3.1.48 \(\int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^3} \, dx\) [48]

Optimal. Leaf size=139 \[ \frac {2 \sqrt {a} \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^3 f}+\frac {2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^3 f}-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 a c^3 f}+\frac {2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^2 c^3 f} \]

[Out]

-2/3*cot(f*x+e)^3*(a+a*sec(f*x+e))^(3/2)/a/c^3/f+2/5*cot(f*x+e)^5*(a+a*sec(f*x+e))^(5/2)/a^2/c^3/f+2*arctan(a^
(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))*a^(1/2)/c^3/f+2*cot(f*x+e)*(a+a*sec(f*x+e))^(1/2)/c^3/f

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Rubi [A]
time = 0.12, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3989, 3972, 331, 209} \begin {gather*} \frac {2 \cot ^5(e+f x) (a \sec (e+f x)+a)^{5/2}}{5 a^2 c^3 f}+\frac {2 \sqrt {a} \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^3 f}-\frac {2 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 a c^3 f}+\frac {2 \cot (e+f x) \sqrt {a \sec (e+f x)+a}}{c^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sec[e + f*x]]/(c - c*Sec[e + f*x])^3,x]

[Out]

(2*Sqrt[a]*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c^3*f) + (2*Cot[e + f*x]*Sqrt[a + a*Sec[e
+ f*x]])/(c^3*f) - (2*Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(3*a*c^3*f) + (2*Cot[e + f*x]^5*(a + a*Sec[e
+ f*x])^(5/2))/(5*a^2*c^3*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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^3} \, dx &=-\frac {\int \cot ^6(e+f x) (a+a \sec (e+f x))^{7/2} \, dx}{a^3 c^3}\\ &=\frac {2 \text {Subst}\left (\int \frac {1}{x^6 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^2 c^3 f}\\ &=\frac {2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^2 c^3 f}-\frac {2 \text {Subst}\left (\int \frac {1}{x^4 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a c^3 f}\\ &=-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 a c^3 f}+\frac {2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^2 c^3 f}+\frac {2 \text {Subst}\left (\int \frac {1}{x^2 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^3 f}\\ &=\frac {2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^3 f}-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 a c^3 f}+\frac {2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^2 c^3 f}-\frac {(2 a) \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 )}{c^3 f}\\ &=\frac {2 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^3 f}+\frac {2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^3 f}-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 a c^3 f}+\frac {2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a^2 c^3 f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.28, size = 78, normalized size = 0.56 \begin {gather*} -\frac {2 \sqrt {\cos (e+f x)} \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sec (e+f x))} \tan \left (\frac {1}{2} (e+f x)\right )}{5 c^3 f (-1+\cos (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sec[e + f*x]]/(c - c*Sec[e + f*x])^3,x]

[Out]

(-2*Sqrt[Cos[e + f*x]]*Hypergeometric2F1[-5/2, -5/2, -3/2, 2*Sin[(e + f*x)/2]^2]*Sqrt[a*(1 + Sec[e + f*x])]*Ta
n[(e + f*x)/2])/(5*c^3*f*(-1 + Cos[e + f*x])^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(310\) vs. \(2(123)=246\).
time = 0.27, size = 311, normalized size = 2.24

method	result	size
default	\(\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )+1\right ) \left (15 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {2}-30 \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {2}+15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-46 \left (\cos ^{3}\left (f x +e \right )\right )+70 \left (\cos ^{2}\left (f x +e \right )\right )-30 \cos \left (f x +e \right )\right )}{15 c^{3} f \sin \left (f x +e \right )^{3} \left (-1+\cos \left (f x +e \right )\right )}\)	\(311\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

1/15/c^3/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(cos(f*x+e)+1)*(15*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(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))*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)-30*si
n(f*x+e)*cos(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(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^(1/2)+15*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*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-46*cos(f*x+e)^3+70*cos(f*x+e)^2-30*cos(f*x+e))
/sin(f*x+e)^3/(-1+cos(f*x+e))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

-integrate(sqrt(a*sec(f*x + e) + a)/(c*sec(f*x + e) - c)^3, x)

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Fricas [A]
time = 3.81, size = 441, normalized size = 3.17 \begin {gather*} \left [\frac {15 \, {\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, {\left (23 \, \cos \left (f x + e\right )^{3} - 35 \, \cos \left (f x + e\right )^{2} + 15 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{30 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}, \frac {15 \, {\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {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 )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (23 \, \cos \left (f x + e\right )^{3} - 35 \, \cos \left (f x + e\right )^{2} + 15 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/30*(15*(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)*sqrt(-a)*log(-(8*a*cos(f*x + e)^3 - 4*(2*cos(f*x + e)^2 - cos(
f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - 7*a*cos(f*x + e) + a)/(cos(f*x + e)
+ 1))*sin(f*x + e) + 4*(23*cos(f*x + e)^3 - 35*cos(f*x + e)^2 + 15*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos
(f*x + e)))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e)), 1/15*(15*(cos(f*x + e)^2 - 2
*cos(f*x + e) + 1)*sqrt(a)*arctan(2*sqrt(a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/
(2*a*cos(f*x + e)^2 + a*cos(f*x + e) - a))*sin(f*x + e) + 2*(23*cos(f*x + e)^3 - 35*cos(f*x + e)^2 + 15*cos(f*
x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*
x + e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\sqrt {a \sec {\left (e + f x \right )} + a}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**(1/2)/(c-c*sec(f*x+e))**3,x)

[Out]

-Integral(sqrt(a*sec(e + f*x) + a)/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1), x)/c**3

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (123) = 246\).
time = 1.18, size = 391, normalized size = 2.81 \begin {gather*} -\frac {\sqrt {2} {\left (\frac {15 \, \sqrt {2} \sqrt {-a} a \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{c^{3} {\left | a \right |}} + \frac {105 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{8} \sqrt {-a} a - 300 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{6} \sqrt {-a} a^{2} + 430 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{4} \sqrt {-a} a^{3} - 260 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} \sqrt {-a} a^{4} + 73 \, \sqrt {-a} a^{5}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} - a\right )}^{5} c^{3}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{30 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^3,x, algorithm="giac")

[Out]

-1/30*sqrt(2)*(15*sqrt(2)*sqrt(-a)*a*log(abs(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2
+ a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))
^2 + 4*sqrt(2)*abs(a) - 6*a))/(c^3*abs(a)) + (105*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e
)^2 + a))^8*sqrt(-a)*a - 300*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^6*sqrt(-a)*
a^2 + 430*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^4*sqrt(-a)*a^3 - 260*(sqrt(-a)
*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(-a)*a^4 + 73*sqrt(-a)*a^5)/(((sqrt(-a)*tan
(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 - a)^5*c^3))*sgn(cos(f*x + e))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(1/2)/(c - c/cos(e + f*x))^3,x)

[Out]

int((a + a/cos(e + f*x))^(1/2)/(c - c/cos(e + f*x))^3, x)

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