3.6.85 \(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\) [585]

Optimal. Leaf size=255 \[ \frac {\left (a A b-3 a^2 B+2 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b^2\right ) d}+\frac {(A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{(a-b) b^2 (a+b)^2 d}-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \]

[Out]

(A*a*b-3*B*a^2+2*B*b^2)*(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))/
b^2/(a^2-b^2)/d+(A*b-B*a)*(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)
)/b/(a^2-b^2)/d+(A*a^2*b-3*A*b^3-3*B*a^3+5*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-b)/b^2/(a+b)^2/d-(A*a*b-3*B*a^2+2*B*b^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/
cos(d*x+c)^(1/2)+a*(A*b-B*a)*sin(d*x+c)/b/(a^2-b^2)/d/(b+a*cos(d*x+c))/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.62, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3033, 3079, 3134, 3138, 2719, 3081, 2720, 2884} \begin {gather*} \frac {(A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}+\frac {\left (-3 a^2 B+a A b+2 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}-\frac {\left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)}+\frac {\left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d (a-b) (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^2),x]

[Out]

((a*A*b - 3*a^2*B + 2*b^2*B)*EllipticE[(c + d*x)/2, 2])/(b^2*(a^2 - b^2)*d) + ((A*b - a*B)*EllipticF[(c + d*x)
/2, 2])/(b*(a^2 - b^2)*d) + ((a^2*A*b - 3*A*b^3 - 3*a^3*B + 5*a*b^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2,
2])/((a - b)*b^2*(a + b)^2*d) - ((a*A*b - 3*a^2*B + 2*b^2*B)*Sin[c + d*x])/(b^2*(a^2 - b^2)*d*Sqrt[Cos[c + d*x
]]) + (a*(A*b - a*B)*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(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 3079

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c +
d*Sin[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*
(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n
, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 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)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx &=\int \frac {B+A \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2} \, dx\\ &=\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}-\frac {\int \frac {\frac {1}{2} \left (a A b-3 a^2 B+2 b^2 B\right )+b (A b-a B) \cos (c+d x)-\frac {1}{2} a (A b-a B) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}-\frac {2 \int \frac {\frac {1}{4} \left (-a^2 A b+2 A b^3+3 a^3 B-4 a b^2 B\right )-\frac {1}{2} b \left (a A b-2 a^2 B+b^2 B\right ) \cos (c+d x)-\frac {1}{4} a \left (a A b-3 a^2 B+2 b^2 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}+\frac {2 \int \frac {\frac {1}{4} a \left (a^2 A b-2 A b^3-3 a^3 B+4 a b^2 B\right )+\frac {1}{4} a^2 b (A b-a B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a b^2 \left (a^2-b^2\right )}+\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b^2\right ) d}-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}+\frac {(A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 b \left (a^2-b^2\right )}+\frac {\left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b^2\right ) d}+\frac {(A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{(a-b) b^2 (a+b)^2 d}-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 14.43, size = 317, normalized size = 1.24 \begin {gather*} \frac {-\frac {\frac {2 \left (-3 a^2 A b+4 A b^3+9 a^3 B-10 a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}-\frac {8 b \left (a A b-2 a^2 B+b^2 B\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a (a+b)}+\frac {2 \left (-a A b+3 a^2 B-2 b^2 B\right ) \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 b \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}+4 \sqrt {\cos (c+d x)} \left (\frac {a^2 (-A b+a B) \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x))}+2 B \tan (c+d x)\right )}{4 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^2),x]

[Out]

(-(((2*(-3*a^2*A*b + 4*A*b^3 + 9*a^3*B - 10*a*b^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) - (8*b
*(a*A*b - 2*a^2*B + b^2*B)*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/
(a*(a + b)) + (2*(-(a*A*b) + 3*a^2*B - 2*b^2*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])/(a*b*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b))) + 4*Sqrt[Cos[c + d*x]]*((a^2*(-(A*b) + a*B)*Sin
[c + d*x])/((a^2 - b^2)*(b + a*Cos[c + d*x])) + 2*B*Tan[c + d*x]))/(4*b^2*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(849\) vs. \(2(329)=658\).
time = 6.09, size = 850, normalized size = 3.33

method	result	size
default	\(\text {Expression too large to display}\)	\(850\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(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*a^2*B/b^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))+2*B/b^2/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(A*b-B*a)/b*(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)*Ellipt
icF(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*c
os(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))))/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((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(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((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)**(5/2)/(a+b*sec(d*x+c))**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5009 deep

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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))/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^2*cos(d*x + c)^(5/2)), 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 )}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\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((A + B/cos(c + d*x))/(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^2),x)

[Out]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^2), x)

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