3.6.31 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\) [531]

Optimal. Leaf size=171 \[ \frac {2 A \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {(43 A-3 B-5 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-3 B-5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \]

[Out]

2*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/32*(43*A-3*B-5*C)*arctan(1/2*a^(1/2)*tan(d*x
+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(A-B+C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(11
*A-3*B-5*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)

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Rubi [A]
time = 0.19, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4137, 4007, 4005, 3859, 209, 3880} \begin {gather*} -\frac {(43 A-3 B-5 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {2 A \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac {(11 A-3 B-5 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d) - ((43*A - 3*B - 5*C)*ArcTan[(Sqrt[a
]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Tan[c + d*x])/(4*d*
(a + a*Sec[c + d*x])^(5/2)) - ((11*A - 3*B - 5*C)*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3880

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

Rule 4005

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

Rule 4007

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

Rule 4137

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x]
+ Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*(2*m + 1) + (b*B*(m + 1) - a*(A*(m + 1) -
C*m))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {\int \frac {-4 a A+\frac {1}{2} a (3 A-3 B-5 C) \sec (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-3 B-5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {8 a^2 A-\frac {1}{4} a^2 (11 A-3 B-5 C) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-3 B-5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {A \int \sqrt {a+a \sec (c+d x)} \, dx}{a^3}-\frac {(43 A-3 B-5 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-3 B-5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(2 A) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^2 d}+\frac {(43 A-3 B-5 C) \text {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=\frac {2 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {(43 A-3 B-5 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-3 B-5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 28.22, size = 16171, normalized size = 94.57 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1096\) vs. \(2(146)=292\).
time = 0.13, size = 1097, normalized size = 6.42

method	result	size
default	\(\text {Expression too large to display}\)	\(1097\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/32/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(32*A*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+43*A*cos(d*x+c)^2*sin
(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)+64*A*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*cos(d*x+c)-3*B*cos(d*x+c)^2*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-5*C*cos(d*x+c)^2*s
in(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)+86*A*sin(d*x+c)*cos(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1
)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+32*A*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d
*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-6*B*sin(d*x+c)*cos(d*x+c)*ln
(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)-10*C*sin(d*x+c)*cos(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c)
)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+43*A*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1
)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-30*A*cos(d*x+c)^3-3*B*ln(-(-(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+14*B*cos
(d*x+c)^3-5*C*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*sin(d*x+c)+2*C*cos(d*x+c)^3+8*A*cos(d*x+c)^2-8*B*cos(d*x+c)^2+8*C*cos(d*x+c)^2+22*A*cos(d
*x+c)-6*B*cos(d*x+c)-10*C*cos(d*x+c))/(1+cos(d*x+c))^2/sin(d*x+c)/a^3

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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)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (146) = 292\).
time = 55.93, size = 706, normalized size = 4.13 \begin {gather*} \left [\frac {\sqrt {2} {\left ({\left (43 \, A - 3 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (43 \, A - 3 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (43 \, A - 3 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 43 \, A - 3 \, B - 5 \, C\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 64 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (15 \, A - 7 \, B - C\right )} \cos \left (d x + c\right )^{2} + {\left (11 \, A - 3 \, B - 5 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac {\sqrt {2} {\left ({\left (43 \, A - 3 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (43 \, A - 3 \, B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (43 \, A - 3 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 43 \, A - 3 \, B - 5 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 64 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left ({\left (15 \, A - 7 \, B - C\right )} \cos \left (d x + c\right )^{2} + {\left (11 \, A - 3 \, B - 5 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/64*(sqrt(2)*((43*A - 3*B - 5*C)*cos(d*x + c)^3 + 3*(43*A - 3*B - 5*C)*cos(d*x + c)^2 + 3*(43*A - 3*B - 5*C)
*cos(d*x + c) + 43*A - 3*B - 5*C)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos
(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 6
4*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) -
4*((15*A - 7*B - C)*cos(d*x + c)^2 + (11*A - 3*B - 5*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))
*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/32*(sqrt(2)*(
(43*A - 3*B - 5*C)*cos(d*x + c)^3 + 3*(43*A - 3*B - 5*C)*cos(d*x + c)^2 + 3*(43*A - 3*B - 5*C)*cos(d*x + c) +
43*A - 3*B - 5*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x
+ c))) - 64*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*sqrt(a)*arctan(sqrt((a*cos(d*x + c
) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*((15*A - 7*B - C)*cos(d*x + c)^2 + (11*A - 3*B -
5*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(
d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]

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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 )} + C \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)/(a*(sec(c + d*x) + 1))**(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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