3.604 \(\int \frac{(a+b \sec (c+d x))^4}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=207 \[ \frac{8 a b \left (a^2+3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{2 b^2 \left (a^2-5 b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 \left (30 a^2 b^2+3 a^4-5 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{28 a^3 b \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.36943, antiderivative size = 207, 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 = {3841, 4074, 4047, 3771, 2641, 4046, 2639} \[ -\frac{2 b^2 \left (a^2-5 b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}+\frac{8 a b \left (a^2+3 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 \left (30 a^2 b^2+3 a^4-5 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{28 a^3 b \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^4/Sec[c + d*x]^(5/2),x]

[Out]

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

Rule 3841

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

Rule 4074

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[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, 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}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+b \sec (c+d x)) \left (7 a^2 b+\frac{3}{2} a \left (a^2+5 b^2\right ) \sec (c+d x)-\frac{1}{2} b \left (a^2-5 b^2\right ) \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{28 a^3 b \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{4}{15} \int \frac{-\frac{3}{4} a^2 \left (3 a^2+29 b^2\right )-5 a b \left (a^2+3 b^2\right ) \sec (c+d x)+\frac{3}{4} b^2 \left (a^2-5 b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{28 a^3 b \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{4}{15} \int \frac{-\frac{3}{4} a^2 \left (3 a^2+29 b^2\right )+\frac{3}{4} b^2 \left (a^2-5 b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (4 a b \left (a^2+3 b^2\right )\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{28 a^3 b \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 b^2 \left (a^2-5 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{1}{5} \left (-3 a^4-30 a^2 b^2+5 b^4\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (4 a b \left (a^2+3 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 (a^2+3 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{28 a^3 b \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 b^2 \left (a^2-5 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{1}{5} \left (\left (-3 a^4-30 a^2 b^2+5 b^4\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (3 a^4+30 a^2 b^2-5 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 (a^2+3 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{28 a^3 b \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 b^2 \left (a^2-5 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.58226, size = 138, normalized size = 0.67 \[ \frac{\sqrt{\sec (c+d x)} \left (80 a b \left (a^2+3 b^2\right ) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+2 \sin (c+d x) \left (40 a^3 b \cos (c+d x)+3 a^4 \cos (2 (c+d x))+3 a^4+30 b^4\right )+12 \left (30 a^2 b^2+3 a^4-5 b^4\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^4/Sec[c + d*x]^(5/2),x]

[Out]

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

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Maple [B]  time = 1.752, size = 619, normalized size = 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)^(5/2),x)

[Out]

-2/15*(-24*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+8*
(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(3*a+10*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*
(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*a^4+20*a^3*b+15*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+
1/2*c)+20*a^3*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(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)+60*a*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*
x+1/2*c)^2-1)^(1/2)*EllipticF(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)
-9*a^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-90*a^2*b^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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*(-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))*b^4)/(-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*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))^4/sec(d*x+c)^(5/2),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^{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}}{\sec \left (d x + c\right )^{\frac{5}{2}}}, 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)^(5/2),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)
/sec(d*x + c)^(5/2), 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 )^{4}}{\sec ^{\frac{5}{2}}{\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))**4/sec(d*x+c)**(5/2),x)

[Out]

Integral((a + b*sec(c + d*x))**4/sec(c + d*x)**(5/2), 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 )}^{4}}{\sec \left (d x + c\right )^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^4/sec(d*x + c)^(5/2), x)