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

Optimal. Leaf size=220 \[ \frac{(10 A-5 B+2 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}+\frac{(10 A-5 B+2 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A-4 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (\sec (c+d x)+1)}-\frac{(7 A-4 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2} \]

[Out]

-(((7*A - 4*B + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d)) + ((10*A - 5*B +
2*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + ((10*A - 5*B + 2*C)*Sin[c +
d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - ((7*A - 4*B + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d
*x])) - ((A - B + C)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.402946, antiderivative size = 220, 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 = {4084, 4020, 3787, 3769, 3771, 2641, 2639} \[ \frac{(10 A-5 B+2 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A-4 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (\sec (c+d x)+1)}+\frac{(10 A-5 B+2 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(7 A-4 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((7*A - 4*B + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d)) + ((10*A - 5*B +
2*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + ((10*A - 5*B + 2*C)*Sin[c +
d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]) - ((7*A - 4*B + C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d
*x])) - ((A - B + C)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^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 4020

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[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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 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]

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{3}{2} a (3 A-B+C)-\frac{1}{2} a (5 A-5 B-C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(7 A-4 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{3}{2} a^2 (10 A-5 B+2 C)-\frac{3}{2} a^2 (7 A-4 B+C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{(7 A-4 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{(7 A-4 B+C) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2}+\frac{(10 A-5 B+2 C) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}\\ &=\frac{(10 A-5 B+2 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A-4 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{(10 A-5 B+2 C) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2}-\frac{\left ((7 A-4 B+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}\\ &=-\frac{(7 A-4 B+C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{(10 A-5 B+2 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A-4 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\left ((10 A-5 B+2 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}\\ &=-\frac{(7 A-4 B+C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{(10 A-5 B+2 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}+\frac{(10 A-5 B+2 C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A-4 B+C) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 6.74433, size = 762, normalized size = 3.46 \[ -\frac{2 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-14 \sqrt{2} A \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )-40 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+8 \sqrt{2} B \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )+20 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-2 \sqrt{2} C \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )-8 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-2 \sqrt{\sec (c+d x)} \left (\sec \left (\frac{c}{2}\right ) (A-B+C) \sin \left (\frac{d x}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right )+\tan \left (\frac{c}{2}\right ) (A-B+C) \sec ^2\left (\frac{1}{2} (c+d x)\right )-2 \sec \left (\frac{c}{2}\right ) (10 A-7 B+4 C) \sin \left (\frac{d x}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right )+3 \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \cos (d x) ((2 A-B) \cos (2 c)+5 A-3 B+C)-2 \tan \left (\frac{c}{2}\right ) (10 A-7 B+4 C)-12 (2 A-B) \cos (c) \sin (d x)+2 A \sin (2 c) \cos (2 d x)+2 A \cos (2 c) \sin (2 d x)\right )\right )}{3 a^2 d (\sec (c+d x)+1)^2 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Cos[(c + d*x)/2]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-14*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2
*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 +
E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (8*Sqrt[2]*B*Sqrt[E^(I*(c +
d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((
2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) - (2*Sqrt[2]*C
*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c
+ d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x
) - 40*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + 20*B*Sqrt[Cos[c + d*x]]*EllipticF[(
c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 8*C*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 2*Sq
rt[Sec[c + d*x]]*(3*(5*A - 3*B + C + (2*A - B)*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2] + 2*A*Cos[2*d*x]*Sin[2*c]
- 2*(10*A - 7*B + 4*C)*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x)/2] + (A - B + C)*Sec[c/2]*Sec[(c + d*x)/2]^3*Sin[(d
*x)/2] - 12*(2*A - B)*Cos[c]*Sin[d*x] + 2*A*Cos[2*c]*Sin[2*d*x] - 2*(10*A - 7*B + 4*C)*Tan[c/2] + (A - B + C)*
Sec[(c + d*x)/2]^2*Tan[c/2])))/(3*a^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^2
)

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Maple [A]  time = 3.004, size = 472, normalized size = 2.2 \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)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x)

[Out]

-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(10*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*B*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*C*EllipticF(cos(1/2*d*x+1/2*c
),2^(1/2))+3*C*EllipticE(cos(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+2*(2*sin(1/2*d*x
+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(10*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+21*A*EllipticE(cos
(1/2*d*x+1/2*c),2^(1/2))-5*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+
2*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+16*A*s
in(1/2*d*x+1/2*c)^8+(-76*A+24*B-12*C)*sin(1/2*d*x+1/2*c)^6+(84*A-34*B+16*C)*sin(1/2*d*x+1/2*c)^4+(-25*A+11*B-5
*C)*sin(1/2*d*x+1/2*c)^2)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/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)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\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 )}}{a^{2} \sec \left (d x + c\right )^{4} + 2 \, a^{2} \sec \left (d x + c\right )^{3} + a^{2} \sec \left (d x + c\right )^{2}}, x\right ) \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)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(a^2*sec(d*x + c)^4 + 2*a^2*sec(d*x + c)^3
+ a^2*sec(d*x + c)^2), x)

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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)/sec(d*x+c)**(3/2)/(a+a*sec(d*x+c))**2,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 (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \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)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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