3.1322 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\)

Optimal. Leaf size=158 \[ \frac{2 C \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 b d}+\frac{2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d (a+b)}-\frac{2 (b B-a C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac{2 (b B-a C) \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)}}+\frac{2 C \sin (c+d x)}{3 b d \cos ^{\frac{3}{2}}(c+d x)} \]

[Out]

(-2*(b*B - a*C)*EllipticE[(c + d*x)/2, 2])/(b^2*d) + (2*C*EllipticF[(c + d*x)/2, 2])/(3*b*d) + (2*(A*b^2 - a*(
b*B - a*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(b^2*(a + b)*d) + (2*C*Sin[c + d*x])/(3*b*d*Cos[c + d*x
]^(3/2)) + (2*(b*B - a*C)*Sin[c + d*x])/(b^2*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.934632, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4112, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d (a+b)}-\frac{2 (b B-a C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac{2 (b B-a C) \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)}}+\frac{2 C F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b d}+\frac{2 C \sin (c+d x)}{3 b d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B - a*C)*EllipticE[(c + d*x)/2, 2])/(b^2*d) + (2*C*EllipticF[(c + d*x)/2, 2])/(3*b*d) + (2*(A*b^2 - a*(
b*B - a*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(b^2*(a + b)*d) + (2*C*Sin[c + d*x])/(3*b*d*Cos[c + d*x
]^(3/2)) + (2*(b*B - a*C)*Sin[c + d*x])/(b^2*d*Sqrt[Cos[c + d*x]])

Rule 4112

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx &=\int \frac{C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (b+a \cos (c+d x))} \, dx\\ &=\frac{2 C \sin (c+d x)}{3 b d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \int \frac{\frac{3}{2} (b B-a C)+\frac{1}{2} b (3 A+C) \cos (c+d x)+\frac{1}{2} a C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{3 b}\\ &=\frac{2 C \sin (c+d x)}{3 b d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (b B-a C) \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)}}+\frac{4 \int \frac{\frac{1}{4} \left (b^2 (3 A+C)-3 a (b B-a C)\right )-\frac{1}{4} b (3 b B-4 a C) \cos (c+d x)-\frac{3}{4} a (b B-a C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 b^2}\\ &=\frac{2 C \sin (c+d x)}{3 b d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (b B-a C) \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)}}-\frac{4 \int \frac{-\frac{1}{4} a \left (b^2 (3 A+C)-3 a (b B-a C)\right )-\frac{1}{4} a^2 b C \cos (c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a b^2}-\frac{(b B-a C) \int \sqrt{\cos (c+d x)} \, dx}{b^2}\\ &=-\frac{2 (b B-a C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac{2 C \sin (c+d x)}{3 b d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (b B-a C) \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)}}+\frac{C \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 b}+\left (A-\frac{a (b B-a C)}{b^2}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx\\ &=-\frac{2 (b B-a C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac{2 C F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b d}+\frac{2 \left (A-\frac{a (b B-a C)}{b^2}\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{(a+b) d}+\frac{2 C \sin (c+d x)}{3 b d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 (b B-a C) \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.48184, size = 269, normalized size = 1.7 \[ \frac{-\frac{6 (b B-a C) \sin (c+d x) \left (2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )-\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a \sqrt{\sin ^2(c+d x)}}+\frac{b \left (8 a b C-6 b^2 B\right ) \left (2 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{2 b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}+\frac{2 b \left (9 a^2 C-9 a b B+6 A b^2+2 b^2 C\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{12 b (b B-a C) \sin (c+d x)}{\sqrt{\cos (c+d x)}}+\frac{4 b^2 C \sin (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)}}{6 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*(6*A*b^2 - 9*a*b*B + 9*a^2*C + 2*b^2*C)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (b*(-6*b^2*
B + 8*a*b*C)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)))/a + (4*b
^2*C*Sin[c + d*x])/Cos[c + d*x]^(3/2) + (12*b*(b*B - a*C)*Sin[c + d*x])/Sqrt[Cos[c + d*x]] - (6*(b*B - a*C)*(-
2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] - (a^2
- 2*b^2)*EllipticPi[-(a/b), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*Sqrt[Sin[c + d*x]^2]))/(6*b^3*
d)

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Maple [B]  time = 7.112, size = 472, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C/b*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2
*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*
x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
)-2*(A*b^2-B*a*b+C*a^2)/b^2/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2*(B*b-C*a)/b^2
*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2
)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d
*x+1/2*c)*sin(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)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2
*d*x+1/2*c)^2-1)^(1/2)/d

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

[Out]

Timed out

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Fricas [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+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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