3.1.68 \(\int \frac {c-c \sec (e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx\) [68]

Optimal. Leaf size=87 \[ \frac {2 c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} f}-\frac {2 \sqrt {2} c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} f} \]

[Out]

2*c*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/f/a^(1/2)-2*c*arctan(1/2*a^(1/2)*tan(f*x+e)*2^(1/2)/(a+a
*sec(f*x+e))^(1/2))*2^(1/2)/f/a^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3989, 3972, 492, 209} \begin {gather*} \frac {2 c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {a} f}-\frac {2 \sqrt {2} c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {a} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - c*Sec[e + f*x])/Sqrt[a + a*Sec[e + f*x]],x]

[Out]

(2*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(Sqrt[a]*f) - (2*Sqrt[2]*c*ArcTan[(Sqrt[a]*Tan[e
+ f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*f)

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 492

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(-a)*(e^n/(b*c -
a*d)), Int[(e*x)^(m - n)/(a + b*x^n), x], x] + Dist[c*(e^n/(b*c - a*d)), Int[(e*x)^(m - n)/(c + d*x^n), x], x
] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1]

Rule 3972

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

Rule 3989

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[((-a)*c)^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !(IntegerQ[n] && GtQ[m - n, 0])

Rubi steps

\begin {align*} \int \frac {c-c \sec (e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx &=-\left ((a c) \int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{3/2}} \, dx\right )\\ &=\frac {(2 a c) \text {Subst}\left (\int \frac {x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}\\ &=-\frac {(2 c) \text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}+\frac {(4 c) \text {Subst}\left (\int \frac {1}{2+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}\\ &=\frac {2 c \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} f}-\frac {2 \sqrt {2} c \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.28, size = 82, normalized size = 0.94 \begin {gather*} \frac {2 c \left (\text {ArcTan}\left (\sqrt {-1+\sec (e+f x)}\right )-\sqrt {2} \text {ArcTan}\left (\frac {\sqrt {-1+\sec (e+f x)}}{\sqrt {2}}\right )\right ) \cot \left (\frac {1}{2} (e+f x)\right ) \sqrt {-1+\sec (e+f x)}}{f \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sec[e + f*x])/Sqrt[a + a*Sec[e + f*x]],x]

[Out]

(2*c*(ArcTan[Sqrt[-1 + Sec[e + f*x]]] - Sqrt[2]*ArcTan[Sqrt[-1 + Sec[e + f*x]]/Sqrt[2]])*Cot[(e + f*x)/2]*Sqrt
[-1 + Sec[e + f*x]])/(f*Sqrt[a*(1 + Sec[e + f*x])])

________________________________________________________________________________________

Maple [A]
time = 0.18, size = 144, normalized size = 1.66

method	result	size
default	\(-\frac {c \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (\sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )+2 \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right )\right )}{f a}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-c/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(2^(1/2)*arctanh(1/2*(-2*cos(f*x
+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))+2*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)
-cos(f*x+e)+1)/sin(f*x+e)))/a

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

________________________________________________________________________________________

Fricas [A]
time = 4.27, size = 321, normalized size = 3.69 \begin {gather*} \left [\frac {\sqrt {2} a c \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \, \cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) - \sqrt {-a} c \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right )}{a f}, \frac {2 \, {\left (\sqrt {2} \sqrt {a} c \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - \sqrt {a} c \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right )\right )}}{a f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[(sqrt(2)*a*c*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*cos(f*x + e)*sin(f*
x + e) + 3*cos(f*x + e)^2 + 2*cos(f*x + e) - 1)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) - sqrt(-a)*c*log((2*a*c
os(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e)
- a)/(cos(f*x + e) + 1)))/(a*f), 2*(sqrt(2)*sqrt(a)*c*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*c
os(f*x + e)/(sqrt(a)*sin(f*x + e))) - sqrt(a)*c*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(s
qrt(a)*sin(f*x + e))))/(a*f)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - c \left (\int \frac {\sec {\left (e + f x \right )}}{\sqrt {a \sec {\left (e + f x \right )} + a}}\, dx + \int \left (- \frac {1}{\sqrt {a \sec {\left (e + f x \right )} + a}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))/(a+a*sec(f*x+e))**(1/2),x)

[Out]

-c*(Integral(sec(e + f*x)/sqrt(a*sec(e + f*x) + a), x) + Integral(-1/sqrt(a*sec(e + f*x) + a), x))

________________________________________________________________________________________

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((c-c*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c-\frac {c}{\cos \left (e+f\,x\right )}}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))/(a + a/cos(e + f*x))^(1/2),x)

[Out]

int((c - c/cos(e + f*x))/(a + a/cos(e + f*x))^(1/2), x)

________________________________________________________________________________________