3.6.83 \(\int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx\) [583]

Optimal. Leaf size=203 \[ \frac {(A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-A b^2-a b B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (3 a^2 A b-A b^3-a^3 B-a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 (a-b) (a+b)^2 d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))} \]

[Out]

(A*b-B*a)*(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/(a^2-b^2)/d+
(2*A*a^2-A*b^2-B*a*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^
2/(a^2-b^2)/d-(3*A*a^2*b-A*b^3-B*a^3-B*a*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^2/(a-b)/(a+b)^2/d-(A*b-B*a)*sin(d*x+c)*cos(d*x+c)^(1/2)/(a^2-b^2)/d/(b+a*co
s(d*x+c))

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Rubi [A]
time = 0.39, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3033, 3078, 3138, 2719, 3081, 2720, 2884} \begin {gather*} \frac {\left (2 a^2 A-a b B-A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {(A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\left (a^3 (-B)+3 a^2 A b-a b^2 B-A b^3\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (a-b) (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^2),x]

[Out]

((A*b - a*B)*EllipticE[(c + d*x)/2, 2])/(a*(a^2 - b^2)*d) + ((2*a^2*A - A*b^2 - a*b*B)*EllipticF[(c + d*x)/2,
2])/(a^2*(a^2 - b^2)*d) - ((3*a^2*A*b - A*b^3 - a^3*B - a*b^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a
^2*(a - b)*(a + b)^2*d) - ((A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*(b + a*Cos[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 3033

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3078

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*a - A*b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e
+ f*x])^n/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n - 1)*Simp[c*(a*A - b*B)*(m + 1) + d*n*(A*b - a*B) + (d*(a*A - b*B)*(m + 1) - c*(A*b - a*B)*
(m + 2))*Sin[e + f*x] - d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 3081

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3138

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

Rubi steps

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

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Mathematica [A]
time = 12.47, size = 260, normalized size = 1.28 \begin {gather*} \frac {\frac {4 (-A b+a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x))}-\frac {\frac {2 (-A b+a B) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {(4 a A-4 b 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 )}{a}+\frac {2 (A b-a B) \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)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(-a+b) (a+b)}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^2),x]

[Out]

((4*(-(A*b) + a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(b + a*Cos[c + d*x])) - ((2*(-(A*b) + a*B)*El
lipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + ((4*a*A - 4*b*B)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*Ellipt
icPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)))/a + (2*(A*b - 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[C
os[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/((-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. \(801\) vs. \(2(279)=558\).
time = 5.37, size = 802, normalized size = 3.95

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 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 )}{a^{2} \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 {2 b \left (A b -B a \right ) \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^{2}}-\frac {2 \left (-2 A b +B a \right ) \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 )}{a \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}\)	\(802\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),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*A/a^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)
)+2*b*(A*b-B*a)/a^2*(1/b*a^2/(a^2-b^2)*cos(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)
/(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*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)))-2*(-2*A*b+B*a)/a/(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]
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((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^2*sqrt(cos(d*x + c))), x)

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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((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),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 {A + B \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2} \sqrt {\cos {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^2*sqrt(cos(d*x + c))), x)

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

[Out]

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

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