3.603 \(\int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=208 \[ -\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {4 a b \left (a^2-6 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d}+\frac {8 a b \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d} \]

[Out]

-2/3*b^2*(a^2-b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/3*a^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(1/2)-4/3*
a*b*(a^2-6*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+8*a*b*(a^2-b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*
EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/3*(a^4+18*a^2*b^2+b^4)*(cos(1/2*d*
x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.35, antiderivative size = 208, normalized size of antiderivative = 1.00, 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, 4076, 4047, 3771, 2641, 4046, 2639} \[ -\frac {2 b^2 \left (a^2-b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {4 a b \left (a^2-6 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 \left (18 a^2 b^2+a^4+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {8 a b \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))^2}{3 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^4/Sec[c + d*x]^(3/2),x]

[Out]

(8*a*b*(a^2 - b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*(a^4 + 18*a^2*b^2 +
b^4)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) - (4*a*b*(a^2 - 6*b^2)*Sqrt[Sec[c
+ d*x]]*Sin[c + d*x])/(3*d) - (2*b^2*(a^2 - b^2)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + (2*a^2*(a + b*Sec[c
+ d*x])^2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

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]

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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {(a+b \sec (c+d x))^4}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2}{3} \int \frac {(a+b \sec (c+d x)) \left (5 a^2 b+\frac {1}{2} a \left (a^2+9 b^2\right ) \sec (c+d x)-\frac {3}{2} b \left (a^2-b^2\right ) \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=-\frac {2 b^2 \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {4}{9} \int \frac {\frac {15 a^3 b}{2}+\frac {3}{4} \left (a^4+18 a^2 b^2+b^4\right ) \sec (c+d x)-\frac {3}{2} a b \left (a^2-6 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\\ &=-\frac {2 b^2 \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {4}{9} \int \frac {\frac {15 a^3 b}{2}-\frac {3}{2} a b \left (a^2-6 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (a^4+18 a^2 b^2+b^4\right ) \int \sqrt {\sec (c+d x)} \, dx\\ &=-\frac {4 a b \left (a^2-6 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\left (4 a b \left (a^2-b^2\right )\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (\left (a^4+18 a^2 b^2+b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (a^4+18 a^2 b^2+b^4\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 {4 a b \left (a^2-6 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\left (4 a b \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {8 a b \left (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)}}{d}+\frac {2 \left (a^4+18 a^2 b^2+b^4\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 {4 a b \left (a^2-6 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a^2 (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.24, size = 130, normalized size = 0.62 \[ \frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \left (a^4 \cos (2 (c+d x))+a^4+24 a b^3 \cos (c+d x)+2 b^4\right )}{\cos ^{\frac {3}{2}}(c+d x)}+24 a b \left (a^2-b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \left (a^4+18 a^2 b^2+b^4\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^4/Sec[c + d*x]^(3/2),x]

[Out]

(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(24*a*b*(a^2 - b^2)*EllipticE[(c + d*x)/2, 2] + 2*(a^4 + 18*a^2*b^2 + b
^4)*EllipticF[(c + d*x)/2, 2] + ((a^4 + 2*b^4 + 24*a*b^3*Cos[c + d*x] + a^4*Cos[2*(c + d*x)])*Sin[c + d*x])/Co
s[c + d*x]^(3/2)))/(3*d)

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{4} \sec \left (d x + c\right )^{4} + 4 \, a b^{3} \sec \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \sec \left (d x + c\right )^{2} + 4 \, a^{3} b \sec \left (d x + c\right ) + a^{4}}{\sec \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^4/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 5.02, size = 777, normalized size = 3.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(-8*(-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*(a^3+6*b^3)*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)*(a^4+12*a*b^3+b^4)*sin(1/2*d*x+1/2*c)^2*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)*(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))*a^4+18*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2+EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))*b^4-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b+12*EllipticE(cos(1/2*d*x+1/2*c),2
^(1/2))*a*b^3)*sin(1/2*d*x+1/2*c)^2+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)*Elliptic
F(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)+18*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+s
in(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)*(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))*b^4-12*(-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))*a^3*b+12*(-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))*a*b^3)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{4}}{\sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^4/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^4}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^4/(1/cos(c + d*x))^(3/2),x)

[Out]

int((a + b/cos(c + d*x))^4/(1/cos(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{4}}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**4/sec(d*x+c)**(3/2),x)

[Out]

Integral((a + b*sec(c + d*x))**4/sec(c + d*x)**(3/2), x)

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