3.594 \(\int \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=189 \[ \frac{2 a \left (a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{d}+\frac{6 b \left (5 a^2+b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}-\frac{6 b \left (5 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d} \]

[Out]

(-6*b*(5*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*a*(a^2 + b^2)*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (6*b*(5*a^2 + b^2)*Sqrt[Sec[c + d*x]]*Sin
[c + d*x])/(5*d) + (8*a*b^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*b^2*Sec[c + d*x]^(3/2)*(a + b*Sec[c +
d*x])*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.201691, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3842, 4047, 3768, 3771, 2639, 4046, 2641} \[ \frac{6 b \left (5 a^2+b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 a \left (a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}-\frac{6 b \left (5 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 b^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^3,x]

[Out]

(-6*b*(5*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*a*(a^2 + b^2)*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (6*b*(5*a^2 + b^2)*Sqrt[Sec[c + d*x]]*Sin
[c + d*x])/(5*d) + (8*a*b^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*b^2*Sec[c + d*x]^(3/2)*(a + b*Sec[c +
d*x])*Sin[c + d*x])/(5*d)

Rule 3842

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[1/(d*(m + n - 1)), In
t[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a*b^2*d*n + b*(b^2*d*(m + n - 2) +
3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2*d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] &&  !Integ
erQ[m])

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 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 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 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 \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^3 \, dx &=\frac{2 b^2 \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac{2}{5} \int \sqrt{\sec (c+d x)} \left (\frac{1}{2} a \left (5 a^2+b^2\right )+\frac{3}{2} b \left (5 a^2+b^2\right ) \sec (c+d x)+6 a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 b^2 \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\frac{2}{5} \int \sqrt{\sec (c+d x)} \left (\frac{1}{2} a \left (5 a^2+b^2\right )+6 a b^2 \sec ^2(c+d x)\right ) \, dx+\frac{1}{5} \left (3 b \left (5 a^2+b^2\right )\right ) \int \sec ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{6 b \left (5 a^2+b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{8 a b^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 b^2 \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\left (a \left (a^2+b^2\right )\right ) \int \sqrt{\sec (c+d x)} \, dx-\frac{1}{5} \left (3 b \left (5 a^2+b^2\right )\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{6 b \left (5 a^2+b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{8 a b^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 b^2 \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}+\left (a \left (a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{5} \left (3 b \left (5 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{6 b \left (5 a^2+b^2\right ) \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 \left (a^2+b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{6 b \left (5 a^2+b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{8 a b^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 b^2 \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.42845, size = 134, normalized size = 0.71 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (5 a \left (a^2+b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-3 b \left (5 a^2+b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{b \sin (c+d x) \left (3 \left (5 a^2+b^2\right ) \cos (2 (c+d x))+5 \left (3 a^2+b^2\right )+10 a b \cos (c+d x)\right )}{2 \cos ^{\frac{5}{2}}(c+d x)}\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^3,x]

[Out]

(2*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-3*b*(5*a^2 + b^2)*EllipticE[(c + d*x)/2, 2] + 5*a*(a^2 + b^2)*Ellip
ticF[(c + d*x)/2, 2] + (b*(5*(3*a^2 + b^2) + 10*a*b*Cos[c + d*x] + 3*(5*a^2 + b^2)*Cos[2*(c + d*x)])*Sin[c + d
*x])/(2*Cos[c + d*x]^(5/2))))/(5*d)

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Maple [B]  time = 4.544, size = 738, 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)^(1/2)*(a+b*sec(d*x+c))^3,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*a^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))+
6*a*b^2*(-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^3/(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*EllipticE(cos(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*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(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)+6*a^2*b*(-(-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)^(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*(-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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{3} \sqrt{\sec \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^3*sqrt(sec(d*x + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}\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)^(1/2)*(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral((b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3)*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)**(1/2)*(a+b*sec(d*x+c))**3,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 )}^{3} \sqrt{\sec \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)*(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^3*sqrt(sec(d*x + c)), x)