3.342 \(\int \frac{1}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=85 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac{b \tan (e+f x)}{a f (a-b) \sqrt{a+b \tan ^2(e+f x)}} \]

[Out]

ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(3/2)*f) - (b*Tan[e + f*x])/(a*(a - b)*
f*Sqrt[a + b*Tan[e + f*x]^2])

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Rubi [A]  time = 0.0651576, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3661, 382, 377, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac{b \tan (e+f x)}{a f (a-b) \sqrt{a+b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x]^2)^(-3/2),x]

[Out]

ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(3/2)*f) - (b*Tan[e + f*x])/(a*(a - b)*
f*Sqrt[a + b*Tan[e + f*x]^2])

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

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 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])

Rubi steps

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

Mathematica [C]  time = 6.12497, size = 214, normalized size = 2.52 \[ \frac{4 \sin (e+f x) \cos ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)} \left (a (a-b) \tan ^2(e+f x) \text{Hypergeometric2F1}\left (2,2,\frac{7}{2},\frac{(a-b) \sin ^2(e+f x)}{a}\right )+\frac{15 \left (3 a+2 b \tan ^2(e+f x)\right ) \left (a \sqrt{\frac{(a-b) \sin ^2(2 (e+f x)) \left (a+b \tan ^2(e+f x)\right )}{a^2}}-2 \sin ^{-1}\left (\sqrt{\frac{(a-b) \sin ^2(e+f x)}{a}}\right ) \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )\right )}{\left (\frac{(a-b) \sin ^2(2 (e+f x)) \left (a+b \tan ^2(e+f x)\right )}{a^2}\right )^{3/2}}\right )}{15 a^4 f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*Cos[e + f*x]^3*Sin[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2]*(a*(a - b)*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Sin
[e + f*x]^2)/a]*Tan[e + f*x]^2 + (15*(3*a + 2*b*Tan[e + f*x]^2)*(-2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*(
a*Cos[e + f*x]^2 + b*Sin[e + f*x]^2) + a*Sqrt[((a - b)*Sin[2*(e + f*x)]^2*(a + b*Tan[e + f*x]^2))/a^2]))/(((a
- b)*Sin[2*(e + f*x)]^2*(a + b*Tan[e + f*x]^2))/a^2)^(3/2)))/(15*a^4*f)

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Maple [A]  time = 0., size = 104, normalized size = 1.2 \begin{align*} -{\frac{b\tan \left ( fx+e \right ) }{a \left ( a-b \right ) f}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}+{\frac{1}{f \left ( a-b \right ) ^{2}{b}^{2}}\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\tan \left ( fx+e \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*tan(f*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)^(1/2)+1/f/(a-b)^2*(b^4*(a-b))^(1/2)/b^2*arctan(b^2*(a-b)/(b^4*(a-b)
)^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.27543, size = 722, normalized size = 8.49 \begin{align*} \left [\frac{{\left (a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \sqrt{-a + b} \log \left (-\frac{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} + 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a}{\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{2 \,{\left ({\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}, \frac{{\left (a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \sqrt{a - b} \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{a - b} \tan \left (f x + e\right )}\right ) - \sqrt{b \tan \left (f x + e\right )^{2} + a}{\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/2*((a*b*tan(f*x + e)^2 + a^2)*sqrt(-a + b)*log(-((a - 2*b)*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*sq
rt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) - 2*sqrt(b*tan(f*x + e)^2 + a)*(a*b - b^2)*tan(f*x + e))/((
a^3*b - 2*a^2*b^2 + a*b^3)*f*tan(f*x + e)^2 + (a^4 - 2*a^3*b + a^2*b^2)*f), ((a*b*tan(f*x + e)^2 + a^2)*sqrt(a
- b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) - sqrt(b*tan(f*x + e)^2 + a)*(a*b - b^2)*
tan(f*x + e))/((a^3*b - 2*a^2*b^2 + a*b^3)*f*tan(f*x + e)^2 + (a^4 - 2*a^3*b + a^2*b^2)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \tan ^{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(1/(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \tan \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(1/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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