3.369 \(\int \tan ^2(e+f x) (a+b \tan ^2(e+f x))^p \, dx\)
Optimal. Leaf size=83 \[ \frac {\tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac {3}{2};1,-p;\frac {5}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )}{3 f} \]
[Out]
1/3*AppellF1(3/2,1,-p,5/2,-tan(f*x+e)^2,-b*tan(f*x+e)^2/a)*tan(f*x+e)^3*(a+b*tan(f*x+e)^2)^p/f/((1+b*tan(f*x+e
)^2/a)^p)
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Rubi [A] time = 0.10, antiderivative size = 83, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used =
{3670, 511, 510} \[ \frac {\tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac {3}{2};1,-p;\frac {5}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )}{3 f} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^p,x]
[Out]
(AppellF1[3/2, 1, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Tan[e + f*x]^3*(a + b*Tan[e + f*x]^2)^p)/
(3*f*(1 + (b*Tan[e + f*x]^2)/a)^p)
Rule 510
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rule 511
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && !(IntegerQ[
p] || GtQ[a, 0])
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 \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\left (\left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (1+\frac {b x^2}{a}\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {F_1\left (\frac {3}{2};1,-p;\frac {5}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}}{3 f}\\ \end {align*}
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Mathematica [B] time = 15.10, size = 1992, normalized size = 24.00 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^p,x]
[Out]
(Tan[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(2*p)*(Hypergeometric2F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2)/a)]/(1 + (b
*Tan[e + f*x]^2)/a)^p + (3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Cos[e + f*x]^
2)/(-3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1
, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]) + a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[
e + f*x]^2])*Tan[e + f*x]^2)))/(f*(2*b*p*Sec[e + f*x]^2*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^(-1 + p)*(Hyperg
eometric2F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2)/a)]/(1 + (b*Tan[e + f*x]^2)/a)^p + (3*a*AppellF1[1/2, -p, 1, 3/
2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Cos[e + f*x]^2)/(-3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]
^2)/a), -Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]) +
a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]^2)) + Sec[e + f*x]^2*(a +
b*Tan[e + f*x]^2)^p*(Hypergeometric2F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2)/a)]/(1 + (b*Tan[e + f*x]^2)/a)^p + (
3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Cos[e + f*x]^2)/(-3*a*AppellF1[1/2, -p
, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^
2)/a), -Tan[e + f*x]^2]) + a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]
^2)) + Tan[e + f*x]*(a + b*Tan[e + f*x]^2)^p*((-2*b*p*Hypergeometric2F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2)/a)]
*Sec[e + f*x]^2*Tan[e + f*x]*(1 + (b*Tan[e + f*x]^2)/a)^(-1 - p))/a - (6*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[
e + f*x]^2)/a), -Tan[e + f*x]^2]*Cos[e + f*x]*Sin[e + f*x])/(-3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^
2)/a), -Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]) + a
*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]^2) + (3*a*Cos[e + f*x]^2*((
2*b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3*a
) - (2*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/3))/(-
3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2
, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]) + a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f
*x]^2])*Tan[e + f*x]^2) + (Csc[e + f*x]*Sec[e + f*x]*(-Hypergeometric2F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2)/a)
] + (1 + (b*Tan[e + f*x]^2)/a)^p))/(1 + (b*Tan[e + f*x]^2)/a)^p - (3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e +
f*x]^2)/a), -Tan[e + f*x]^2]*Cos[e + f*x]^2*(4*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -T
an[e + f*x]^2]) + a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Sec[e + f*x]^2*Tan[e
+ f*x] - 3*a*((2*b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan
[e + f*x])/(3*a) - (2*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e
+ f*x])/3) + 2*Tan[e + f*x]^2*(-(b*p*((-6*AppellF1[5/2, 1 - p, 2, 7/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]
^2]*Sec[e + f*x]^2*Tan[e + f*x])/5 - (6*b*(1 - p)*AppellF1[5/2, 2 - p, 1, 7/2, -((b*Tan[e + f*x]^2)/a), -Tan[e
+ f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(5*a))) + a*((6*b*p*AppellF1[5/2, 1 - p, 2, 7/2, -((b*Tan[e + f*x]^2)/
a), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(5*a) - (12*AppellF1[5/2, -p, 3, 7/2, -((b*Tan[e + f*x]^2)/a
), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/5))))/(-3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a)
, -Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]) + a*Appe
llF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]^2)^2)))
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fricas [F] time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^2*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")
[Out]
integral((b*tan(f*x + e)^2 + a)^p*tan(f*x + e)^2, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^2*(a+b*tan(f*x+e)^2)^p,x, algorithm="giac")
[Out]
integrate((b*tan(f*x + e)^2 + a)^p*tan(f*x + e)^2, x)
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maple [F] time = 0.90, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{2}\left (f x +e \right )\right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(f*x+e)^2*(a+b*tan(f*x+e)^2)^p,x)
[Out]
int(tan(f*x+e)^2*(a+b*tan(f*x+e)^2)^p,x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^2*(a+b*tan(f*x+e)^2)^p,x, algorithm="maxima")
[Out]
integrate((b*tan(f*x + e)^2 + a)^p*tan(f*x + e)^2, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^2\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(e + f*x)^2*(a + b*tan(e + f*x)^2)^p,x)
[Out]
int(tan(e + f*x)^2*(a + b*tan(e + f*x)^2)^p, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{p} \tan ^{2}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)**2*(a+b*tan(f*x+e)**2)**p,x)
[Out]
Integral((a + b*tan(e + f*x)**2)**p*tan(e + f*x)**2, x)
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