3.168 \(\int (a+b \tan (e+f x))^m (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=178 \[ \frac{(A-i B-C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}+\frac{(i A-B-i C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)}+\frac{C (a+b \tan (e+f x))^{m+1}}{b f (m+1)} \]

[Out]

(C*(a + b*Tan[e + f*x])^(1 + m))/(b*f*(1 + m)) + ((A - I*B - C)*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*A - B - I*C)*Hypergeometric2F
1[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)*f*(1 + m))

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Rubi [A]  time = 0.184451, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {3630, 3539, 3537, 68} \[ \frac{(A-i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}+\frac{(i A-B-i C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)}+\frac{C (a+b \tan (e+f x))^{m+1}}{b f (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^m*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]

[Out]

(C*(a + b*Tan[e + f*x])^(1 + m))/(b*f*(1 + m)) + ((A - I*B - C)*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*A - B - I*C)*Hypergeometric2F
1[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)*f*(1 + m))

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

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 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac{C (a+b \tan (e+f x))^{1+m}}{b f (1+m)}+\int (a+b \tan (e+f x))^m (A-C+B \tan (e+f x)) \, dx\\ &=\frac{C (a+b \tan (e+f x))^{1+m}}{b f (1+m)}+\frac{1}{2} (A-i B-C) \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx+\frac{1}{2} (A+i B-C) \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx\\ &=\frac{C (a+b \tan (e+f x))^{1+m}}{b f (1+m)}+\frac{(i A+B-i C) \operatorname{Subst}\left (\int \frac{(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 f}+\frac{(i (-A-i B+C)) \operatorname{Subst}\left (\int \frac{(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 f}\\ &=\frac{C (a+b \tan (e+f x))^{1+m}}{b f (1+m)}-\frac{(i A+B-i C) \, _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{(i A-B-i C) \, _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)}\\ \end{align*}

Mathematica [A]  time = 0.209102, size = 135, normalized size = 0.76 \[ \frac{(a+b \tan (e+f x))^{m+1} \left (-\frac{i (A-i B-C) \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a-i b}\right )}{a-i b}+\frac{i (A+i B-C) \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a+i b}\right )}{a+i b}+\frac{2 C}{b}\right )}{2 f (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^m*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]

[Out]

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

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Maple [F]  time = 0.39, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m} \left ( A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^m*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x)

[Out]

int((a+b*tan(f*x+e))^m*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\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*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(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 (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\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*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

integral((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(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 (A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\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*(A+B*tan(f*x+e)+C*tan(f*x+e)**2),x)

[Out]

Integral((a + b*tan(e + f*x))**m*(A + B*tan(e + f*x) + C*tan(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\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*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="giac")

[Out]

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(b*tan(f*x + e) + a)^m, x)