3.354 \(\int \frac {\tan ^2(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=128 \[ \frac {(2 a+b) \tan (e+f x)}{3 a f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}+\frac {\tan (e+f x)}{3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}} \]

[Out]

-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(5/2)/f+1/3*(2*a+b)*tan(f*x+e)/a/(a-b)^2/f/(a+b
*tan(f*x+e)^2)^(1/2)+1/3*tan(f*x+e)/(a-b)/f/(a+b*tan(f*x+e)^2)^(3/2)

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Rubi [A]  time = 0.15, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3670, 471, 527, 12, 377, 203} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}+\frac {(2 a+b) \tan (e+f x)}{3 a f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}+\frac {\tan (e+f x)}{3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(5/2)*f)) + Tan[e + f*x]/(3*(a - b)*f
*(a + b*Tan[e + f*x]^2)^(3/2)) + ((2*a + b)*Tan[e + f*x])/(3*a*(a - b)^2*f*Sqrt[a + b*Tan[e + f*x]^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

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 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((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[(((d*ff*x)/c)^m*(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, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \frac {\tan ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x)}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\operatorname {Subst}\left (\int \frac {1-2 x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 (a-b) f}\\ &=\frac {\tan (e+f x)}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(2 a+b) \tan (e+f x)}{3 a (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {3 a}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a (a-b)^2 f}\\ &=\frac {\tan (e+f x)}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(2 a+b) \tan (e+f x)}{3 a (a-b)^2 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)^2 f}\\ &=\frac {\tan (e+f x)}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(2 a+b) \tan (e+f x)}{3 a (a-b)^2 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)^2 f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}+\frac {\tan (e+f x)}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(2 a+b) \tan (e+f x)}{3 a (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 7.77, size = 365, normalized size = 2.85 \[ \frac {\cos ^4(e+f x) \cot (e+f x) \left (12 (a-b)^3 \tan ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, _2F_1\left (2,2;\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sqrt {\frac {\sin ^2(e+f x) \cos ^2(e+f x) \left (a^2+a b \left (\tan ^2(e+f x)-1\right )-b^2 \tan ^2(e+f x)\right )}{a^2}}+35 a \sec ^2(e+f x) \left (5 a+2 b \tan ^2(e+f x)\right ) \left (a \sec ^2(e+f x) \left (a \left (\tan ^2(e+f x)-3\right )-4 b \tan ^2(e+f x)\right ) \sqrt {\frac {\sin ^2(e+f x) \cos ^2(e+f x) \left (a^2+a b \left (\tan ^2(e+f x)-1\right )-b^2 \tan ^2(e+f x)\right )}{a^2}}+3 \sin ^{-1}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \left (a+b \tan ^2(e+f x)\right )^2\right )\right )}{315 a^4 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)} \left (\frac {b \tan ^2(e+f x)}{a}+1\right ) \sqrt {\frac {(a-b) \sin ^2(e+f x) \cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

(Cos[e + f*x]^4*Cot[e + f*x]*(12*(a - b)^3*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*
x]^6*(a + b*Tan[e + f*x]^2)*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e +
f*x]^2)))/a^2] + 35*a*Sec[e + f*x]^2*(5*a + 2*b*Tan[e + f*x]^2)*(3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*(a
+ b*Tan[e + f*x]^2)^2 + a*Sec[e + f*x]^2*(-4*b*Tan[e + f*x]^2 + a*(-3 + Tan[e + f*x]^2))*Sqrt[(Cos[e + f*x]^2
*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e + f*x]^2)))/a^2])))/(315*a^4*(a - b)^2*f*Sqrt[a +
b*Tan[e + f*x]^2]*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2]*(1 + (b*Tan[e + f*x
]^2)/a))

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fricas [B]  time = 0.55, size = 529, normalized size = 4.13 \[ \left [-\frac {3 \, {\left (a b^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{2} b \tan \left (f x + e\right )^{2} + a^{3}\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, {\left ({\left (2 \, a^{2} b - a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{3} - a^{2} b\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{4} b^{2} - 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} - a b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} f\right )}}, -\frac {3 \, {\left (a b^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{2} b \tan \left (f x + e\right )^{2} + a^{3}\right )} \sqrt {a - b} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) - {\left ({\left (2 \, a^{2} b - a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{3} - a^{2} b\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{3 \, {\left ({\left (a^{4} b^{2} - 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} - a b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(3*(a*b^2*tan(f*x + e)^4 + 2*a^2*b*tan(f*x + e)^2 + a^3)*sqrt(-a + b)*log(-((a - 2*b)*tan(f*x + e)^2 + 2
*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) - 2*((2*a^2*b - a*b^2 - b^3)*
tan(f*x + e)^3 + 3*(a^3 - a^2*b)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a^4*b^2 - 3*a^3*b^3 + 3*a^2*b^4 -
a*b^5)*f*tan(f*x + e)^4 + 2*(a^5*b - 3*a^4*b^2 + 3*a^3*b^3 - a^2*b^4)*f*tan(f*x + e)^2 + (a^6 - 3*a^5*b + 3*a
^4*b^2 - a^3*b^3)*f), -1/3*(3*(a*b^2*tan(f*x + e)^4 + 2*a^2*b*tan(f*x + e)^2 + a^3)*sqrt(a - b)*arctan(-sqrt(b
*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) - ((2*a^2*b - a*b^2 - b^3)*tan(f*x + e)^3 + 3*(a^3 - a^2*b)*t
an(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a^4*b^2 - 3*a^3*b^3 + 3*a^2*b^4 - a*b^5)*f*tan(f*x + e)^4 + 2*(a^5*
b - 3*a^4*b^2 + 3*a^3*b^3 - a^2*b^4)*f*tan(f*x + e)^2 + (a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*f)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (f x + e\right )^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^2/(b*tan(f*x + e)^2 + a)^(5/2), x)

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maple [B]  time = 0.26, size = 232, normalized size = 1.81 \[ \frac {\tan \left (f x +e \right )}{3 f a \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {2 \tan \left (f x +e \right )}{3 f \,a^{2} \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}+\frac {b \tan \left (f x +e \right )}{3 a \left (a -b \right ) f \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {2 b \tan \left (f x +e \right )}{3 f \left (a -b \right ) a^{2} \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}+\frac {b \tan \left (f x +e \right )}{f \left (a -b \right )^{2} a \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \left (a -b \right )^{3} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x)

[Out]

1/3/f*tan(f*x+e)/a/(a+b*tan(f*x+e)^2)^(3/2)+2/3/f/a^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2)+1/3*b*tan(f*x+e)/a/(
a-b)/f/(a+b*tan(f*x+e)^2)^(3/2)+2/3/f*b/(a-b)/a^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2)+1/f*b/(a-b)^2*tan(f*x+e)
/a/(a+b*tan(f*x+e)^2)^(1/2)-1/f/(a-b)^3*(b^4*(a-b))^(1/2)/b^2*arctan((a-b)*b^2/(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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^2/(a + b*tan(e + f*x)^2)^(5/2),x)

[Out]

int(tan(e + f*x)^2/(a + b*tan(e + f*x)^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**2/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

Integral(tan(e + f*x)**2/(a + b*tan(e + f*x)**2)**(5/2), x)

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