3.1.17 \(\int (c \sec (a+b x))^{7/2} \, dx\) [17]

Optimal. Leaf size=98 \[ -\frac {6 c^4 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b \sqrt {\cos (a+b x)} \sqrt {c \sec (a+b x)}}+\frac {6 c^3 \sqrt {c \sec (a+b x)} \sin (a+b x)}{5 b}+\frac {2 c (c \sec (a+b x))^{5/2} \sin (a+b x)}{5 b} \]

[Out]

2/5*c*(c*sec(b*x+a))^(5/2)*sin(b*x+a)/b-6/5*c^4*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticE(sin(
1/2*a+1/2*b*x),2^(1/2))/b/cos(b*x+a)^(1/2)/(c*sec(b*x+a))^(1/2)+6/5*c^3*sin(b*x+a)*(c*sec(b*x+a))^(1/2)/b

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Rubi [A]
time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3853, 3856, 2719} \begin {gather*} -\frac {6 c^4 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b \sqrt {\cos (a+b x)} \sqrt {c \sec (a+b x)}}+\frac {6 c^3 \sin (a+b x) \sqrt {c \sec (a+b x)}}{5 b}+\frac {2 c \sin (a+b x) (c \sec (a+b x))^{5/2}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sec[a + b*x])^(7/2),x]

[Out]

(-6*c^4*EllipticE[(a + b*x)/2, 2])/(5*b*Sqrt[Cos[a + b*x]]*Sqrt[c*Sec[a + b*x]]) + (6*c^3*Sqrt[c*Sec[a + b*x]]
*Sin[a + b*x])/(5*b) + (2*c*(c*Sec[a + b*x])^(5/2)*Sin[a + b*x])/(5*b)

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 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]

Rubi steps

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

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Mathematica [A]
time = 0.18, size = 62, normalized size = 0.63 \begin {gather*} \frac {c (c \sec (a+b x))^{5/2} \left (-12 \cos ^{\frac {5}{2}}(a+b x) E\left (\left .\frac {1}{2} (a+b x)\right |2\right )+7 \sin (a+b x)+3 \sin (3 (a+b x))\right )}{10 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*Sec[a + b*x])^(7/2),x]

[Out]

(c*(c*Sec[a + b*x])^(5/2)*(-12*Cos[a + b*x]^(5/2)*EllipticE[(a + b*x)/2, 2] + 7*Sin[a + b*x] + 3*Sin[3*(a + b*
x)]))/(10*b)

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Maple [C] Result contains complex when optimal does not.
time = 42.48, size = 354, normalized size = 3.61

method	result	size
default	\(\frac {2 \left (-1+\cos \left (b x +a \right )\right )^{2} \left (3 i \sin \left (b x +a \right ) \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {\frac {1}{\cos \left (b x +a \right )+1}}\, \sqrt {\frac {\cos \left (b x +a \right )}{\cos \left (b x +a \right )+1}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}, i\right )-3 i \sin \left (b x +a \right ) \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {\frac {1}{\cos \left (b x +a \right )+1}}\, \sqrt {\frac {\cos \left (b x +a \right )}{\cos \left (b x +a \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}, i\right )+3 i \sin \left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {\frac {1}{\cos \left (b x +a \right )+1}}\, \sqrt {\frac {\cos \left (b x +a \right )}{\cos \left (b x +a \right )+1}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}, i\right )-3 i \sin \left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {\frac {1}{\cos \left (b x +a \right )+1}}\, \sqrt {\frac {\cos \left (b x +a \right )}{\cos \left (b x +a \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}, i\right )-3 \left (\cos ^{3}\left (b x +a \right )\right )+2 \left (\cos ^{2}\left (b x +a \right )\right )+1\right ) \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {7}{2}}}{5 b \sin \left (b x +a \right )^{5}}\)	\(354\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sec(b*x+a))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/5/b*(-1+cos(b*x+a))^2*(3*I*sin(b*x+a)*cos(b*x+a)^3*(1/(cos(b*x+a)+1))^(1/2)*(cos(b*x+a)/(cos(b*x+a)+1))^(1/2
)*EllipticE(I*(-1+cos(b*x+a))/sin(b*x+a),I)-3*I*sin(b*x+a)*cos(b*x+a)^3*(1/(cos(b*x+a)+1))^(1/2)*(cos(b*x+a)/(
cos(b*x+a)+1))^(1/2)*EllipticF(I*(-1+cos(b*x+a))/sin(b*x+a),I)+3*I*sin(b*x+a)*cos(b*x+a)^2*(1/(cos(b*x+a)+1))^
(1/2)*(cos(b*x+a)/(cos(b*x+a)+1))^(1/2)*EllipticE(I*(-1+cos(b*x+a))/sin(b*x+a),I)-3*I*sin(b*x+a)*cos(b*x+a)^2*
(1/(cos(b*x+a)+1))^(1/2)*(cos(b*x+a)/(cos(b*x+a)+1))^(1/2)*EllipticF(I*(-1+cos(b*x+a))/sin(b*x+a),I)-3*cos(b*x
+a)^3+2*cos(b*x+a)^2+1)*cos(b*x+a)*(cos(b*x+a)+1)^2*(c/cos(b*x+a))^(7/2)/sin(b*x+a)^5

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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((c*sec(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

integrate((c*sec(b*x + a))^(7/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.48, size = 125, normalized size = 1.28 \begin {gather*} \frac {-3 i \, \sqrt {2} c^{\frac {7}{2}} \cos \left (b x + a\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 i \, \sqrt {2} c^{\frac {7}{2}} \cos \left (b x + a\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + 2 \, {\left (3 \, c^{3} \cos \left (b x + a\right )^{2} + c^{3}\right )} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sin \left (b x + a\right )}{5 \, b \cos \left (b x + a\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sec(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

1/5*(-3*I*sqrt(2)*c^(7/2)*cos(b*x + a)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) + I*si
n(b*x + a))) + 3*I*sqrt(2)*c^(7/2)*cos(b*x + a)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x +
a) - I*sin(b*x + a))) + 2*(3*c^3*cos(b*x + a)^2 + c^3)*sqrt(c/cos(b*x + a))*sin(b*x + a))/(b*cos(b*x + a)^2)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sec(b*x+a))**(7/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3061 deep

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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((c*sec(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate((c*sec(b*x + a))^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{7/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/cos(a + b*x))^(7/2),x)

[Out]

int((c/cos(a + b*x))^(7/2), x)

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