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

Optimal. Leaf size=172 \[ \frac{2 a (A+B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 a (A+B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 a (5 A+3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}-\frac{2 a (5 A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a B \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.164512, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3997, 3787, 3768, 3771, 2639, 2641} \[ \frac{2 a (A+B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 a (5 A+3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 a (A+B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 a (5 A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a B \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.643106, size = 168, normalized size = 0.98 \[ \frac{a \sec ^2\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1) (A+B \sec (c+d x)) \left (5 (A+B) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-3 (5 A+3 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+15 A \sin (c+d x)+5 A \tan (c+d x)+9 B \sin (c+d x)+5 B \tan (c+d x)+3 B \tan (c+d x) \sec (c+d x)\right )}{15 d \sec ^{\frac{3}{2}}(c+d x) (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[(c + d*x)/2]^2*(1 + Sec[c + d*x])*(A + B*Sec[c + d*x])*(-3*(5*A + 3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c
+ d*x)/2, 2] + 5*(A + B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 15*A*Sin[c + d*x] + 9*B*Sin[c + d*x] +
5*A*Tan[c + d*x] + 5*B*Tan[c + d*x] + 3*B*Sec[c + d*x]*Tan[c + d*x]))/(15*d*(B + A*Cos[c + d*x])*Sec[c + d*x]
^(3/2))

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Maple [B]  time = 5.197, size = 662, 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)),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*(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)))-2/5*B/(8*sin(1/2*d*x+1/2*c)^6-12*sin(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)*s
in(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)+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*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(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),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 (B 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)),x, algorithm="fricas")

[Out]

integral((B*a*sec(d*x + c)^3 + (A + B)*a*sec(d*x + c)^2 + A*a*sec(d*x + c))*sqrt(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(sec(d*x+c)**(3/2)*(a+a*sec(d*x+c))*(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{\left (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)),x, algorithm="giac")

[Out]

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