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

Optimal. Leaf size=217 \[ \frac{2 a (7 A+7 B+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 a (7 A+7 B+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{2 a (5 A+3 (B+C)) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}-\frac{2 a (5 A+3 (B+C)) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a (B+C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 a C \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{7 d} \]

[Out]

(-2*a*(5*A + 3*(B + C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*a*(7*A + 7
*B + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a*(5*A + 3*(B + C))*Sqr
t[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a*(7*A + 7*B + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*a*(B
+ C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (2*a*C*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d)

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Rubi [A]  time = 0.256471, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4076, 4047, 3768, 3771, 2641, 4046, 2639} \[ \frac{2 a (7 A+7 B+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{2 a (5 A+3 (B+C)) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 a (7 A+7 B+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{2 a (5 A+3 (B+C)) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a (B+C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 a C \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(5*A + 3*(B + C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*a*(7*A + 7
*B + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a*(5*A + 3*(B + C))*Sqr
t[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a*(7*A + 7*B + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*a*(B
+ C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (2*a*C*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d)

Rule 4076

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_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x
])^n)/(f*(n + 2)), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, x] &&  !LtQ[n, -1]

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 \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 a C \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2}{7} \int \sec ^{\frac{3}{2}}(c+d x) \left (\frac{7 a A}{2}+\frac{1}{2} a (7 A+7 B+5 C) \sec (c+d x)+\frac{7}{2} a (B+C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 a C \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2}{7} \int \sec ^{\frac{3}{2}}(c+d x) \left (\frac{7 a A}{2}+\frac{7}{2} a (B+C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{7} (a (7 A+7 B+5 C)) \int \sec ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{2 a (7 A+7 B+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 a (B+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a C \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{21} (a (7 A+7 B+5 C)) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{5} (a (5 A+3 (B+C))) \int \sec ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{2 a (5 A+3 (B+C)) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a (7 A+7 B+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 a (B+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a C \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}-\frac{1}{5} (a (5 A+3 (B+C))) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (a (7 A+7 B+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a (7 A+7 B+5 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 a (5 A+3 (B+C)) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a (7 A+7 B+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 a (B+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a C \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}-\frac{1}{5} \left (a (5 A+3 (B+C)) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 a (5 A+3 (B+C)) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 a (7 A+7 B+5 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 a (5 A+3 (B+C)) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a (7 A+7 B+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 a (B+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a C \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [C]  time = 6.69553, size = 527, normalized size = 2.43 \[ \frac{2 a \csc (c) e^{-i d x} \cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (7 \sqrt{2} \left (-1+e^{2 i c}\right ) e^{2 i d x} (5 A+3 (B+C)) \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+10 \sin (c) e^{i d x} (7 A+7 B+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-\frac{\left (-1+e^{2 i c}\right ) e^{-i (c-d x)} \sqrt{\sec (c+d x)} \left (35 A \left (3 e^{i (c+d x)}+e^{2 i (c+d x)}+3 e^{3 i (c+d x)}-1\right ) \left (1+e^{2 i (c+d x)}\right )^2+7 B \left (3 e^{i (c+d x)}-5 e^{2 i (c+d x)}+27 e^{3 i (c+d x)}+5 e^{4 i (c+d x)}+33 e^{5 i (c+d x)}+5 e^{6 i (c+d x)}+9 e^{7 i (c+d x)}-5\right )+C \left (21 e^{i (c+d x)}-85 e^{2 i (c+d x)}+189 e^{3 i (c+d x)}+85 e^{4 i (c+d x)}+231 e^{5 i (c+d x)}+25 e^{6 i (c+d x)}+63 e^{7 i (c+d x)}-25\right )\right )}{\left (1+e^{2 i (c+d x)}\right )^3}\right )}{105 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Cos[c + d*x]^2*Csc[c]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(7*Sqrt[2]*(5*A + 3*(B + C))*E^((2*I)*d*x)*
(-1 + E^((2*I)*c))*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometri
c2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] - ((-1 + E^((2*I)*c))*(35*A*(1 + E^((2*I)*(c + d*x)))^2*(-1 + 3*E^(I
*(c + d*x)) + E^((2*I)*(c + d*x)) + 3*E^((3*I)*(c + d*x))) + 7*B*(-5 + 3*E^(I*(c + d*x)) - 5*E^((2*I)*(c + d*x
)) + 27*E^((3*I)*(c + d*x)) + 5*E^((4*I)*(c + d*x)) + 33*E^((5*I)*(c + d*x)) + 5*E^((6*I)*(c + d*x)) + 9*E^((7
*I)*(c + d*x))) + C*(-25 + 21*E^(I*(c + d*x)) - 85*E^((2*I)*(c + d*x)) + 189*E^((3*I)*(c + d*x)) + 85*E^((4*I)
*(c + d*x)) + 231*E^((5*I)*(c + d*x)) + 25*E^((6*I)*(c + d*x)) + 63*E^((7*I)*(c + d*x))))*Sqrt[Sec[c + d*x]])/
(E^(I*(c - d*x))*(1 + E^((2*I)*(c + d*x)))^3) + 10*(7*A + 7*B + 5*C)*E^(I*d*x)*Sqrt[Cos[c + d*x]]*EllipticF[(c
+ d*x)/2, 2]*Sqrt[Sec[c + d*x]]*Sin[c]))/(105*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [B]  time = 8.038, size = 850, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4/5*(1/2*C+1/2*B)/(8*sin(1/2*d*x+1/2*c)^6-12*si
n(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1
/2*d*x+1/2*c)-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)-8*sin(1/2*d*x+1/2*c)^2*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)+4*(1/2*A+1/2*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*C*(-1/56*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)
^4-5/42*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+5
/21*(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*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)
^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(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*si
n(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(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))*(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 a \sec \left (d x + c\right )^{4} +{\left (B + C\right )} a \sec \left (d x + c\right )^{3} +{\left (A + B\right )} a \sec \left (d x + c\right )^{2} + A a \sec \left (d x + c\right )\right )} \sqrt{\sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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