3.61 \(\int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=74 \[ \frac {2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^2 f}-\frac {8 a \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 c^2 f} \]

[Out]

2*a^(5/2)*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c^2/f-8/3*a*cot(f*x+e)^3*(a+a*sec(f*x+e))^(3/2)/c^
2/f

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Rubi [A]  time = 0.17, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3904, 3887, 461, 203} \[ \frac {2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^2 f}-\frac {8 a \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 c^2 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^2,x]

[Out]

(2*a^(5/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c^2*f) - (8*a*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 461

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

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))^{5/2}}{(c-c \sec (e+f x))^2} \, dx &=\frac {\int \cot ^4(e+f x) (a+a \sec (e+f x))^{9/2} \, dx}{a^2 c^2}\\ &=-\frac {(2 a) \operatorname {Subst}\left (\int \frac {\left (2+a x^2\right )^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 {(2 a) \operatorname {Subst}\left (\int \left (\frac {4}{x^4}+\frac {a^2}{1+a x^2}\right ) \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^2 f}\\ &=-\frac {8 a \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^2 f}-\frac {\left (2 a^3\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^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^2 f}-\frac {8 a \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 = 4.22, size = 102, normalized size = 1.38 \[ -\frac {\cos ^{\frac {5}{2}}(e+f x) \csc ^3\left (\frac {1}{2} (e+f x)\right ) \sec ^5\left (\frac {1}{2} (e+f x)\right ) (a (\sec (e+f x)+1))^{5/2} \left (4 \cos ^{\frac {3}{2}}(e+f x)-3 (1-\cos (e+f x))^{3/2} \sin ^{-1}\left (\sqrt {1-\cos (e+f x)}\right )\right )}{24 c^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^2,x]

[Out]

-1/24*(Cos[e + f*x]^(5/2)*(-3*ArcSin[Sqrt[1 - Cos[e + f*x]]]*(1 - Cos[e + f*x])^(3/2) + 4*Cos[e + f*x]^(3/2))*
Csc[(e + f*x)/2]^3*Sec[(e + f*x)/2]^5*(a*(1 + Sec[e + f*x]))^(5/2))/(c^2*f)

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fricas [B]  time = 0.57, size = 339, normalized size = 4.58 \[ \left [\frac {16 \, a^{2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + 3 \, {\left (a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right )}{6 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}, \frac {8 \, a^{2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + 3 \, {\left (a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right )}{3 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

[1/6*(16*a^2*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)^2 + 3*(a^2*cos(f*x + e) - a^2)*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))*si
n(f*x + e) - 7*a*cos(f*x + e) + a)/(cos(f*x + e) + 1))*sin(f*x + e))/((c^2*f*cos(f*x + e) - c^2*f)*sin(f*x + e
)), 1/3*(8*a^2*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)^2 + 3*(a^2*cos(f*x + e) - a^2)*sqrt(a)*arc
tan(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))/((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))^(5/2)/(c-c*sec(f*x+e))^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 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)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)-128*sqrt(2)*a^3*sqrt(-a)*(-a)^2*(1/96*(
3*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^4+a^2)/a^2/c^2/((sqrt(-a*tan(1/2*(f*x+ex
p(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2-a)^3+1/128*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))/sqrt(2)/a^2/c^2/abs(a))*sign(cos(f*x+exp(1)))/f

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maple [B]  time = 1.42, size = 351, normalized size = 4.74 \[ -\frac {\left (3 \sqrt {2}\, \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )-3 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {2}-3 \sqrt {2}\, \cos \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )+8 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+3 \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {2}\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, a^{2}}{3 c^{2} f \left (-1+\cos \left (f x +e \right )\right )^{2} \left (1+\cos \left (f x +e \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^2,x)

[Out]

-1/3/c^2/f*(3*2^(1/2)*cos(f*x+e)^3*(-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*cos(f*x+e)^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))*2^(1/2)-3*2^(1/2)*cos(f*x+e)*(-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))+8*cos(f*x+
e)^2*sin(f*x+e)+3*(-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))*2^(1/2))*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)/(-1+cos(f*x+e))^2/(1+cos(f*x+e))*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+a*sec(f*x+e))^(5/2)/(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 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(5/2)/(c - c/cos(e + f*x))^2,x)

[Out]

int((a + a/cos(e + f*x))^(5/2)/(c - c/cos(e + f*x))^2, x)

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sympy [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))**(5/2)/(c-c*sec(f*x+e))**2,x)

[Out]

Timed out

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