3.65 \(\int (b \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=178 \[ \frac{2 b B \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sqrt{b \sec (c+d x)}}{3 d}-\frac{2 b^2 (5 A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 b (5 A+3 C) \sin (c+d x) \sqrt{b \sec (c+d x)}}{5 d}+\frac{2 B \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}+\frac{2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d} \]

[Out]

(-2*b^2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b*B*Sqrt[Cos
[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (2*b*(5*A + 3*C)*Sqrt[b*Sec[c + d*x]]*Sin[c
+ d*x])/(5*d) + (2*B*(b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + (2*C*(b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(
5*d)

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Rubi [A]  time = 0.161057, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4047, 3768, 3771, 2641, 4046, 2639} \[ -\frac{2 b^2 (5 A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 b (5 A+3 C) \sin (c+d x) \sqrt{b \sec (c+d x)}}{5 d}+\frac{2 B \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}+\frac{2 b B \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \sec (c+d x)}}{3 d}+\frac{2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b*B*Sqrt[Cos
[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*d) + (2*b*(5*A + 3*C)*Sqrt[b*Sec[c + d*x]]*Sin[c
+ d*x])/(5*d) + (2*B*(b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + (2*C*(b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(
5*d)

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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

Mathematica [C]  time = 6.50493, size = 640, normalized size = 3.6 \[ \frac{2 \sqrt{2} A \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{3 d \sec ^{\frac{7}{2}}(c+d x) (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{2 \sqrt{2} C \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{5 d \sec ^{\frac{7}{2}}(c+d x) (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{4 B \cos ^{\frac{7}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{3 d (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{\cos ^3(c+d x) (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{4 (5 A+3 C) \csc (c) \cos (d x)}{5 d}+\frac{4 \sec (c) \sec (c+d x) (5 B \sin (d x)+3 C \sin (c))}{15 d}+\frac{4 B \tan (c)}{3 d}+\frac{4 C \sec (c) \sin (d x) \sec ^2(c+d x)}{5 d}\right )}{A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*B*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2]*(b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^
2))/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (2*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))])*(b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] +
C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2)) + (2*S
qrt[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))])*(
b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(5*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*
Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2)) + (Cos[c + d*x]^3*(b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c +
d*x]^2)*((4*(5*A + 3*C)*Cos[d*x]*Csc[c])/(5*d) + (4*C*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(5*d) + (4*Sec[c]*Sec[c
+ d*x]*(3*C*Sin[c] + 5*B*Sin[d*x]))/(15*d) + (4*B*Tan[c])/(3*d)))/(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2
*d*x])

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Maple [C]  time = 0.3, size = 832, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/d*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^2*(-15*I*A*sin(d*x+c)*cos(d*x+c)^3*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*B*sin(d*x+c)*cos(d*x+c)^2*(1/(cos(d*x+c)
+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-9*I*C*sin(d*x+c)*cos(d*
x+c)^3*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*
I*B*sin(d*x+c)*cos(d*x+c)^3*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x
+c))/sin(d*x+c),I)-9*I*C*sin(d*x+c)*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*El
lipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+15*I*A*sin(d*x+c)*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+9*I*C*sin(d*x+c)*cos(d*x+c)^3*(1/(cos(d*x+c)+1))
^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+15*I*A*sin(d*x+c)*cos(d*x+c
)^3*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-15*I*
A*sin(d*x+c)*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c
))/sin(d*x+c),I)+9*I*C*sin(d*x+c)*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elli
pticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-15*A*cos(d*x+c)^3-5*B*cos(d*x+c)^3-9*C*cos(d*x+c)^3+15*A*cos(d*x+c)^2+6*
C*cos(d*x+c)^2+5*B*cos(d*x+c)+3*C)*(b/cos(d*x+c))^(3/2)/sin(d*x+c)^5/cos(d*x+c)

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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((b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*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 ({\left (C b \sec \left (d x + c\right )^{3} + B b \sec \left (d x + c\right )^{2} + A b \sec \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b*sec(d*x + c)^3 + B*b*sec(d*x + c)^2 + A*b*sec(d*x + c))*sqrt(b*sec(d*x + c)), 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((b*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*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{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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