3.602 \(\int \frac{(a+b \sec (c+d x))^4}{\sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=209 \[ \frac{8 a b \left (3 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 b^2 \left (29 a^2+3 b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 \left (-30 a^2 b^2+5 a^4-3 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{28 a b^3 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b^2 \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d} \]

[Out]

(2*(5*a^4 - 30*a^2*b^2 - 3*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (8*a*
b*(3*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*b^2*(29*a^2 + 3*b^
2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (28*a*b^3*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*b^2*Sqrt[Se
c[c + d*x]]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.349171, antiderivative size = 209, 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, 4076, 4047, 3771, 2641, 4046, 2639} \[ \frac{2 b^2 \left (29 a^2+3 b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}+\frac{8 a b \left (3 a^2+b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 \left (-30 a^2 b^2+5 a^4-3 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{28 a b^3 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b^2 \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^4/Sqrt[Sec[c + d*x]],x]

[Out]

(2*(5*a^4 - 30*a^2*b^2 - 3*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (8*a*
b*(3*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*b^2*(29*a^2 + 3*b^
2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (28*a*b^3*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*d) + (2*b^2*Sqrt[Se
c[c + d*x]]*(a + b*Sec[c + d*x])^2*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 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 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 \frac{(a+b \sec (c+d x))^4}{\sqrt{\sec (c+d x)}} \, dx &=\frac{2 b^2 \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{2}{5} \int \frac{(a+b \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)+7 a b^2 \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{28 a b^3 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 b^2 \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{4}{15} \int \frac{\frac{3}{4} a^2 \left (5 a^2-b^2\right )+5 a b \left (3 a^2+b^2\right ) \sec (c+d x)+\frac{3}{4} b^2 \left (29 a^2+3 b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{28 a b^3 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 b^2 \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{4}{15} \int \frac{\frac{3}{4} a^2 \left (5 a^2-b^2\right )+\frac{3}{4} b^2 \left (29 a^2+3 b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (4 a b \left (3 a^2+b^2\right )\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 b^2 \left (29 a^2+3 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{28 a b^3 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 b^2 \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{1}{5} \left (5 a^4-30 a^2 b^2-3 b^4\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (4 a b \left (3 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{8 a b \left (3 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)}}{3 d}+\frac{2 b^2 \left (29 a^2+3 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{28 a b^3 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 b^2 \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac{1}{5} \left (\left (5 a^4-30 a^2 b^2-3 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (5 a^4-30 a^2 b^2-3 b^4\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{8 a b \left (3 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)}}{3 d}+\frac{2 b^2 \left (29 a^2+3 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{28 a b^3 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac{2 b^2 \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 2.12493, size = 146, normalized size = 0.7 \[ \frac{\sec ^{\frac{5}{2}}(c+d x) \left (b \left (80 a \left (3 a^2+b^2\right ) \cos ^{\frac{5}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+2 b \sin (c+d x) \left (9 \left (10 a^2+b^2\right ) \cos (2 (c+d x))+15 \left (6 a^2+b^2\right )+40 a b \cos (c+d x)\right )\right )+12 \left (-30 a^2 b^2+5 a^4-3 b^4\right ) \cos ^{\frac{5}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^4/Sqrt[Sec[c + d*x]],x]

[Out]

(Sec[c + d*x]^(5/2)*(12*(5*a^4 - 30*a^2*b^2 - 3*b^4)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + b*(80*a*(3
*a^2 + b^2)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 2*b*(15*(6*a^2 + b^2) + 40*a*b*Cos[c + d*x] + 9*(10
*a^2 + b^2)*Cos[2*(c + d*x)])*Sin[c + d*x])))/(30*d)

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Maple [B]  time = 4.679, size = 907, normalized size = 4.3 \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))^4/sec(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^4*(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))
-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*a^4*(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))+8*a^3*b*(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)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))+8*a*b^3*(-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*si
n(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^4/(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)+12*a^2*b^2*(-(-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 \frac{{\left (b \sec \left (d x + c\right ) + a\right )}^{4}}{\sqrt{\sec \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))^4/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^4/sqrt(sec(d*x + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{4} \sec \left (d x + c\right )^{4} + 4 \, a b^{3} \sec \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \sec \left (d x + c\right )^{2} + 4 \, a^{3} b \sec \left (d x + c\right ) + a^{4}}{\sqrt{\sec \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))^4/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b^4*sec(d*x + c)^4 + 4*a*b^3*sec(d*x + c)^3 + 6*a^2*b^2*sec(d*x + c)^2 + 4*a^3*b*sec(d*x + c) + a^4)
/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((a+b*sec(d*x+c))**4/sec(d*x+c)**(1/2),x)

[Out]

Timed out

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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 )}^{4}}{\sqrt{\sec \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))^4/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^4/sqrt(sec(d*x + c)), x)