3.200 \(\int \frac{1}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\)

Optimal. Leaf size=168 \[ -\frac{5 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a d}-\frac{\sin (c+d x)}{d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)}+\frac{7 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}+\frac{21 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a d} \]

[Out]

(21*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a*d) - (5*Sqrt[Cos[c + d*x]]*EllipticF
[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) + (7*Sin[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)) - (5*Sin[c + d*x])/
(3*a*d*Sqrt[Sec[c + d*x]]) - Sin[c + d*x]/(d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x]))

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Rubi [A]  time = 0.129467, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3819, 3787, 3769, 3771, 2639, 2641} \[ -\frac{\sin (c+d x)}{d \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)}+\frac{7 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{5 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{21 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a d} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x]

[Out]

(21*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a*d) - (5*Sqrt[Cos[c + d*x]]*EllipticF
[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) + (7*Sin[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)) - (5*Sin[c + d*x])/
(3*a*d*Sqrt[Sec[c + d*x]]) - Sin[c + d*x]/(d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x]))

Rule 3819

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(f*(a + b*Csc[e + f*x])), x] - Dist[1/a^2, Int[(d*Csc[e + f*x])^n*(a*(n - 1) - b*n*Csc[
e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 3787

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

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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]

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]

Rubi steps

\begin{align*} \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx &=-\frac{\sin (c+d x)}{d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac{\int \frac{-\frac{7 a}{2}+\frac{5}{2} a \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{a^2}\\ &=-\frac{\sin (c+d x)}{d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac{5 \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{2 a}+\frac{7 \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{2 a}\\ &=\frac{7 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac{5 \int \sqrt{\sec (c+d x)} \, dx}{6 a}+\frac{21 \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{10 a}\\ &=\frac{7 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))}-\frac{\left (5 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a}+\frac{\left (21 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{10 a}\\ &=\frac{21 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 a d}-\frac{5 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a d}+\frac{7 \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{5 \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{d \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 2.77552, size = 347, normalized size = 2.07 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (-\sqrt{\sec (c+d x)} \left (18 (11 \cos (2 c)+17) \csc (c) \cos (d x)+4 \left (10 \sin (2 c) \cos (2 d x)-3 \sin (3 c) \cos (3 d x)-99 \cos (c) \sin (d x)+10 \cos (2 c) \sin (2 d x)-3 \cos (3 c) \sin (3 d x)-30 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right )-30 \tan \left (\frac{c}{2}\right )\right )\right )+\frac{8 i \sqrt{2} e^{-i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (63 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+25 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+63 \left (1+e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}\right )}{60 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x]

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(((8*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(63*(1 + E^((
2*I)*(c + d*x))) + 63*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((
2*I)*(c + d*x))] + 25*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4,
1/2, 5/4, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) - Sqrt[Sec[c + d*x]]*(18*(17 + 11*Cos[2
*c])*Cos[d*x]*Csc[c] + 4*(10*Cos[2*d*x]*Sin[2*c] - 3*Cos[3*d*x]*Sin[3*c] - 30*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d
*x)/2] - 99*Cos[c]*Sin[d*x] + 10*Cos[2*c]*Sin[2*d*x] - 3*Cos[3*c]*Sin[3*d*x] - 30*Tan[c/2]))))/(60*a*d*(1 + Se
c[c + d*x]))

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Maple [A]  time = 1.399, size = 229, normalized size = 1.4 \begin{align*} -{\frac{1}{15\,ad}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( -\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 25\,{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +63\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) +48\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-56\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-30\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+23\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15/a*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1
)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(25*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+63*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2)))+48*sin(1/2*d*x+1/2*c)^8-56*sin(1/2*d*x+1/2*c)^6-30*sin(1/2*d*x+1/2*c)^4+23*sin(1/2*d*x+1/2*c)^2)/co
s(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)/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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sec \left (d x + c\right ) + a\right )} \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(1/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate(1/((a*sec(d*x + c) + a)*sec(d*x + c)^(5/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\sec \left (d x + c\right )}}{a \sec \left (d x + c\right )^{4} + a \sec \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral(sqrt(sec(d*x + c))/(a*sec(d*x + c)^4 + a*sec(d*x + c)^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\sec ^{\frac{7}{2}}{\left (c + d x \right )} + \sec ^{\frac{5}{2}}{\left (c + d x \right )}}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c)),x)

[Out]

Integral(1/(sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x)/a

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sec \left (d x + c\right ) + a\right )} \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(1/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)*sec(d*x + c)^(5/2)), x)