3.1305 \(\int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=214 \[ \frac{d^2 (3 b c-a d) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+1)}+\frac{d^3 (a+b \tan (e+f x))^{m+2}}{b^2 f (m+2)}+\frac{(d+i c)^3 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (a-i b)}-\frac{(-d+i c)^3 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)} \]

[Out]

(d^2*(3*b*c - a*d)*(a + b*Tan[e + f*x])^(1 + m))/(b^2*f*(1 + m)) + ((I*c + d)^3*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)*f*(1 + m)) - ((I*c - d)^3*Hype
rgeometric2F1[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)) + (d^3*(a + b*Tan[e + f*x])^(2 + m))/(b^2*f*(2 + m))

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Rubi [A]  time = 0.474896, antiderivative size = 234, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3566, 3630, 3539, 3537, 68} \[ -\frac{d^2 (a d-b c (2 m+5)) (a+b \tan (e+f x))^{m+1}}{b^2 f (m+1) (m+2)}+\frac{d^2 (c+d \tan (e+f x)) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}+\frac{(c-i d)^3 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}-\frac{(-d+i c)^3 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^3,x]

[Out]

-((d^2*(a*d - b*c*(5 + 2*m))*(a + b*Tan[e + f*x])^(1 + m))/(b^2*f*(1 + m)*(2 + m))) + ((c - I*d)^3*Hypergeomet
ric2F1[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)^3*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)*f*(1 + m)) + (d^2*(a + b*Tan[e + f*x])^(1 + m)*(c + d*Tan[e + f*x]))/(b*f*(2 + m))

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

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 (c+d \tan (e+f x))^3 \, dx &=\frac{d^2 (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))}{b f (2+m)}+\frac{\int (a+b \tan (e+f x))^m \left (b c^3 (2+m)-d^2 (a d+b c (1+m))+b d \left (3 c^2-d^2\right ) (2+m) \tan (e+f x)-d^2 (a d-b c (5+2 m)) \tan ^2(e+f x)\right ) \, dx}{b (2+m)}\\ &=-\frac{d^2 (a d-b c (5+2 m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{d^2 (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))}{b f (2+m)}+\frac{\int (a+b \tan (e+f x))^m \left (b c \left (c^2-3 d^2\right ) (2+m)+b d \left (3 c^2-d^2\right ) (2+m) \tan (e+f x)\right ) \, dx}{b (2+m)}\\ &=-\frac{d^2 (a d-b c (5+2 m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{d^2 (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))}{b f (2+m)}+\frac{1}{2} (c-i d)^3 \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx+\frac{1}{2} (c+i d)^3 \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx\\ &=-\frac{d^2 (a d-b c (5+2 m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{d^2 (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))}{b f (2+m)}+\frac{(i c-d)^3 \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)^3 \operatorname{Subst}\left (\int \frac{(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 f}\\ &=-\frac{d^2 (a d-b c (5+2 m)) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m) (2+m)}+\frac{(i c+d)^3 \, _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 c-d)^3 \, _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{d^2 (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))}{b f (2+m)}\\ \end{align*}

Mathematica [A]  time = 1.90874, size = 189, normalized size = 0.88 \[ \frac{(a+b \tan (e+f x))^{m+1} \left (\frac{2 d^2 (b c (2 m+5)-a d)}{b (m+1)}-\frac{i b (m+2) (c-i d)^3 \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{(m+1) (a-i b)}+\frac{i b (m+2) (c+i d)^3 \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{(m+1) (a+i b)}+2 d^2 (c+d \tan (e+f x))\right )}{2 b f (m+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^3,x]

[Out]

((a + b*Tan[e + f*x])^(1 + m)*((2*d^2*(-(a*d) + b*c*(5 + 2*m)))/(b*(1 + m)) - (I*b*(c - I*d)^3*(2 + m)*Hyperge
ometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - I*b)])/((a - I*b)*(1 + m)) + (I*b*(c + I*d)^3*(2 + m)*Hy
pergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a + I*b)])/((a + I*b)*(1 + m)) + 2*d^2*(c + d*Tan[e + f
*x])))/(2*b*f*(2 + m))

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Maple [F]  time = 0.327, 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 ) ^{3}\, 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))^3,x)

[Out]

int((a+b*tan(f*x+e))^m*(c+d*tan(f*x+e))^3,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 )}^{3}{\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))^3,x, algorithm="maxima")

[Out]

integrate((d*tan(f*x + e) + c)^3*(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^{3} \tan \left (f x + e\right )^{3} + 3 \, c d^{2} \tan \left (f x + e\right )^{2} + 3 \, c^{2} d \tan \left (f x + e\right ) + c^{3}\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))^3,x, algorithm="fricas")

[Out]

integral((d^3*tan(f*x + e)^3 + 3*c*d^2*tan(f*x + e)^2 + 3*c^2*d*tan(f*x + e) + c^3)*(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 )^{3}\, 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))**3,x)

[Out]

Integral((a + b*tan(e + f*x))**m*(c + d*tan(e + f*x))**3, 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 )}^{3}{\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))^3,x, algorithm="giac")

[Out]

integrate((d*tan(f*x + e) + c)^3*(b*tan(f*x + e) + a)^m, x)