\(\int \frac {\sec (e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [356]

Optimal result

Integrand size = 23, antiderivative size = 78 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{(a+b)^{3/2} f}+\frac {b \sin (e+f x)}{a (a+b) f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/(a+b)^(3/2)/f+b*sin(f*x+e)/a/(a+b)/f/(a+b*sin(f*x+e)^
2)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3269, 390, 385, 212} \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{f (a+b)^{3/2}}+\frac {b \sin (e+f x)}{a f (a+b) \sqrt {a+b \sin ^2(e+f x)}} \]

[In]

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

[Out]

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

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 3269

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {b \sin (e+f x)}{a (a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f} \\ & = \frac {b \sin (e+f x)}{a (a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{1-(a+b) x^2} \, dx,x,\frac {\sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{(a+b) f} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{(a+b)^{3/2} f}+\frac {b \sin (e+f x)}{a (a+b) f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 7.52 (sec) , antiderivative size = 480, normalized size of antiderivative = 6.15 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sec (e+f x) \tan (e+f x) \left (-45 \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right )-\frac {30 b \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x)}{a}-\frac {45 (a+b) \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \tan ^2(e+f x)}{a}-\frac {30 b (a+b) \arcsin \left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{a^2}+4 \operatorname {Hypergeometric2F1}\left (2,2,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}+\frac {4 b \operatorname {Hypergeometric2F1}\left (2,2,\frac {7}{2},-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2}}{a}+45 \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}+\frac {30 b \sin ^2(e+f x) \sqrt {-\frac {(a+b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{a^2}}}{a}\right )}{15 a f \sqrt {a+b \sin ^2(e+f x)} \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2}} \]

[In]

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

[Out]

(Sec[e + f*x]*Tan[e + f*x]*(-45*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]] - (30*b*ArcSin[Sqrt[-(((a + b)*Tan
[e + f*x]^2)/a)]]*Sin[e + f*x]^2)/a - (45*(a + b)*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Tan[e + f*x]^2)/
a - (30*b*(a + b)*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2*Tan[e + f*x]^2)/a^2 + 4*Hypergeom
etric2F1[2, 2, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)
*Tan[e + f*x]^2)/a))^(5/2) + (4*b*Hypergeometric2F1[2, 2, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*S
qrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(5/2))/a + 45*Sqrt[-(((a + b)*S
ec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2)/a^2)] + (30*b*Sin[e + f*x]^2*Sqrt[-(((a + b)*Sec[e + f*x]
^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2)/a^2)])/a))/(15*a*f*Sqrt[a + b*Sin[e + f*x]^2]*Sqrt[(Sec[e + f*x]^2*(
a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(396\) vs. \(2(70)=140\).

Time = 1.14 (sec) , antiderivative size = 397, normalized size of antiderivative = 5.09

method	result	size
default	\(\frac {a b \left (-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a +b}\, \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, b \sin \left (f x +e \right )-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2}-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2}+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b}{2 \sqrt {a +b}\, a \left (-\left (\cos ^{2}\left (f x +e \right )\right ) a b -\left (\cos ^{2}\left (f x +e \right )\right ) b^{2}+a^{2}+2 a b +b^{2}\right ) f}\)	\(397\)

[In]

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

[Out]

1/2/(a+b)^(1/2)/a/(-cos(f*x+e)^2*a*b-cos(f*x+e)^2*b^2+a^2+2*a*b+b^2)*(a*b*(-ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(
a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))+ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f
*x+e)+a)))*cos(f*x+e)^2+2*(a+b)^(1/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b*sin(f*x+e)-ln(2/(1+sin(f*x+e))
*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+
e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b+ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*
a^2+ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (70) = 140\).

Time = 0.39 (sec) , antiderivative size = 453, normalized size of antiderivative = 5.81 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (a b \cos \left (f x + e\right )^{2} - a^{2} - a b\right )} \sqrt {a + b} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}, -\frac {{\left (a b \cos \left (f x + e\right )^{2} - a^{2} - a b\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (70) = 140\).

Time = 0.33 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.95 \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {2 \, b \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} + \sqrt {b \sin \left (f x + e\right )^{2} + a} a b} + \frac {\operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}} + \frac {\operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}}}{2 \, f} \]

[In]

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

[Out]

1/2*(2*b*sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + a)*a^2 + sqrt(b*sin(f*x + e)^2 + a)*a*b) + arcsinh(b*sin(f*x +
e)/(sqrt(a*b)*(sin(f*x + e) + 1)) - a/(sqrt(a*b)*(sin(f*x + e) + 1)))/(a + b)^(3/2) + arcsinh(-b*sin(f*x + e)/
(sqrt(a*b)*(sin(f*x + e) - 1)) - a/(sqrt(a*b)*(sin(f*x + e) - 1)))/(a + b)^(3/2))/f

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{\cos \left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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