3.54 \(\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=102 \[ \frac {2 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^2 f}-\frac {4 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 c^2 f}+\frac {2 a \cot (e+f x) \sqrt {a \sec (e+f x)+a}}{c^2 f} \]

[Out]

2*a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c^2/f-4/3*cot(f*x+e)^3*(a+a*sec(f*x+e))^(3/2)/c^2/
f+2*a*cot(f*x+e)*(a+a*sec(f*x+e))^(1/2)/c^2/f

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Rubi [A]  time = 0.16, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3904, 3887, 453, 325, 203} \[ \frac {2 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^2 f}-\frac {4 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 c^2 f}+\frac {2 a \cot (e+f x) \sqrt {a \sec (e+f x)+a}}{c^2 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(3/2)/(c - c*Sec[e + f*x])^2,x]

[Out]

(2*a^(3/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c^2*f) + (2*a*Cot[e + f*x]*Sqrt[a + a*Sec
[e + f*x]])/(c^2*f) - (4*Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(3*c^2*f)

Rule 203

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

Rule 325

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 3887

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 3904

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

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Mathematica [A]  time = 0.64, size = 113, normalized size = 1.11 \[ -\frac {2 a \sqrt {\cos (e+f x)} \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \left (\sqrt {\cos (e+f x)} (5 \cos (e+f x)-3)-6 \sqrt {2} \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin ^3\left (\frac {1}{2} (e+f x)\right )\right )}{3 c^2 f (\cos (e+f x)-1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(3/2)/(c - c*Sec[e + f*x])^2,x]

[Out]

(-2*a*Sqrt[Cos[e + f*x]]*Sqrt[a*(1 + Sec[e + f*x])]*(Sqrt[Cos[e + f*x]]*(-3 + 5*Cos[e + f*x]) - 6*Sqrt[2]*ArcS
in[Sqrt[2]*Sin[(e + f*x)/2]]*Sin[(e + f*x)/2]^3)*Tan[(e + f*x)/2])/(3*c^2*f*(-1 + Cos[e + f*x])^2)

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fricas [A]  time = 0.55, size = 351, normalized size = 3.44 \[ \left [\frac {3 \, {\left (a \cos \left (f x + e\right ) - a\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, {\left (5 \, a \cos \left (f x + e\right )^{2} - 3 \, a \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{6 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}, \frac {3 \, {\left (a \cos \left (f x + e\right ) - a\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (5 \, a \cos \left (f x + e\right )^{2} - 3 \, a \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{3 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(a*cos(f*x + e) - a)*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*(5*a*cos(f*x + e)^2 - 3*a*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^2*f*cos(f*x + e) - c^
2*f)*sin(f*x + e)), 1/3*(3*(a*cos(f*x + e) - a)*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*(5*a*cos(f*x + e)^2 -
3*a*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^2*f*cos(f*x + e) - c^2*f)*sin(f*x + e))]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)4*(1/196608*(-2621
44*a^4*sqrt(-a)*sign(cos(f*x+exp(1)))-393216*a^2*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2
*(f*x+exp(1))))^4*sign(cos(f*x+exp(1)))+393216*a^3*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1
/2*(f*x+exp(1))))^2*sign(cos(f*x+exp(1))))/sqrt(2)/c^2/((sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(
f*x+exp(1))))^2-a)^3-1/4*a^2*sqrt(-a)*sign(cos(f*x+exp(1)))*ln(abs(2*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(
-a)*tan(1/2*(f*x+exp(1))))^2-4*sqrt(2)*abs(a)-6*a)/abs(2*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*
(f*x+exp(1))))^2+4*sqrt(2)*abs(a)-6*a))/c^2/abs(a))/f

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maple [B]  time = 1.55, size = 215, normalized size = 2.11 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \left (-3 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )+3 \sqrt {2}\, \sin \left (f x +e \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}+10 \left (\cos ^{2}\left (f x +e \right )\right )-6 \cos \left (f x +e \right )\right ) a}{3 c^{2} f \sin \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/c^2/f*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*(-3*cos(f*x+e)*sin(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))
^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))+3*2^(1/2)*sin(f*x+e)*ar
ctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(
1/2)+10*cos(f*x+e)^2-6*cos(f*x+e))/sin(f*x+e)/(-1+cos(f*x+e))*a

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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