3.4.55 \(\int \frac {1}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\) [355]

Optimal. Leaf size=134 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}-\frac {b \tan (e+f x)}{3 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {(5 a-2 b) b \tan (e+f x)}{3 a^2 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}} \]

[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*(5*a-2*b)*b*tan(f*x+e)/a^2/(a-b)^2/f
/(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)

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Rubi [A]
time = 0.07, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3742, 425, 541, 12, 385, 209} \begin {gather*} -\frac {b (5 a-2 b) \tan (e+f x)}{3 a^2 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}+\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}-\frac {b \tan (e+f x)}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(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) - (b*Tan[e + f*x])/(3*a*(a - b
)*f*(a + b*Tan[e + f*x]^2)^(3/2)) - ((5*a - 2*b)*b*Tan[e + f*x])/(3*a^2*(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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 425

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[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 541

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 3742

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])

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 7.03, size = 1331, normalized size = 9.93 \begin {gather*} \frac {\cos (e+f x) \sin (e+f x) \left (1575 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right )-\frac {3150 (a-b) \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \sin ^2(e+f x)}{a}+\frac {1575 (a-b)^2 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \sin ^4(e+f x)}{a^2}+\frac {2100 b \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^2(e+f x)}{a}-\frac {4200 (a-b) b \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \sin ^2(e+f x) \tan ^2(e+f x)}{a^2}+\frac {2100 (a-b)^2 b \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \sin ^4(e+f x) \tan ^2(e+f x)}{a^3}+\frac {840 b^2 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \tan ^4(e+f x)}{a^2}-\frac {1680 (a-b) b^2 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \sin ^2(e+f x) \tan ^4(e+f x)}{a^3}+\frac {840 (a-b)^2 b^2 \text {ArcSin}\left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \sin ^4(e+f x) \tan ^4(e+f x)}{a^4}+2100 \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{3/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}+96 \, _2F_1\left (2,2;\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{7/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}+24 \, _3F_2\left (2,2,2;1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{7/2} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}+\frac {2800 b \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{3/2} \tan ^2(e+f x) \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}{a}+\frac {168 b \, _2F_1\left (2,2;\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{7/2} \tan ^2(e+f x) \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}{a}+\frac {48 b \, _3F_2\left (2,2,2;1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{7/2} \tan ^2(e+f x) \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}{a}+\frac {1120 b^2 \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{3/2} \tan ^4(e+f x) \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}{a^2}+\frac {72 b^2 \, _2F_1\left (2,2;\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{7/2} \tan ^4(e+f x) \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}{a^2}+\frac {24 b^2 \, _3F_2\left (2,2,2;1,\frac {9}{2};\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{7/2} \tan ^4(e+f x) \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}{a^2}-1575 \sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}-\frac {2100 b \tan ^2(e+f x) \sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}{a}-\frac {840 b^2 \tan ^4(e+f x) \sqrt {\frac {(a-b) \cos ^2(e+f x) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}{a^2}\right )}{315 a^2 f \left (\frac {(a-b) \sin ^2(e+f x)}{a}\right )^{5/2} \sqrt {a+b \tan ^2(e+f x)} \sqrt {\frac {\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}} \left (1+\frac {b \tan ^2(e+f x)}{a}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[e + f*x]*Sin[e + f*x]*(1575*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]] - (3150*(a - b)*ArcSin[Sqrt[((a - b)
*Sin[e + f*x]^2)/a]]*Sin[e + f*x]^2)/a + (1575*(a - b)^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sin[e + f*x]
^4)/a^2 + (2100*b*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x]^2)/a - (4200*(a - b)*b*ArcSin[Sqrt[((a
- b)*Sin[e + f*x]^2)/a]]*Sin[e + f*x]^2*Tan[e + f*x]^2)/a^2 + (2100*(a - b)^2*b*ArcSin[Sqrt[((a - b)*Sin[e +
f*x]^2)/a]]*Sin[e + f*x]^4*Tan[e + f*x]^2)/a^3 + (840*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Tan[e + f*x
]^4)/a^2 - (1680*(a - b)*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sin[e + f*x]^2*Tan[e + f*x]^4)/a^3 + (84
0*(a - b)^2*b^2*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*Sin[e + f*x]^4*Tan[e + f*x]^4)/a^4 + 2100*(((a - b)*S
in[e + f*x]^2)/a)^(3/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a] + 96*Hypergeometric2F1[2, 2, 9/2, ((a
- b)*Sin[e + f*x]^2)/a]*(((a - b)*Sin[e + f*x]^2)/a)^(7/2)*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a] + 2
4*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*(((a - b)*Sin[e + f*x]^2)/a)^(7/2)*Sqrt[(
Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a] + (2800*b*(((a - b)*Sin[e + f*x]^2)/a)^(3/2)*Tan[e + f*x]^2*Sqrt[(Co
s[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a])/a + (168*b*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin[e + f*x]^2)/a]*(
((a - b)*Sin[e + f*x]^2)/a)^(7/2)*Tan[e + f*x]^2*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a])/a + (48*b*Hy
pergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*(((a - b)*Sin[e + f*x]^2)/a)^(7/2)*Tan[e + f*
x]^2*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a])/a + (1120*b^2*(((a - b)*Sin[e + f*x]^2)/a)^(3/2)*Tan[e +
f*x]^4*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a])/a^2 + (72*b^2*Hypergeometric2F1[2, 2, 9/2, ((a - b)*S
in[e + f*x]^2)/a]*(((a - b)*Sin[e + f*x]^2)/a)^(7/2)*Tan[e + f*x]^4*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2
))/a])/a^2 + (24*b^2*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Sin[e + f*x]^2)/a]*(((a - b)*Sin[e + f*x]
^2)/a)^(7/2)*Tan[e + f*x]^4*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a])/a^2 - 1575*Sqrt[((a - b)*Cos[e +
f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2] - (2100*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])/a - (840*b^2*Tan[e + f*x]^4*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^
2*(a + b*Tan[e + f*x]^2))/a^2])/a^2))/(315*a^2*f*(((a - b)*Sin[e + f*x]^2)/a)^(5/2)*Sqrt[a + b*Tan[e + f*x]^2]
*Sqrt[(Cos[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a]*(1 + (b*Tan[e + f*x]^2)/a))

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Maple [A]
time = 0.10, size = 163, normalized size = 1.22

method	result	size
derivativedivides	\(\frac {-\frac {b \left (\frac {\tan \left (f x +e \right )}{3 a \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {2 \tan \left (f x +e \right )}{3 a^{2} \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{a -b}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{\left (a -b \right )^{3} b^{2}}-\frac {b \tan \left (f x +e \right )}{\left (a -b \right )^{2} a \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}}{f}\)	\(163\)
default	\(\frac {-\frac {b \left (\frac {\tan \left (f x +e \right )}{3 a \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {2 \tan \left (f x +e \right )}{3 a^{2} \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{a -b}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{\left (a -b \right )^{3} b^{2}}-\frac {b \tan \left (f x +e \right )}{\left (a -b \right )^{2} a \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}}{f}\)	\(163\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (126) = 252\).
time = 1.37, size = 581, normalized size = 4.34 \begin {gather*} \left [-\frac {3 \, {\left (a^{2} b^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{3} b \tan \left (f x + e\right )^{2} + a^{4}\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 (5 \, a^{2} b^{2} - 7 \, a b^{3} + 2 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (2 \, a^{3} b - 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} f\right )}}, \frac {3 \, {\left (a^{2} b^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{3} b \tan \left (f x + e\right )^{2} + a^{4}\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 (5 \, a^{2} b^{2} - 7 \, a b^{3} + 2 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (2 \, a^{3} b - 3 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{3 \, {\left ({\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(3*(a^2*b^2*tan(f*x + e)^4 + 2*a^3*b*tan(f*x + e)^2 + a^4)*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*((5*a^2*b^2 - 7*a*b^3 +
2*b^4)*tan(f*x + e)^3 + 3*(2*a^3*b - 3*a^2*b^2 + a*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a^5*b^2 -
3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5)*f*tan(f*x + e)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*tan(f*x + e
)^2 + (a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*f), 1/3*(3*(a^2*b^2*tan(f*x + e)^4 + 2*a^3*b*tan(f*x + e)^2 + a^4)
*sqrt(a - b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) - ((5*a^2*b^2 - 7*a*b^3 + 2*b^4)*t
an(f*x + e)^3 + 3*(2*a^3*b - 3*a^2*b^2 + a*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a^5*b^2 - 3*a^4*b^
3 + 3*a^3*b^4 - a^2*b^5)*f*tan(f*x + e)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*tan(f*x + e)^2 + (a^
7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tan(e + f*x)^2)^(5/2),x)

[Out]

int(1/(a + b*tan(e + f*x)^2)^(5/2), x)

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