3.1307 \(\int (a+b \tan (e+f x))^m (c+d \tan (e+f x)) \, dx\)

Optimal. Leaf size=143 \[ \frac{(c-i d) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}+\frac{(-d+i c) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)} \]

[Out]

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

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Rubi [A]  time = 0.134209, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3539, 3537, 68} \[ \frac{(c-i d) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}+\frac{(-d+i c) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3539

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

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

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

Mathematica [A]  time = 0.170099, size = 120, normalized size = 0.84 \[ \frac{i (a+b \tan (e+f x))^{m+1} \left (\frac{(c+i d) \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{a+i b}-\frac{(c-i d) \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{a-i b}\right )}{2 f (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.496, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m} \left ( c+d\tan \left ( fx+e \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \tan \left (f x + e\right ) + c\right )}{\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*(c+d*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*tan(f*x + e) + c)*(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 (d \tan \left (f x + e\right ) + c\right )}{\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*(c+d*tan(f*x+e)),x, algorithm="fricas")

[Out]

integral((d*tan(f*x + e) + c)*(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} \left (c + d \tan{\left (e + f x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(e + f*x))**m*(c + d*tan(e + f*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d \tan \left (f x + e\right ) + c\right )}{\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*(c+d*tan(f*x+e)),x, algorithm="giac")

[Out]

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