3.2.72 \(\int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx\) [172]

Optimal. Leaf size=187 \[ -\frac {12 a^2 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {8 a^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{7 d}+\frac {12 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {4 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \]

[Out]

8/7*a^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d+4/5*a^2*sec(d*x+c)^(5/2)*sin(d*x+c)/d+2/7*a^2*sec(d*x+c)^(7/2)*sin(d*x+c
)/d+12/5*a^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d-12/5*a^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+8/7*a^2*(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 = 0.11, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3873, 3853, 3856, 2719, 4131, 2720} \begin {gather*} \frac {2 a^2 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}+\frac {4 a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {8 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{7 d}+\frac {12 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {8 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{7 d}-\frac {12 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*a^2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (8*a^2*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(7*d) + (12*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (8*a^2*Se
c[c + d*x]^(3/2)*Sin[c + d*x])/(7*d) + (4*a^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (2*a^2*Sec[c + d*x]^(7/
2)*Sin[c + d*x])/(7*d)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3853

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 3856

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 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4131

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]

Rubi steps

\begin {align*} \int \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2 \, dx &=\left (2 a^2\right ) \int \sec ^{\frac {7}{2}}(c+d x) \, dx+\int \sec ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {4 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{5} \left (6 a^2\right ) \int \sec ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{7} \left (12 a^2\right ) \int \sec ^{\frac {5}{2}}(c+d x) \, dx\\ &=\frac {12 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {4 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{7} \left (4 a^2\right ) \int \sqrt {\sec (c+d x)} \, dx-\frac {1}{5} \left (6 a^2\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {12 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {4 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{7} \left (4 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (6 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=-\frac {12 a^2 \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^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{7 d}+\frac {12 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {4 a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^2 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 2.17, size = 285, normalized size = 1.52 \begin {gather*} \frac {a^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^2 \left (-\frac {2 i \sqrt {2} e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^2(c+d x) \left (21 e^{i c} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )+e^{i d x} \left (10 \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )+7 e^{i (c+d x)} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right )\right )}{-1+e^{2 i c}}+\frac {42 \cos (d x) \csc (c)+(15+14 \cos (c+d x)+10 \cos (2 (c+d x))) \sec ^2(c+d x) \tan (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)}\right )}{70 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(((-2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*
Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^2*(21*E^(I*c)*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x)
)] + E^(I*d*x)*(10*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))] + 7*E^(I*(c + d*x
))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])))/(E^(I*d*x)*(-1 + E^((2*I)*c))) + (42*Cos[d*x]*Csc
[c] + (15 + 14*Cos[c + d*x] + 10*Cos[2*(c + d*x)])*Sec[c + d*x]^2*Tan[c + d*x])/Sec[c + d*x]^(3/2)))/(70*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(438\) vs. \(2(211)=422\).
time = 0.10, size = 439, normalized size = 2.35

method	result	size
default	\(-\frac {a^{2} \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{28 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{4}}-\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{7 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}+\frac {124 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{35 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{5 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{3}}-\frac {24 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}-\frac {12 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(439\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1/28*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)^4-4/7*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+s
in(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+124/35*(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))-1/5
*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)^3-24/5*sin
(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)-12/5*(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)*(Ell
ipticF(cos(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*cos(1/2*d*x+1
/2*c)^2-1)^(1/2)/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.60, size = 215, normalized size = 1.15 \begin {gather*} -\frac {2 \, {\left (10 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 10 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (42 \, a^{2} \cos \left (d x + c\right )^{3} + 20 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) + 5 \, a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{35 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/35*(10*I*sqrt(2)*a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 10*I*sqrt(2
)*a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*I*sqrt(2)*a^2*cos(d*x + c)
^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I*sqrt(2)*a^2*cos(d*
x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (42*a^2*cos(d*x +
c)^3 + 20*a^2*cos(d*x + c)^2 + 14*a^2*cos(d*x + c) + 5*a^2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^
3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(5/2),x)

[Out]

int((a + a/cos(c + d*x))^2*(1/cos(c + d*x))^(5/2), x)

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