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

Optimal. Leaf size=201 \[ \frac {(11 A+3 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{6 a d \sqrt {a \sec (c+d x)+a}}+\frac {(7 A+3 C) \sin (c+d x)}{6 a d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}} \]

[Out]

1/4*(11*A+3*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))/a^(3/2)/d*2^(1/
2)-1/2*(A+C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)/sec(d*x+c)^(1/2)+1/6*(7*A+3*C)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2
)/(a+a*sec(d*x+c))^(1/2)-1/6*(19*A+3*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.51, antiderivative size = 201, 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 = {4085, 4022, 4013, 3808, 206} \[ \frac {(11 A+3 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{6 a d \sqrt {a \sec (c+d x)+a}}+\frac {(7 A+3 C) \sin (c+d x)}{6 a d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*A + 3*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2
]*a^(3/2)*d) - ((A + C)*Sin[c + d*x])/(2*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)) + ((7*A + 3*C)*Sin[c
+ d*x])/(6*a*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((19*A + 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/
(6*a*d*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 4022

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[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[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)]

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}-\frac {\int \frac {-\frac {1}{2} a (7 A+3 C)+2 a A \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(7 A+3 C) \sin (c+d x)}{6 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {\int \frac {\frac {1}{4} a^2 (19 A+3 C)-\frac {1}{2} a^2 (7 A+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{3 a^3}\\ &=-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(7 A+3 C) \sin (c+d x)}{6 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {(19 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}}+\frac {(11 A+3 C) \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 {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(7 A+3 C) \sin (c+d x)}{6 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {(19 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}}-\frac {(11 A+3 C) \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 {(11 A+3 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 )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(7 A+3 C) \sin (c+d x)}{6 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {(19 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 3.03, size = 316, normalized size = 1.57 \[ \frac {(\sec (c+d x)+1)^{3/2} \left (A+C \sec ^2(c+d x)\right ) \left (\frac {2 \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right ) \sqrt {\sec (c+d x)+1} \sec ^3\left (\frac {1}{2} (c+d x)\right ) (12 A \cos (c+d x)-2 A \cos (2 (c+d x))+17 A+3 C)}{\sec ^{\frac {3}{2}}(c+d x)}+3 \sqrt {2} (11 A+3 C) \cos ^2(c+d x) \sqrt {\tan ^2(c+d x)} \cot (c+d x) \left (\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)-2 \sqrt {2} \sqrt {\tan ^2(c+d x)} \sqrt {\sec (c+d x)+1} \sqrt {\sec (c+d x)}+1\right )-\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)+2 \sqrt {2} \sqrt {\tan ^2(c+d x)} \sqrt {\sec (c+d x)+1} \sqrt {\sec (c+d x)}+1\right )\right )\right )}{24 d (a (\sec (c+d x)+1))^{3/2} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2)*((2*(17*A + 3*C + 12*A*Cos[c + d*x] - 2*A*Cos[2*(c + d*x)])*S
ec[(c + d*x)/2]^3*Sqrt[1 + Sec[c + d*x]]*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2]))/Sec[c + d*x]^(3/2) + 3*Sqr
t[2]*(11*A + 3*C)*Cos[c + d*x]^2*Cot[c + d*x]*(Log[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 - 2*Sqrt[2]*Sqrt[Sec[
c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[Tan[c + d*x]^2]] - Log[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 + 2*Sqrt[2]
*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[Tan[c + d*x]^2]])*Sqrt[Tan[c + d*x]^2]))/(24*d*(A + 2*C + A*Co
s[2*(c + d*x)])*(a*(1 + Sec[c + d*x]))^(3/2))

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fricas [A]  time = 0.48, size = 456, normalized size = 2.27 \[ \left [\frac {3 \, \sqrt {2} {\left ({\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 11 \, A + 3 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, {\left (4 \, A \cos \left (d x + c\right )^{3} - 12 \, A \cos \left (d x + c\right )^{2} - {\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {3 \, \sqrt {2} {\left ({\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 11 \, A + 3 \, C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (4 \, A \cos \left (d x + c\right )^{3} - 12 \, A \cos \left (d x + c\right )^{2} - {\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/24*(3*sqrt(2)*((11*A + 3*C)*cos(d*x + c)^2 + 2*(11*A + 3*C)*cos(d*x + c) + 11*A + 3*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*c
os(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(4*A*cos(d*x + c)^3 - 12*A*cos(d*x + c)^2 - (19*
A + 3*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x
+ c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), -1/12*(3*sqrt(2)*((11*A + 3*C)*cos(d*x + c)^2 + 2*(11*A + 3*C)*cos(d*
x + c) + 11*A + 3*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)^3 - 12*A*cos(d*x + c)^2 - (19*A + 3*C)*cos(d*x + c))*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(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 \[ \int \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 3.38, size = 306, normalized size = 1.52 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (33 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+9 C \sin \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+33 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, A \sin \left (d x +c \right )+8 A \left (\cos ^{3}\left (d x +c \right )\right )+9 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-32 A \left (\cos ^{2}\left (d x +c \right )\right )-14 A \cos \left (d x +c \right )-6 C \cos \left (d x +c \right )+38 A +6 C \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{12 d \sin \left (d x +c \right )^{3} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(33*A*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/
2))*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)+9*C*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(
1/2))*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+33*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+
c)))^(1/2)*A*sin(d*x+c)+8*A*cos(d*x+c)^3+9*C*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c
)))^(1/2)*sin(d*x+c)-32*A*cos(d*x+c)^2-14*A*cos(d*x+c)-6*C*cos(d*x+c)+38*A+6*C)*cos(d*x+c)^2*(1/cos(d*x+c))^(3
/2)/sin(d*x+c)^3/a^2

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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)/sec(d*x+c)^(3/2)/(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 = -0.00 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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