3.249 \(\int \sec ^2(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=111 \[ \frac {3 (a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 f}+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 f}+\frac {3 (a+b)^2 \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{8 \sqrt {b} f} \]

[Out]

3/8*(a+b)^2*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f/b^(1/2)+3/8*(a+b)*(a+b+b*tan(f*x+e)^2)^(1
/2)*tan(f*x+e)/f+1/4*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/f

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4146, 195, 217, 206} \[ \frac {3 (a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 f}+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 f}+\frac {3 (a+b)^2 \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{8 \sqrt {b} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b)^2*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(8*Sqrt[b]*f) + (3*(a + b)*Tan[e
+ f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(8*f) + (Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(4*f)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

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

Rule 217

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

Rule 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int \sec ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b+b x^2\right )^{3/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac {(3 (a+b)) \operatorname {Subst}\left (\int \sqrt {a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {3 (a+b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}+\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac {\left (3 (a+b)^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {3 (a+b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}+\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac {\left (3 (a+b)^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 f}\\ &=\frac {3 (a+b)^2 \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 \sqrt {b} f}+\frac {3 (a+b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}+\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.25, size = 84, normalized size = 0.76 \[ \frac {(a+b)^2 \sin (2 (e+f x)) \sqrt {a+b \sec ^2(e+f x)} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {b \sin ^2(e+f x)}{-a \sin ^2(e+f x)+a+b}\right )}{f (a \cos (2 (e+f x))+a+2 b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.95, size = 390, normalized size = 3.51 \[ \left [\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {b} \cos \left (f x + e\right )^{3} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, {\left ({\left (5 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{32 \, b f \cos \left (f x + e\right )^{3}}, \frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} + 2 \, {\left ({\left (5 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{16 \, b f \cos \left (f x + e\right )^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(a^2 + 2*a*b + b^2)*sqrt(b)*cos(f*x + e)^3*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*co
s(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^
2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) + 4*((5*a*b + 3*b^2)*cos(f*x + e)^2 + 2*b^2)*sqrt((a*cos(f*x + e)^2 +
b)/cos(f*x + e)^2)*sin(f*x + e))/(b*f*cos(f*x + e)^3), 1/16*(3*(a^2 + 2*a*b + b^2)*sqrt(-b)*arctan(-1/2*((a -
b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)
^2 + b^2)*sin(f*x + e)))*cos(f*x + e)^3 + 2*((5*a*b + 3*b^2)*cos(f*x + e)^2 + 2*b^2)*sqrt((a*cos(f*x + e)^2 +
b)/cos(f*x + e)^2)*sin(f*x + e))/(b*f*cos(f*x + e)^3)]

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^2, x)

________________________________________________________________________________________

maple [C]  time = 1.50, size = 1768, normalized size = 15.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/f*sin(f*x+e)*(6*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a
+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*Ell
ipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(
-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*cos(f*x+e)^4*sin(f*x+e)*a^2+1
2*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*
a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticPi((-1+cos(f*
x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1
/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*cos(f*x+e)^4*sin(f*x+e)*a*b+6*2^(1/2)*((I*a^(1/
2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos
(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)
*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/
2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*cos(f*x+e)^4*sin(f*x+e)*b^2-3*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+
e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b
^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))
^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*cos(f*x+e)^4*sin(f
*x+e)*a^2-6*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/
2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((
-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a
^2+6*a*b-b^2)/(a+b)^2)^(1/2))*cos(f*x+e)^4*sin(f*x+e)*a*b-3*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b
^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f
*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x
+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*cos(f*x+e)^4*sin(f*x+e)*b^2+5*((
2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^5*a^2+3*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*
a*b-5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^4*a^2-3*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b)
)^(1/2)*a*b+7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^3*a*b+3*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-
b)/(a+b))^(1/2)*b^2-7*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a*b-3*cos(f*x+e)^2*((2*I*a^(1/2)*b^
(1/2)+a-b)/(a+b))^(1/2)*b^2+2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)*b^2-2*((2*I*a^(1/2)*b^(1/2)+a
-b)/(a+b))^(1/2)*b^2)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)/(-1+cos(f*x+e))/(b+a*cos(f*x+e)^2)^2/cos(f*x+e)/
((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.34, size = 104, normalized size = 0.94 \[ \frac {\frac {3 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + 3 \, {\left (a + b\right )} \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right ) + 3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*(a + b)*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) + 3*(a + b)*sqrt(b)*arcsinh(b*tan(f*x + e)/sq
rt((a + b)*b)) + 2*(b*tan(f*x + e)^2 + a + b)^(3/2)*tan(f*x + e) + 3*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*ta
n(f*x + e))/f

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \sec ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sec(e + f*x)**2)**(3/2)*sec(e + f*x)**2, x)

________________________________________________________________________________________