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

Optimal. Leaf size=149 \[ \frac{2 a \left (a^2+b^2\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{d}-\frac{6 b \left (5 a^2+b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{6 b \left (5 a^2+b^2\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{8 a b^2 \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 b^2 \sin (c+d x) (a+b \sec (c+d x))}{5 d \cos ^{\frac{3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.245708, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4264, 3842, 4047, 3768, 3771, 2639, 4046, 2641} \[ \frac{2 a \left (a^2+b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}-\frac{6 b \left (5 a^2+b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{6 b \left (5 a^2+b^2\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{8 a b^2 \sin (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 b^2 \sin (c+d x) (a+b \sec (c+d x))}{5 d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

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

Mathematica [A]  time = 0.8762, size = 125, normalized size = 0.84 \[ \frac{10 a \left (a^2+b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+3 \left (5 a^2 b+b^3\right ) \sin (2 (c+d x))-6 b \left (5 a^2+b^2\right ) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+10 a b^2 \sin (c+d x)+2 b^3 \tan (c+d x)}{5 d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*b*(5*a^2 + b^2)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*a*(a^2 + b^2)*Cos[c + d*x]^(3/2)*Ellipti
cF[(c + d*x)/2, 2] + 10*a*b^2*Sin[c + d*x] + 3*(5*a^2*b + b^3)*Sin[2*(c + d*x)] + 2*b^3*Tan[c + d*x])/(5*d*Cos
[c + d*x]^(3/2))

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Maple [B]  time = 4.534, size = 738, normalized size = 5. \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))^3/cos(d*x+c)^(1/2),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)*EllipticE(cos(1/2*d*x+1/2*
c),2^(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*(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))*(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)+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(-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))^3/cos(d*x+c)^(1/2),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{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}}{\sqrt{\cos \left (d x + c\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^3/cos(d*x+c)^(1/2),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(cos(d*x + c)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \sec{\left (c + d x \right )}\right )^{3}}{\sqrt{\cos{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**3/cos(d*x+c)**(1/2),x)

[Out]

Integral((a + b*sec(c + d*x))**3/sqrt(cos(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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