3.1308 \(\int (a+b \tan (e+f x))^m \, dx\)

Optimal. Leaf size=167 \[ \frac{b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} f (m+1) \left (a-\sqrt{-b^2}\right )}-\frac{b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} f (m+1) \left (a+\sqrt{-b^2}\right )} \]

[Out]

(b*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - Sqrt[-b^2])]*(a + b*Tan[e + f*x])^(1 + m))/(2*
Sqrt[-b^2]*(a - Sqrt[-b^2])*f*(1 + m)) - (b*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a + Sqrt[
-b^2])]*(a + b*Tan[e + f*x])^(1 + m))/(2*Sqrt[-b^2]*(a + Sqrt[-b^2])*f*(1 + m))

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Rubi [A]  time = 0.251154, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3485, 712, 68} \[ \frac{b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} f (m+1) \left (a-\sqrt{-b^2}\right )}-\frac{b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+\sqrt{-b^2}}\right )}{2 \sqrt{-b^2} f (m+1) \left (a+\sqrt{-b^2}\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^m,x]

[Out]

(b*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - Sqrt[-b^2])]*(a + b*Tan[e + f*x])^(1 + m))/(2*
Sqrt[-b^2]*(a - Sqrt[-b^2])*f*(1 + m)) - (b*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a + Sqrt[
-b^2])]*(a + b*Tan[e + f*x])^(1 + m))/(2*Sqrt[-b^2]*(a + Sqrt[-b^2])*f*(1 + m))

Rule 3485

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

Rule 712

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int (a+b \tan (e+f x))^m \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^m}{b^2+x^2} \, dx,x,b \tan (e+f x)\right )}{f}\\ &=\frac{b \operatorname{Subst}\left (\int \left (\frac{\sqrt{-b^2} (a+x)^m}{2 b^2 \left (\sqrt{-b^2}-x\right )}+\frac{\sqrt{-b^2} (a+x)^m}{2 b^2 \left (\sqrt{-b^2}+x\right )}\right ) \, dx,x,b \tan (e+f x)\right )}{f}\\ &=-\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^m}{\sqrt{-b^2}-x} \, dx,x,b \tan (e+f x)\right )}{2 \sqrt{-b^2} f}-\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^m}{\sqrt{-b^2}+x} \, dx,x,b \tan (e+f x)\right )}{2 \sqrt{-b^2} f}\\ &=\frac{b \, _2F_1\left (1,1+m;2+m;\frac{a+b \tan (e+f x)}{a-\sqrt{-b^2}}\right ) (a+b \tan (e+f x))^{1+m}}{2 \sqrt{-b^2} \left (a-\sqrt{-b^2}\right ) f (1+m)}-\frac{b \, _2F_1\left (1,1+m;2+m;\frac{a+b \tan (e+f x)}{a+\sqrt{-b^2}}\right ) (a+b \tan (e+f x))^{1+m}}{2 \sqrt{-b^2} \left (a+\sqrt{-b^2}\right ) f (1+m)}\\ \end{align*}

Mathematica [C]  time = 0.141003, size = 118, normalized size = 0.71 \[ \frac{(a+b \tan (e+f x))^{m+1} \left ((a+i b) \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )-(a-i b) \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )\right )}{2 f (m+1) (a+i b) (b+i a)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^m,x]

[Out]

(((a + I*b)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - I*b)] - (a - I*b)*Hypergeometric2F1[1
, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a + I*b)])*(a + b*Tan[e + f*x])^(1 + m))/(2*(a + I*b)*(I*a + b)*f*(1 + m
))

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Maple [F]  time = 0.185, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e) + a)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \tan \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*tan(f*x + e) + a)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(e + f*x))**m, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e) + a)^m, x)