3.9 \(\int \sec ^{\frac{7}{2}}(a+b x) \, dx\)

Optimal. Leaf size=85 \[ \frac{2 \sin (a+b x) \sec ^{\frac{5}{2}}(a+b x)}{5 b}+\frac{6 \sin (a+b x) \sqrt{\sec (a+b x)}}{5 b}-\frac{6 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.0370816, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3768, 3771, 2639} \[ \frac{2 \sin (a+b x) \sec ^{\frac{5}{2}}(a+b x)}{5 b}+\frac{6 \sin (a+b x) \sqrt{\sec (a+b x)}}{5 b}-\frac{6 \sqrt{\cos (a+b x)} \sqrt{\sec (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^(7/2),x]

[Out]

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

Rule 3768

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

Mathematica [A]  time = 0.174773, size = 59, normalized size = 0.69 \[ \frac{\sec ^{\frac{5}{2}}(a+b x) \left (7 \sin (a+b x)+3 \sin (3 (a+b x))-12 \cos ^{\frac{5}{2}}(a+b x) E\left (\left .\frac{1}{2} (a+b x)\right |2\right )\right )}{10 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^(7/2),x]

[Out]

(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 [B]  time = 2.199, size = 358, normalized size = 4.2 \begin{align*}{\frac{2}{5\,b}\sqrt{- \left ( -2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}} \left ( 12\,\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}-24\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}\cos \left ( 1/2\,bx+a/2 \right ) -12\,\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+24\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}\cos \left ( 1/2\,bx+a/2 \right ) +3\,\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) -8\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}\cos \left ( 1/2\,bx+a/2 \right ) \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}} \left ( 8\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}-12\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+6\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) ^{-1} \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-3}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^(7/2),x)

[Out]

2/5*(-(-2*cos(1/2*b*x+1/2*a)^2+1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/(8*sin(1/2*b*x+1/2*a)^6-12*sin(1/2*b*x+1/2*a)^4+
6*sin(1/2*b*x+1/2*a)^2-1)/sin(1/2*b*x+1/2*a)^3*(12*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^(1/
2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))*sin(1/2*b*x+1/2*a)^4-24*sin(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)-12*(2
*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))*sin(1/2*b*x+
1/2*a)^2+24*sin(1/2*b*x+1/2*a)^4*cos(1/2*b*x+1/2*a)+3*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*(sin(1/2*b*x+1/2*a)^2)^
(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))-8*sin(1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a))*(-2*sin(1/2*b*x+1/2*a)^
4+sin(1/2*b*x+1/2*a)^2)^(1/2)/(2*cos(1/2*b*x+1/2*a)^2-1)^(1/2)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec \left (b x + a\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(b*x + a)^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sec \left (b x + a\right )^{\frac{7}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sec(b*x + a)^(7/2), 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(sec(b*x+a)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec \left (b x + a\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(b*x + a)^(7/2), x)