3.395 \(\int \frac{1}{\cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=155 \[ \frac{\text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{2 a^3 d}+\frac{9 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{9 \sin (c+d x)}{10 d \sqrt{\cos (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}-\frac{2 \sin (c+d x)}{5 a d \cos ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^3} \]

[Out]

(9*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + EllipticF[(c + d*x)/2, 2]/(2*a^3*d) - Sin[c + d*x]/(5*d*Cos[c + d*x
]^(5/2)*(a + a*Sec[c + d*x])^3) - (2*Sin[c + d*x])/(5*a*d*Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2) - (9*Sin[
c + d*x])/(10*d*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.390425, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4264, 3816, 4019, 3787, 3771, 2639, 2641} \[ \frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}+\frac{9 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{9 \sin (c+d x)}{10 d \sqrt{\cos (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}-\frac{2 \sin (c+d x)}{5 a d \cos ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

(9*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + EllipticF[(c + d*x)/2, 2]/(2*a^3*d) - Sin[c + d*x]/(5*d*Cos[c + d*x
]^(5/2)*(a + a*Sec[c + d*x])^3) - (2*Sin[c + d*x])/(5*a*d*Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2) - (9*Sin[
c + d*x])/(10*d*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Sec[c + d*x]))

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 3816

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(d^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2))/(f*(2*m + 1)), x] + Dist[d^2/(a*b*(2*m + 1)), In
t[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m]
)

Rule 4019

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/
(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[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 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}{\cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{7}{2}}(c+d x)}{(a+a \sec (c+d x))^3} \, dx\\ &=-\frac{\sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{3 a}{2}-\frac{9}{2} a \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{2 \sin (c+d x)}{5 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)} \left (3 a^2-\frac{21}{2} a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{\sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{2 \sin (c+d x)}{5 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{9 \sin (c+d x)}{10 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{27 a^3}{4}-\frac{15}{4} a^3 \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a^6}\\ &=-\frac{\sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{2 \sin (c+d x)}{5 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{9 \sin (c+d x)}{10 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx}{4 a^3}+\frac{\left (9 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}\\ &=-\frac{\sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{2 \sin (c+d x)}{5 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{9 \sin (c+d x)}{10 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{4 a^3}+\frac{9 \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac{9 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{2 a^3 d}-\frac{\sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac{2 \sin (c+d x)}{5 a d \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac{9 \sin (c+d x)}{10 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 6.34794, size = 721, normalized size = 4.65 \[ -\frac{2 \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^3(c+d x) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin (c) \left (-\sqrt{\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{d \sqrt{\cot ^2(c)+1} (a \sec (c+d x)+a)^3}+\frac{9 i \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^3(c+d x) \left (\frac{2 e^{2 i d x} \sqrt{e^{-i d x} \left (2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )\right )} \sqrt{i \sin (2 c) e^{2 i d x}+\cos (2 c) e^{2 i d x}+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )}{3 i d \cos (c) \left (1+e^{2 i d x}\right )-3 d \sin (c) \left (-1+e^{2 i d x}\right )}-\frac{2 \sqrt{e^{-i d x} \left (2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )\right )} \sqrt{i \sin (2 c) e^{2 i d x}+\cos (2 c) e^{2 i d x}+1} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )}{d \sin (c) \left (-1+e^{2 i d x}\right )-i d \cos (c) \left (1+e^{2 i d x}\right )}\right )}{10 (a \sec (c+d x)+a)^3}+\frac{\cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (-\frac{2 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \sec ^5\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}-\frac{8 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}-\frac{2 \tan \left (\frac{c}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}-\frac{8 \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}-\frac{36 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}-\frac{36 \csc (c)}{5 d}\right )}{\cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

(((9*I)/10)*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^3*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4
, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))
*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*
Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*S
in[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I
)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c
])))/(a + a*Sec[c + d*x])^3 - (2*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x -
ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^3*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[
-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*Sqrt[1 + Cot[c
]^2]*(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*((-36*Csc[c])/(5*d) - (36*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin
[(d*x)/2])/(5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*Sin[(d*x)/2])/(5*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*Sin
[(d*x)/2])/(5*d) - (8*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(5*d) - (2*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(Cos[c
+ d*x]^(5/2)*(a + a*Sec[c + d*x])^3)

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Maple [A]  time = 1.578, size = 268, normalized size = 1.7 \begin{align*}{\frac{1}{20\,{a}^{3}d}\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 ( 36\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-10\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+18\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -46\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+8\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5}{\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/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x)

[Out]

1/20*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(36*cos(1/2*d*x+1/2*c)^8-10*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+18*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2*d*x+1/2*c),2^(
1/2))-46*cos(1/2*d*x+1/2*c)^6+8*cos(1/2*d*x+1/2*c)^4+cos(1/2*d*x+1/2*c)^2+1)/a^3/cos(1/2*d*x+1/2*c)^5/(-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(-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(1/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\cos \left (d x + c\right )}}{a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right )^{4} \sec \left (d x + c\right ) + a^{3} \cos \left (d x + c\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(sqrt(cos(d*x + c))/(a^3*cos(d*x + c)^4*sec(d*x + c)^3 + 3*a^3*cos(d*x + c)^4*sec(d*x + c)^2 + 3*a^3*c
os(d*x + c)^4*sec(d*x + c) + a^3*cos(d*x + c)^4), 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(1/cos(d*x+c)**(7/2)/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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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 )}^{3} \cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)^3*cos(d*x + c)^(7/2)), x)