3.1163 \(\int \frac {\sqrt {\cos (c+d x)} (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=172 \[ -\frac {(7 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(5 A+C) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}} \]

[Out]

-1/2*(A+C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)/cos(d*x+c)^(1/2)-1/4*(7*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^(3/2)/d*2^(1/2)+1/2*(5*A+C)*
sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.49, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {4265, 4085, 4013, 3808, 206} \[ -\frac {(7 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(5 A+C) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[Cos[c + d*x]]*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

-((7*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]])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Sin[c + d*x])/(2*d*Sqrt[Cos[c + d*x]]*(a + a*Sec[c
+ d*x])^(3/2)) + ((5*A + C)*Sin[c + d*x])/(2*a*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 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))^{3/2}} \, 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))^{3/2}} \, dx\\ &=-\frac {(A+C) \sin (c+d x)}{2 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}{2} a (5 A+C)+a (A-C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(5 A+C) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {\left ((7 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}{4 a}\\ &=-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(5 A+C) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {\left ((7 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 )}{2 a d}\\ &=-\frac {(7 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)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(5 A+C) \sin (c+d x)}{2 a d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.68, size = 114, normalized size = 0.66 \[ -\frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (2 \sin \left (\frac {1}{2} (c+d x)\right ) (4 A \cos (c+d x)+5 A+C)-(7 A-C) (\cos (c+d x)+1) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{2 d \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )-1\right ) \cos ^{\frac {3}{2}}(c+d x) (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[Cos[c + d*x]]*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

-1/2*(Cos[(c + d*x)/2]^3*(-((7*A - C)*ArcTanh[Sin[(c + d*x)/2]]*(1 + Cos[c + d*x])) + 2*(5*A + C + 4*A*Cos[c +
d*x])*Sin[(c + d*x)/2]))/(d*Cos[c + d*x]^(3/2)*(a*(1 + Sec[c + d*x]))^(3/2)*(-1 + Sin[(c + d*x)/2]^2))

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fricas [A]  time = 0.56, size = 406, normalized size = 2.36 \[ \left [-\frac {\sqrt {2} {\left ({\left (7 \, A - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - C\right )} \cos \left (d x + c\right ) + 7 \, A - C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (4 \, A \cos \left (d x + c\right ) + 5 \, A + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac {\sqrt {2} {\left ({\left (7 \, A - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - C\right )} \cos \left (d x + c\right ) + 7 \, A - C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + 2 \, {\left (4 \, A \cos \left (d x + c\right ) + 5 \, A + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \]

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))^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(sqrt(2)*((7*A - C)*cos(d*x + c)^2 + 2*(7*A - C)*cos(d*x + c) + 7*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*(4*A*cos(d*x + c) + 5*A + C)*sqrt((a*cos(d*x + c) + a)/cos(
d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), 1/4*(sqrt(2)
*((7*A - C)*cos(d*x + c)^2 + 2*(7*A - C)*cos(d*x + c) + 7*A - C)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(
d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 2*(4*A*cos(d*x + c) + 5*A + C)*sqrt((a*cos(
d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^
2*d)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*sqrt(cos(d*x + c))/(a*sec(d*x + c) + a)^(3/2), x)

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maple [A]  time = 2.34, size = 235, normalized size = 1.37 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (4 A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+7 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sin \left (d x +c \right )+A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )-C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sin \left (d x +c \right )+C \cos \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-5 A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{2 d \,a^{2} \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )^{3}} \]

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))^(3/2),x)

[Out]

1/2/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(4*A*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)+7*A*arct
an(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)+A*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-C*arctan(1/2*si
n(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)+C*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)-5*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^2/(-2/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^3

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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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

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))^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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))**(3/2),x)

[Out]

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

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