3.1169 \(\int \frac {\sqrt {\cos (c+d x)} (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=219 \[ -\frac {5 (15 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{5/2}} \]

[Out]

-1/4*(A+C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2)-1/16*(13*A-3*C)*sin(d*x+c)/a/d/(a+a*sec(d*x+c)
)^(3/2)/cos(d*x+c)^(1/2)-5/32*(15*A-C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c)
)^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)+1/16*(49*A+C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(
a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.69, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4265, 4085, 4020, 4013, 3808, 206} \[ \frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {5 (15 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(15*A - C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[Cos[
c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A + C)*Sin[c + d*x])/(4*d*Sqrt[Cos[c + d*x]]*(a + a*S
ec[c + d*x])^(5/2)) - ((13*A - 3*C)*Sin[c + d*x])/(16*a*d*Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)) + ((4
9*A + C)*Sin[c + d*x])/(16*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])

Rule 206

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

Rule 3808

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

Rule 4013

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

Rule 4020

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

Rule 4085

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

Rule 4265

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

Rubi steps

\begin {align*} \int \frac {\sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{2} a (9 A+C)+2 a (A-C) \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{4} a^2 (49 A+C)+\frac {1}{2} a^2 (13 A-3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {\left (5 (15 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {\left (5 (15 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=-\frac {5 (15 A-C) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {(13 A-3 C) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(49 A+C) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 2.36, size = 118, normalized size = 0.54 \[ \frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (4 \sin \left (\frac {1}{2} (c+d x)\right ) (5 (17 A+C) \cos (c+d x)+16 A \cos (2 (c+d x))+65 A+C)-40 (15 A-C) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{64 a d \cos ^{\frac {3}{2}}(c+d x) (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(c + d*x)/2]*(-40*(15*A - C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^4 + 4*(65*A + C + 5*(17*A + C)*Co
s[c + d*x] + 16*A*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(64*a*d*Cos[c + d*x]^(3/2)*(a*(1 + Sec[c + d*x]))^(3/2)
)

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fricas [A]  time = 0.54, size = 500, normalized size = 2.28 \[ \left [-\frac {5 \, \sqrt {2} {\left ({\left (15 \, A - C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right ) + 15 \, A - C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (32 \, A \cos \left (d x + c\right )^{2} + 5 \, {\left (17 \, A + C\right )} \cos \left (d x + c\right ) + 49 \, A + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac {5 \, \sqrt {2} {\left ({\left (15 \, A - C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (15 \, A - C\right )} \cos \left (d x + c\right ) + 15 \, A - C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + 2 \, {\left (32 \, A \cos \left (d x + c\right )^{2} + 5 \, {\left (17 \, A + C\right )} \cos \left (d x + c\right ) + 49 \, A + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/64*(5*sqrt(2)*((15*A - C)*cos(d*x + c)^3 + 3*(15*A - C)*cos(d*x + c)^2 + 3*(15*A - C)*cos(d*x + c) + 15*A
- C)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x +
c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(32*A*cos(d*x + c)^2 +
5*(17*A + C)*cos(d*x + c) + 49*A + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(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*(5*sqrt(2)*((15*A - C)*c
os(d*x + c)^3 + 3*(15*A - C)*cos(d*x + c)^2 + 3*(15*A - C)*cos(d*x + c) + 15*A - C)*sqrt(-a)*arctan(sqrt(2)*sq
rt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 2*(32*A*cos(d*x + c)^2 +
5*(17*A + C)*cos(d*x + c) + 49*A + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 2.52, size = 365, normalized size = 1.67 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (32 A \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+53 A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+75 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )+5 C \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-5 C \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )-36 A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+75 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sin \left (d x +c \right )-4 C \cos \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-5 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sin \left (d x +c \right )-49 A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{16 d \,a^{3} \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

-1/16/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(32*A*cos(d*x+c)^3*(-2/(1+cos(d*x+c)))^(1/2)+53*
A*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)+75*A*cos(d*x+c)*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^
(1/2))+5*C*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)-5*C*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d
*x+c)))^(1/2))-36*A*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+75*A*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))
*sin(d*x+c)-4*C*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)-5*C*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(
d*x+c)-49*A*(-2/(1+cos(d*x+c)))^(1/2)-C*(-2/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^(1/2)/a^3/(-2/(1+cos(d*x+c)))^(1
/2)/sin(d*x+c)^5

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^(1/2)*(A + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)^(1/2)*(A + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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