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

Optimal. Leaf size=149 \[ \frac{(3 A+B-5 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{2 C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac{(A-B+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

(2*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) + ((3*A + B - 5*C)*ArcTanh[(Sqrt[a]
*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Se
c[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2))

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Rubi [A]  time = 0.406162, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4084, 4023, 3808, 206, 3801, 215} \[ \frac{(3 A+B-5 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{2 C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac{(A-B+C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) + ((3*A + B - 5*C)*ArcTanh[(Sqrt[a]
*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Se
c[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2))

Rule 4084

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*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4023

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

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])

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 \frac{\sqrt{\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\sqrt{\sec (c+d x)} \left (\frac{1}{2} a (3 A+B-C)+2 a C \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(3 A+B-5 C) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}+\frac{C \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx}{a^2}\\ &=-\frac{(A-B+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(3 A+B-5 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 )}{2 a d}-\frac{(2 C) \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 )}{a^2 d}\\ &=\frac{2 C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac{(3 A+B-5 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 )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.30354, size = 175, normalized size = 1.17 \[ \frac{\cos ^3\left (\frac{1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{A-B+C}{\sin \left (\frac{1}{2} (c+d x)\right )-1}+\frac{A-B+C}{\sin \left (\frac{1}{2} (c+d x)\right )+1}+2 (3 A+B-5 C) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+8 \sqrt{2} C \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{d \sqrt{\sec (c+d x)} (a (\sec (c+d x)+1))^{3/2} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(2*(3*A + B - 5*C)*ArcTanh[Sin[(c + d*x)/2]] + 8*S
qrt[2]*C*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + (A - B + C)/(-1 + Sin[(c + d*x)/2]) + (A - B + C)/(1 + Sin[(c + d
*x)/2])))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sqrt[Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2)
)

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Maple [B]  time = 0.371, size = 384, normalized size = 2.6 \begin{align*}{\frac{\cos \left ( dx+c \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1 \right ) }{4\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}\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 ( -2\,C\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) \sin \left ( dx+c \right ) +2\,C\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) \sin \left ( dx+c \right ) +A\cos \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+3\,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 ) -B\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\cos \left ( dx+c \right ) +B\arctan \left ({\frac{\sin \left ( dx+c \right ) }{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \sin \left ( dx+c \right ) +C\cos \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-5\,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 ) -A\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+B\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \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(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/4/d/a^2*(1/cos(d*x+c))^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*cos(d*x+c)*(-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)))*sin(d*x+c)+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)))*sin(d*x+c)+A*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+3*A*arctan(1/2*sin(d*
x+c)*(-2/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-B*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+B*arctan(1/2*sin(d*x+c)*(-2/
(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+C*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)-5*C*arctan(1/2*sin(d*x+c)*(-2/(cos(d*
x+c)+1))^(1/2))*sin(d*x+c)-A*(-2/(cos(d*x+c)+1))^(1/2)+B*(-2/(cos(d*x+c)+1))^(1/2)-C*(-2/(cos(d*x+c)+1))^(1/2)
)*(-2/(cos(d*x+c)+1))^(1/2)/sin(d*x+c)^3*(cos(d*x+c)^2-1)

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

[Out]

Timed out

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Fricas [B]  time = 0.715002, size = 1656, normalized size = 11.11 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(sqrt(2)*((3*A + B - 5*C)*cos(d*x + c)^2 + 2*(3*A + B - 5*C)*cos(d*x + c) + 3*A + B - 5*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*(A - B + C)*sqrt((a*cos(d*x + c) + a)/cos
(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 4*(C*cos(d*x + c)^2 + 2*C*cos(d*x + c) + C)*sqrt(a)*log((a*cos(d*
x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(a^2*d*cos(d*x + c)^2 + 2*a^2
*d*cos(d*x + c) + a^2*d), -1/4*(sqrt(2)*((3*A + B - 5*C)*cos(d*x + c)^2 + 2*(3*A + B - 5*C)*cos(d*x + c) + 3*A
+ B - 5*C)*sqrt(-a)*arctan(sqrt(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*(A - B + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 4*(C*cos
(d*x + c)^2 + 2*C*cos(d*x + c) + C)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(co
s(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x +
c) + a^2*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(sec(d*x+c)**(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),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 )} \sqrt{\sec \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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