3.46 \(\int \tan ^m(c+d x) (b \tan (c+d x))^n (A+B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=154 \[ \frac{(A-C) \tan ^{m+1}(c+d x) (b \tan (c+d x))^n \text{Hypergeometric2F1}\left (1,\frac{1}{2} (m+n+1),\frac{1}{2} (m+n+3),-\tan ^2(c+d x)\right )}{d (m+n+1)}+\frac{B \tan ^{m+2}(c+d x) (b \tan (c+d x))^n \text{Hypergeometric2F1}\left (1,\frac{1}{2} (m+n+2),\frac{1}{2} (m+n+4),-\tan ^2(c+d x)\right )}{d (m+n+2)}+\frac{C \tan ^{m+1}(c+d x) (b \tan (c+d x))^n}{d (m+n+1)} \]

[Out]

(C*Tan[c + d*x]^(1 + m)*(b*Tan[c + d*x])^n)/(d*(1 + m + n)) + ((A - C)*Hypergeometric2F1[1, (1 + m + n)/2, (3
+ m + n)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m)*(b*Tan[c + d*x])^n)/(d*(1 + m + n)) + (B*Hypergeometric2F1[1
, (2 + m + n)/2, (4 + m + n)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m)*(b*Tan[c + d*x])^n)/(d*(2 + m + n))

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Rubi [A]  time = 0.137775, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {20, 3630, 3538, 3476, 364} \[ \frac{(A-C) \tan ^{m+1}(c+d x) (b \tan (c+d x))^n \, _2F_1\left (1,\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+3);-\tan ^2(c+d x)\right )}{d (m+n+1)}+\frac{B \tan ^{m+2}(c+d x) (b \tan (c+d x))^n \, _2F_1\left (1,\frac{1}{2} (m+n+2);\frac{1}{2} (m+n+4);-\tan ^2(c+d x)\right )}{d (m+n+2)}+\frac{C \tan ^{m+1}(c+d x) (b \tan (c+d x))^n}{d (m+n+1)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^m*(b*Tan[c + d*x])^n*(A + B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

(C*Tan[c + d*x]^(1 + m)*(b*Tan[c + d*x])^n)/(d*(1 + m + n)) + ((A - C)*Hypergeometric2F1[1, (1 + m + n)/2, (3
+ m + n)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m)*(b*Tan[c + d*x])^n)/(d*(1 + m + n)) + (B*Hypergeometric2F1[1
, (2 + m + n)/2, (4 + m + n)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m)*(b*Tan[c + d*x])^n)/(d*(2 + m + n))

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

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 3538

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]

Rule 3476

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \tan ^m(c+d x) (b \tan (c+d x))^n \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx &=\left (\tan ^{-n}(c+d x) (b \tan (c+d x))^n\right ) \int \tan ^{m+n}(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx\\ &=\frac{C \tan ^{1+m}(c+d x) (b \tan (c+d x))^n}{d (1+m+n)}+\left (\tan ^{-n}(c+d x) (b \tan (c+d x))^n\right ) \int \tan ^{m+n}(c+d x) (A-C+B \tan (c+d x)) \, dx\\ &=\frac{C \tan ^{1+m}(c+d x) (b \tan (c+d x))^n}{d (1+m+n)}+\left (B \tan ^{-n}(c+d x) (b \tan (c+d x))^n\right ) \int \tan ^{1+m+n}(c+d x) \, dx+\left ((A-C) \tan ^{-n}(c+d x) (b \tan (c+d x))^n\right ) \int \tan ^{m+n}(c+d x) \, dx\\ &=\frac{C \tan ^{1+m}(c+d x) (b \tan (c+d x))^n}{d (1+m+n)}+\frac{\left (B \tan ^{-n}(c+d x) (b \tan (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{x^{1+m+n}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left ((A-C) \tan ^{-n}(c+d x) (b \tan (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{x^{m+n}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{C \tan ^{1+m}(c+d x) (b \tan (c+d x))^n}{d (1+m+n)}+\frac{(A-C) \, _2F_1\left (1,\frac{1}{2} (1+m+n);\frac{1}{2} (3+m+n);-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x) (b \tan (c+d x))^n}{d (1+m+n)}+\frac{B \, _2F_1\left (1,\frac{1}{2} (2+m+n);\frac{1}{2} (4+m+n);-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x) (b \tan (c+d x))^n}{d (2+m+n)}\\ \end{align*}

Mathematica [A]  time = 0.376412, size = 115, normalized size = 0.75 \[ \frac{\tan ^{m+1}(c+d x) (b \tan (c+d x))^n \left (\frac{(A-C) \text{Hypergeometric2F1}\left (1,\frac{1}{2} (m+n+1),\frac{1}{2} (m+n+3),-\tan ^2(c+d x)\right )}{m+n+1}+\frac{B \tan (c+d x) \text{Hypergeometric2F1}\left (1,\frac{1}{2} (m+n+2),\frac{1}{2} (m+n+4),-\tan ^2(c+d x)\right )}{m+n+2}+\frac{C}{m+n+1}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^m*(b*Tan[c + d*x])^n*(A + B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

(Tan[c + d*x]^(1 + m)*(b*Tan[c + d*x])^n*(C/(1 + m + n) + ((A - C)*Hypergeometric2F1[1, (1 + m + n)/2, (3 + m
+ n)/2, -Tan[c + d*x]^2])/(1 + m + n) + (B*Hypergeometric2F1[1, (2 + m + n)/2, (4 + m + n)/2, -Tan[c + d*x]^2]
*Tan[c + d*x])/(2 + m + n)))/d

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( b\tan \left ( dx+c \right ) \right ) ^{n} \left ( A+B\tan \left ( dx+c \right ) +C \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^m*(b*tan(d*x+c))^n*(A+B*tan(d*x+c)+C*tan(d*x+c)^2),x)

[Out]

int(tan(d*x+c)^m*(b*tan(d*x+c))^n*(A+B*tan(d*x+c)+C*tan(d*x+c)^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \tan \left (d x + c\right )^{2} + B \tan \left (d x + c\right ) + A\right )} \left (b \tan \left (d x + c\right )\right )^{n} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(b*tan(d*x+c))^n*(A+B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*tan(d*x + c)^2 + B*tan(d*x + c) + A)*(b*tan(d*x + c))^n*tan(d*x + c)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \tan \left (d x + c\right )^{2} + B \tan \left (d x + c\right ) + A\right )} \left (b \tan \left (d x + c\right )\right )^{n} \tan \left (d x + c\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(b*tan(d*x+c))^n*(A+B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*tan(d*x + c)^2 + B*tan(d*x + c) + A)*(b*tan(d*x + c))^n*tan(d*x + c)^m, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan{\left (c + d x \right )}\right )^{n} \left (A + B \tan{\left (c + d x \right )} + C \tan ^{2}{\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**m*(b*tan(d*x+c))**n*(A+B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

Integral((b*tan(c + d*x))**n*(A + B*tan(c + d*x) + C*tan(c + d*x)**2)*tan(c + d*x)**m, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \tan \left (d x + c\right )^{2} + B \tan \left (d x + c\right ) + A\right )} \left (b \tan \left (d x + c\right )\right )^{n} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(b*tan(d*x+c))^n*(A+B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*tan(d*x + c)^2 + B*tan(d*x + c) + A)*(b*tan(d*x + c))^n*tan(d*x + c)^m, x)