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

Optimal. Leaf size=78 \[ \frac{b \sin (e+f x)}{a f (a+b) \sqrt{a+b \sin ^2(e+f x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{f (a+b)^{3/2}} \]

[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])

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Rubi [A]  time = 0.101223, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3190, 382, 377, 206} \[ \frac{b \sin (e+f x)}{a f (a+b) \sqrt{a+b \sin ^2(e+f x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{f (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[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 3190

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]

Rule 382

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] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 377

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 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])

Rubi steps

\begin{align*} \int \frac{\sec (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{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{\tanh ^{-1}\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]  time = 7.44037, size = 480, normalized size = 6.15 \[ \frac{\tan (e+f x) \sec (e+f x) \left (-\frac{30 b (a+b) \sin ^2(e+f x) \tan ^2(e+f x) \sin ^{-1}\left (\sqrt{-\frac{(a+b) \tan ^2(e+f x)}{a}}\right )}{a^2}+\frac{30 b \sin ^2(e+f x) \sqrt{-\frac{(a+b) \tan ^2(e+f x) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a^2}}}{a}+45 \sqrt{-\frac{(a+b) \tan ^2(e+f x) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a^2}}+\frac{4 b \sin ^2(e+f x) \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{5/2} \, _2F_1\left (2,2;\frac{7}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt{\frac{\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}}}{a}+4 \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{5/2} \, _2F_1\left (2,2;\frac{7}{2};-\frac{(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt{\frac{\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}}-\frac{30 b \sin ^2(e+f x) \sin ^{-1}\left (\sqrt{-\frac{(a+b) \tan ^2(e+f x)}{a}}\right )}{a}-\frac{45 (a+b) \tan ^2(e+f x) \sin ^{-1}\left (\sqrt{-\frac{(a+b) \tan ^2(e+f x)}{a}}\right )}{a}-45 \sin ^{-1}\left (\sqrt{-\frac{(a+b) \tan ^2(e+f x)}{a}}\right )\right )}{15 a f \sqrt{a+b \sin ^2(e+f x)} \left (-\frac{(a+b) \tan ^2(e+f x)}{a}\right )^{3/2} \sqrt{\frac{\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}}} \]

Warning: Unable to verify antiderivative.

[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))

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Maple [B]  time = 3.955, size = 398, normalized size = 5.1 \begin{align*}{\frac{1}{2\,a \left ( -ab \left ( \cos \left ( fx+e \right ) \right ) ^{2}-{b}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}+{a}^{2}+2\,ab+{b}^{2} \right ) f} \left ( 2\,\sqrt{a+b}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+{\frac{a{b}^{2}+{b}^{3}}{{b}^{2}}}}b\sin \left ( fx+e \right ) -ab \left ( \ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ) -\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ){a}^{2}+\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ) ab-\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ){a}^{2}-\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ) ab \right ){\frac{1}{\sqrt{a+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.16569, size = 1068, normalized size = 13.69 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)