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]

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

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Rubi [A]  time = 0.13, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.22, 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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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d \tan \left (f x + e\right ) + c\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d \tan \left (f x + e\right ) + c\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

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)

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maple [F]  time = 1.65, size = 0, normalized size = 0.00 \[ \int \left (a +b \tan \left (f x +e \right )\right )^{m} \left (c +d \tan \left (f x +e \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d \tan \left (f x + e\right ) + c\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (e + f x \right )}\right )^{m} \left (c + d \tan {\left (e + f x \right )}\right )\, dx \]

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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