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] + Dist[(a*A*m + b*B*(m + 1))/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n
, x], x] /; FreeQ[{a, b, d, e, f, A, B, 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{\sqrt{\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{\int \frac{\sqrt{\sec (c+d x)} \left (-\frac{1}{2} a (7 A-C)+a (A-3 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(9 A-7 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(19 A+3 C) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(9 A-7 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(19 A+3 C) \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 )}{16 a^2 d}\\ &=\frac{(19 A+3 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 )}{16 \sqrt{2} a^{5/2} d}-\frac{(A+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(9 A-7 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 2.84041, size = 308, normalized size = 2. \[ \frac{(\sec (c+d x)+1)^{5/2} \left (A+C \sec ^2(c+d x)\right ) \left (\frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{3}{2} (c+d x)\right )\right ) \sqrt{\sec (c+d x)+1} \sec ^5\left (\frac{1}{2} (c+d x)\right ) ((13 A-3 C) \cos (c+d x)+9 A-7 C)}{\sec ^{\frac{3}{2}}(c+d x)}+\sqrt{2} (19 A+3 C) \cos ^2(c+d x) \sqrt{\tan ^2(c+d x)} \cot (c+d x) \left (\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)-2 \sqrt{2} \sqrt{\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1} \sqrt{\sec (c+d x)}+1\right )-\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)+2 \sqrt{2} \sqrt{\tan ^2(c+d x)} \sqrt{\sec (c+d x)+1} \sqrt{\sec (c+d x)}+1\right )\right )\right )}{64 d (a (\sec (c+d x)+1))^{5/2} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((1 + Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2)*(((9*A - 7*C + (13*A - 3*C)*Cos[c + d*x])*Sec[(c + d*x)/2]^5*
Sqrt[1 + Sec[c + d*x]]*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2]))/Sec[c + d*x]^(3/2) + Sqrt[2]*(19*A + 3*C)*Co
s[c + d*x]^2*Cot[c + d*x]*(Log[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 - 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + S
ec[c + d*x]]*Sqrt[Tan[c + d*x]^2]] - Log[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 + 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*
Sqrt[1 + Sec[c + d*x]]*Sqrt[Tan[c + d*x]^2]])*Sqrt[Tan[c + d*x]^2]))/(64*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a*(
1 + Sec[c + d*x]))^(5/2))

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Maple [B]  time = 0.364, size = 348, normalized size = 2.3 \begin{align*}{\frac{\cos \left ( dx+c \right ) \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{16\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 19\,A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) +13\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+3\,C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) -3\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+19\,A\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sin \left ( dx+c \right ) -4\,A\cos \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+3\,C\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sin \left ( dx+c \right ) -4\,C\cos \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-9\,A\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+7\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \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((A+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/16/d/a^3*(1/cos(d*x+c))^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^2*(19*A*sin(d*x
+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))+13*A*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/2)+3*
C*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-3*C*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^
(1/2)+19*A*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-4*A*cos(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))*sin(d*x+c)-4*C*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)-9*A
*(-2/(cos(d*x+c)+1))^(1/2)+7*C*(-2/(cos(d*x+c)+1))^(1/2))/sin(d*x+c)^5/(-2/(cos(d*x+c)+1))^(1/2)

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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+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.539892, size = 1308, normalized size = 8.49 \begin{align*} \left [\frac{\sqrt{2}{\left ({\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 19 \, A + 3 \, C\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{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 ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - \frac{4 \,{\left ({\left (13 \, A - 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (9 \, A - 7 \, C\right )} \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 )}{\sqrt{\cos \left (d x + c\right )}}}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac{\sqrt{2}{\left ({\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 19 \, A + 3 \, C\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{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 )}}{a \sin \left (d x + c\right )}\right ) + \frac{2 \,{\left ({\left (13 \, A - 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (9 \, A - 7 \, C\right )} \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 )}{\sqrt{\cos \left (d x + c\right )}}}{32 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/64*(sqrt(2)*((19*A + 3*C)*cos(d*x + c)^3 + 3*(19*A + 3*C)*cos(d*x + c)^2 + 3*(19*A + 3*C)*cos(d*x + c) + 19
*A + 3*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(
d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((13*A - 3*C)*cos(
d*x + c)^2 + (9*A - 7*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c))
)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/32*(sqrt(2)*((19*A + 3*C)
*cos(d*x + c)^3 + 3*(19*A + 3*C)*cos(d*x + c)^2 + 3*(19*A + 3*C)*cos(d*x + c) + 19*A + 3*C)*sqrt(-a)*arctan(sq
rt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 2*((13*A - 3*C)*
cos(d*x + c)^2 + (9*A - 7*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x +
c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*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((A+C*sec(d*x+c)**2)*sec(d*x+c)**(1/2)/(a+a*sec(d*x+c))**(5/2),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} + A\right )} \sqrt{\sec \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*sqrt(sec(d*x + c))/(a*sec(d*x + c) + a)^(5/2), x)