\(\int \frac {\sqrt {a+a \sec (c+d x)} (A+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [254]

Optimal result

Integrand size = 37, antiderivative size = 116 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a A \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

[Out]

2*C*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*a^(1/2)/d+2/3*a*A*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*se
c(d*x+c))^(1/2)+2/3*A*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {4172, 4100, 3886, 221} \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 \sqrt {a} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d} \]

[In]

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

[Out]

(2*Sqrt[a]*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a*A*Sqrt[Sec[c + d*x]]*Sin[c + d
*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

Rule 221

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

Rule 3886

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

Rule 4100

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

Rule 4172

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {a A}{2}+\frac {3}{2} a C \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{3 a} \\ & = \frac {2 a A \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+C \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx \\ & = \frac {2 a A \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {(2 C) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d} \\ & = \frac {2 \sqrt {a} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a A \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 a \left (A (2+\cos (c+d x)) \sqrt {1-\sec (c+d x)}-3 C \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{3 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

(2*a*(A*(2 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]] - 3*C*ArcSin[Sqrt[Sec[c + d*x]]]*Sqrt[Sec[c + d*x]])*Tan[c +
d*x])/(3*d*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*Sqrt[a*(1 + Sec[c + d*x])])

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.68

method	result	size
parts	\(\frac {2 A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (\arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-\arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) \cos \left (d x +c \right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(195\)
default	\(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 C \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+3 C \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 A \sin \left (d x +c \right )+3 C \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sec \left (d x +c \right )+3 C \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sec \left (d x +c \right )+4 A \tan \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(303\)

[In]

int((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3*A/d*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*(sin(d*x+c)+2*tan(d*x+c))+C/d*(a*(1+sec(d*x+c
)))^(1/2)*sec(d*x+c)^(1/2)*(arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-arc
tan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2)))*cos(d*x+c)/(cos(d*x+c)+1)/(-1/(co
s(d*x+c)+1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 348, normalized size of antiderivative = 3.00 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {3 \, {\left (C \cos \left (d x + c\right ) + C\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {3 \, {\left (C \cos \left (d x + c\right ) + C\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/6*(3*(C*cos(d*x + c) + C)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*
x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3
+ cos(d*x + c)^2)) + 4*(A*cos(d*x + c)^2 + 2*A*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x
+ c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d), 1/3*(3*(C*cos(d*x + c) + C)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*
cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) + 2
*(A*cos(d*x + c)^2 + 2*A*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c))
)/(d*cos(d*x + c) + d)]

Sympy [F]

\[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (98) = 196\).

Time = 0.50 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.06 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left (3 \, \cos \left (\frac {2}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 \, \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \sin \left (\frac {2}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 2 \, \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} A \sqrt {a} + 3 \, C \sqrt {a} {\left (\log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) + \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )\right )}}{6 \, d} \]

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

1/6*(sqrt(2)*(3*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 3*cos(3/2*
d*x + 3/2*c)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sin(3/2*d*x + 3/2*c) + 3*sin(1/3
*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*A*sqrt(a) + 3*C*sqrt(a)*(log(2*cos(1/2*d*x + 1/2*c)^2 +
2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1
/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c)
+ 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*s
in(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1
/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)))/d

Giac [F]

\[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {a \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((A+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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