3.9.27 \(\int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\) [827]

Optimal. Leaf size=148 \[ -\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\left (a^2-b^2\right ) d}-\frac {b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a \left (a^2-b^2\right ) d}+\frac {\left (a^2+b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a (a-b) (a+b)^2 d}+\frac {a \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))} \]

[Out]

-(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))/(a^2-b^2)/d-b*(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))/a/d/(a^2-b^2)+(a^2+b^2)*(cos(1/2*
d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a/(a-b)/(a+b)^2/d+a*si
n(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.29, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {4349, 3929, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \begin {gather*} -\frac {b F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \left (a^2-b^2\right )}+\frac {\left (a^2+b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a d (a-b) (a+b)^2}+\frac {a \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(EllipticE[(c + d*x)/2, 2]/((a^2 - b^2)*d)) - (b*EllipticF[(c + d*x)/2, 2])/(a*(a^2 - b^2)*d) + ((a^2 + b^2)*
EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a*(a - b)*(a + b)^2*d) + (a*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[Cos[
c + d*x]]*(a + b*Sec[c + d*x]))

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 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

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 3872

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 3929

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

Rule 3934

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4191

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 4349

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]

Rubi steps

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

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Mathematica [A]
time = 12.07, size = 229, normalized size = 1.55 \begin {gather*} \frac {\frac {4 a \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x))}-\frac {2 \left (-\frac {a^2 \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+2 b \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )+\frac {\left (-2 a b E\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right ) \sin (c+d x)}{b \sqrt {\sin ^2(c+d x)}}\right )}{a (a-b) (a+b)}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(b + a*Cos[c + d*x])) - (2*(-((a^2*EllipticPi[(2*a)/(a + b
), (c + d*x)/2, 2])/(a + b)) + 2*b*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2,
2])/(a + b)) + ((-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c +
d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(b*Sqrt[Sin[c +
d*x]^2])))/(a*(a - b)*(a + b)))/(4*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(706\) vs. \(2(224)=448\).
time = 0.29, size = 707, normalized size = 4.78

method	result	size
default	\(-\frac {\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 {2 b \left (\frac {a^{2} \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 )}}{b \left (a^{2}-b^{2}\right ) \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -a +b \right )}-\frac {\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 )}{2 \left (a +b \right ) b \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 {a \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 )}{2 b \left (a^{2}-b^{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 )}}-\frac {a \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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{2 b \left (a^{2}-b^{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 )}}-\frac {a^{3} \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}\, \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right )}{2 b \left (a^{2}-b^{2}\right ) \left (a^{2}-b a \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 )}}+\frac {3 b a \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}\, \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right )}{2 \left (a^{2}-b^{2}\right ) \left (a^{2}-b a \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 )}}\right )}{a}-\frac {2 \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}\, \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right )}{\left (a^{2}-b a \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 )}}\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}\)	\(707\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/a*(1/b*a^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-1/2/(a+b)/b*(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))+1/2/b*a/(a^2-b^2)*(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))-1/2/b*a/(a^2-b^2)*(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)
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^3*(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(
a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(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*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))-2/(a^2-a*b)*
(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),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)] 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(1/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \sec {\left (c + d x \right )}\right )^{2} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((a + b*sec(c + d*x))**2*cos(c + d*x)**(3/2)), x)

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^(3/2)*(a + b/cos(c + d*x))^2),x)

[Out]

int(1/(cos(c + d*x)^(3/2)*(a + b/cos(c + d*x))^2), x)

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