3.7.99 \(\int (d \tan (e+f x))^n (a+b \tan (e+f x)) \, dx\) [699]

Optimal. Leaf size=103 \[ \frac {a \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {b \, _2F_1\left (1,\frac {2+n}{2};\frac {4+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)} \]

[Out]

a*hypergeom([1, 1/2+1/2*n],[3/2+1/2*n],-tan(f*x+e)^2)*(d*tan(f*x+e))^(1+n)/d/f/(1+n)+b*hypergeom([1, 1+1/2*n],
[2+1/2*n],-tan(f*x+e)^2)*(d*tan(f*x+e))^(2+n)/d^2/f/(2+n)

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Rubi [A]
time = 0.06, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3619, 3557, 371} \begin {gather*} \frac {a (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac {b (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac {n+2}{2};\frac {n+4}{2};-\tan ^2(e+f x)\right )}{d^2 f (n+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^n*(a + b*Tan[e + f*x]),x]

[Out]

(a*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) + (b*Hy
pergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(2 + n))/(d^2*f*(2 + n))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3619

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

Rubi steps

\begin {align*} \int (d \tan (e+f x))^n (a+b \tan (e+f x)) \, dx &=a \int (d \tan (e+f x))^n \, dx+\frac {b \int (d \tan (e+f x))^{1+n} \, dx}{d}\\ &=\frac {b \text {Subst}\left (\int \frac {x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}+\frac {(a d) \text {Subst}\left (\int \frac {x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=\frac {a \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {b \, _2F_1\left (1,\frac {2+n}{2};\frac {4+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 99, normalized size = 0.96 \begin {gather*} \frac {\tan (e+f x) (d \tan (e+f x))^n \left (a (2+n) \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};-\tan ^2(e+f x)\right )+b (1+n) \, _2F_1\left (1,\frac {2+n}{2};\frac {4+n}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)\right )}{f (1+n) (2+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^n*(a + b*Tan[e + f*x]),x]

[Out]

(Tan[e + f*x]*(d*Tan[e + f*x])^n*(a*(2 + n)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2] + b*(1
 + n)*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*Tan[e + f*x]))/(f*(1 + n)*(2 + n))

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Maple [F]
time = 0.27, size = 0, normalized size = 0.00 \[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +b \tan \left (f x +e \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e)),x)

[Out]

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e)),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^n*(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)*(d*tan(f*x + e))^n, x)

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Fricas [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((d*tan(f*x+e))^n*(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e) + a)*(d*tan(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**n*(a+b*tan(f*x+e)),x)

[Out]

Integral((d*tan(e + f*x))**n*(a + b*tan(e + f*x)), 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((d*tan(f*x+e))^n*(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)*(d*tan(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^n*(a + b*tan(e + f*x)),x)

[Out]

int((d*tan(e + f*x))^n*(a + b*tan(e + f*x)), x)

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