\(\int \sec ^2(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [249]

Optimal result

Integrand size = 25, antiderivative size = 111 \[ \int \sec ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {3 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 \sqrt {b} f}+\frac {3 (a+b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}+\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 f} \]

[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] (verified)

Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4231, 201, 223, 212} \[ \int \sec ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {3 (a+b)^2 \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{8 \sqrt {b} f}+\frac {3 (a+b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 f}+\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 f} \]

[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 201

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 212

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

Rule 223

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 4231

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*} \text {integral}& = \frac {\text {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)) \text {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 ) \text {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 ) \text {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 \text {arctanh}\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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.46 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \sec ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {(a+b)^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},\frac {b \sin ^2(e+f x)}{a+b-a \sin ^2(e+f x)}\right ) \sqrt {a+b \sec ^2(e+f x)} \sin (2 (e+f x))}{f (a+2 b+a \cos (2 (e+f x)))} \]

[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)]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(950\) vs. \(2(95)=190\).

Time = 11.80 (sec) , antiderivative size = 951, normalized size of antiderivative = 8.57

method	result	size
default	\(\text {Expression too large to display}\)	\(951\)

[In]

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

[Out]

1/16/f/b^(5/2)*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)/(b+a*cos(f*x+e)^2)/(1+cos(
f*x+e))*(6*cos(f*x+e)^2*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(7/2)+6*cos(f*x+e)*sin(f*x+e)
*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(7/2)+10*cos(f*x+e)^2*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x
+e))^2)^(1/2)*b^(5/2)*a+4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(7/2)*sin(f*x+e)+10*cos(f*x+e)*sin(f*x
+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(5/2)*a+3*cos(f*x+e)^3*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+
e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x
+e)+1))*a^2*b^2+6*cos(f*x+e)^3*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a*b^3+3*cos(f*x+e)^3*ln(4*(((b+a*c
os(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-si
n(f*x+e)*a-a-b)/(sin(f*x+e)+1))*b^4+3*cos(f*x+e)^3*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*
cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*a^2*b^2+6*cos
(f*x+e)^3*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1
+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*a*b^3+3*cos(f*x+e)^3*ln(-4*(((b+a*cos(f*x+e)^2)/(1+cos
(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(si
n(f*x+e)-1))*b^4+4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(7/2)*tan(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 390, normalized size of antiderivative = 3.51 \[ \int \sec ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {b} \cos \left (f x + e\right )^{3} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, {\left ({\left (5 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{32 \, b f \cos \left (f x + e\right )^{3}}, \frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} + 2 \, {\left ({\left (5 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{16 \, b f \cos \left (f x + e\right )^{3}}\right ] \]

[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)]

Sympy [F]

\[ \int \sec ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \sec ^{2}{\left (e + f x \right )}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.94 \[ \int \sec ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {\frac {3 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + 3 \, {\left (a + b\right )} \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right ) + 3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )}{8 \, f} \]

[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

Giac [F]

\[ \int \sec ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{2} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \sec ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{{\cos \left (e+f\,x\right )}^2} \,d x \]

[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)