3.49 \(\int \frac{\tan ^m(c+d x) (A+B \tan (c+d x)+C \tan ^2(c+d x))}{\sqrt{a+b \tan (c+d x)}} \, dx\)

Optimal. Leaf size=328 \[ \frac{2 C \tan ^m(c+d x) \sqrt{a+b \tan (c+d x)} \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} \text{Hypergeometric2F1}\left (\frac{1}{2},-m,\frac{3}{2},\frac{b \tan (c+d x)}{a}+1\right )}{b d}-\frac{\left (\sqrt{-b^2} (A-C)+b B\right ) \tan ^m(c+d x) \sqrt{a+b \tan (c+d x)} \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} F_1\left (\frac{1}{2};1,-m;\frac{3}{2};\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}},\frac{b \tan (c+d x)}{a}+1\right )}{b d \left (a-\sqrt{-b^2}\right )}-\frac{\left (b B-\sqrt{-b^2} (A-C)\right ) \tan ^m(c+d x) \sqrt{a+b \tan (c+d x)} \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} F_1\left (\frac{1}{2};1,-m;\frac{3}{2};\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}},\frac{b \tan (c+d x)}{a}+1\right )}{b d \left (a+\sqrt{-b^2}\right )} \]

[Out]

-(((b*B + Sqrt[-b^2]*(A - C))*AppellF1[1/2, 1, -m, 3/2, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2]), 1 + (b*Tan[c +
d*x])/a]*Tan[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]])/(b*(a - Sqrt[-b^2])*d*(-((b*Tan[c + d*x])/a))^m)) - ((b*B -
Sqrt[-b^2]*(A - C))*AppellF1[1/2, 1, -m, 3/2, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2]), 1 + (b*Tan[c + d*x])/a]*T
an[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]])/(b*(a + Sqrt[-b^2])*d*(-((b*Tan[c + d*x])/a))^m) + (2*C*Hypergeometric
2F1[1/2, -m, 3/2, 1 + (b*Tan[c + d*x])/a]*Tan[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]])/(b*d*(-((b*Tan[c + d*x])/a)
)^m)

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Rubi [A]  time = 1.56086, antiderivative size = 328, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {3655, 6720, 1692, 246, 245, 430, 429} \[ -\frac{\left (\sqrt{-b^2} (A-C)+b B\right ) \tan ^m(c+d x) \sqrt{a+b \tan (c+d x)} \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} F_1\left (\frac{1}{2};1,-m;\frac{3}{2};\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}},\frac{b \tan (c+d x)}{a}+1\right )}{b d \left (a-\sqrt{-b^2}\right )}-\frac{\left (b B-\sqrt{-b^2} (A-C)\right ) \tan ^m(c+d x) \sqrt{a+b \tan (c+d x)} \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} F_1\left (\frac{1}{2};1,-m;\frac{3}{2};\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}},\frac{b \tan (c+d x)}{a}+1\right )}{b d \left (a+\sqrt{-b^2}\right )}+\frac{2 C \tan ^m(c+d x) \sqrt{a+b \tan (c+d x)} \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};\frac{b \tan (c+d x)}{a}+1\right )}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*B + Sqrt[-b^2]*(A - C))*AppellF1[1/2, 1, -m, 3/2, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2]), 1 + (b*Tan[c +
d*x])/a]*Tan[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]])/(b*(a - Sqrt[-b^2])*d*(-((b*Tan[c + d*x])/a))^m)) - ((b*B -
Sqrt[-b^2]*(A - C))*AppellF1[1/2, 1, -m, 3/2, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2]), 1 + (b*Tan[c + d*x])/a]*T
an[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]])/(b*(a + Sqrt[-b^2])*d*(-((b*Tan[c + d*x])/a))^m) + (2*C*Hypergeometric
2F1[1/2, -m, 3/2, 1 + (b*Tan[c + d*x])/a]*Tan[c + d*x]^m*Sqrt[a + b*Tan[c + d*x]])/(b*d*(-((b*Tan[c + d*x])/a)
)^m)

Rule 3655

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[((a + b*ff*x)^m*(c + d*ff*x)^n*(A + B*ff*x + C*ff^2*x^2))/(1 + ff^2*x^2), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

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

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^m \left (A+B x+C x^2\right )}{\sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{\left (\frac{-a+x^2}{b}\right )^m \left (A b^2+\left (a-x^2\right ) \left (-b B+C \left (a-x^2\right )\right )\right )}{a^2+b^2-2 a x^2+x^4} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=\frac{\left (2 \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname{Subst}\left (\int \frac{\left (-a+x^2\right )^m \left (A b^2+\left (a-x^2\right ) \left (-b B+C \left (a-x^2\right )\right )\right )}{a^2+b^2-2 a x^2+x^4} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=\frac{\left (2 \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname{Subst}\left (\int \left (C \left (-a+x^2\right )^m+\frac{\left (-a+x^2\right )^m \left (b (A b-a B-b C)+b B x^2\right )}{a^2+b^2-2 a x^2+x^4}\right ) \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=\frac{\left (2 \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname{Subst}\left (\int \frac{\left (-a+x^2\right )^m \left (b (A b-a B-b C)+b B x^2\right )}{a^2+b^2-2 a x^2+x^4} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}+\frac{\left (2 C \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname{Subst}\left (\int \left (-a+x^2\right )^m \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=\frac{\left (2 \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname{Subst}\left (\int \left (\frac{\left (b B-\sqrt{-b^2} (A-C)\right ) \left (-a+x^2\right )^m}{-2 a-2 \sqrt{-b^2}+2 x^2}+\frac{\left (b B+\sqrt{-b^2} (A-C)\right ) \left (-a+x^2\right )^m}{-2 a+2 \sqrt{-b^2}+2 x^2}\right ) \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}+\frac{\left (2 C \tan ^m(c+d x) \left (-\frac{b \tan (c+d x)}{a}\right )^{-m}\right ) \operatorname{Subst}\left (\int \left (1-\frac{x^2}{a}\right )^m \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=\frac{2 C \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};\frac{a+b \tan (c+d x)}{a}\right ) \tan ^m(c+d x) \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} \sqrt{a+b \tan (c+d x)}}{b d}+\frac{\left (2 \left (b B-\sqrt{-b^2} (A-C)\right ) \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname{Subst}\left (\int \frac{\left (-a+x^2\right )^m}{-2 a-2 \sqrt{-b^2}+2 x^2} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}+\frac{\left (2 \left (b B+\sqrt{-b^2} (A-C)\right ) \tan ^m(c+d x) (b \tan (c+d x))^{-m}\right ) \operatorname{Subst}\left (\int \frac{\left (-a+x^2\right )^m}{-2 a+2 \sqrt{-b^2}+2 x^2} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=\frac{2 C \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};\frac{a+b \tan (c+d x)}{a}\right ) \tan ^m(c+d x) \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} \sqrt{a+b \tan (c+d x)}}{b d}+\frac{\left (2 \left (b B-\sqrt{-b^2} (A-C)\right ) \tan ^m(c+d x) \left (-\frac{b \tan (c+d x)}{a}\right )^{-m}\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x^2}{a}\right )^m}{-2 a-2 \sqrt{-b^2}+2 x^2} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}+\frac{\left (2 \left (b B+\sqrt{-b^2} (A-C)\right ) \tan ^m(c+d x) \left (-\frac{b \tan (c+d x)}{a}\right )^{-m}\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x^2}{a}\right )^m}{-2 a+2 \sqrt{-b^2}+2 x^2} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=-\frac{\left (b B+\sqrt{-b^2} (A-C)\right ) F_1\left (\frac{1}{2};1,-m;\frac{3}{2};\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}},\frac{a+b \tan (c+d x)}{a}\right ) \tan ^m(c+d x) \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} \sqrt{a+b \tan (c+d x)}}{b \left (a-\sqrt{-b^2}\right ) d}-\frac{\left (b B-\sqrt{-b^2} (A-C)\right ) F_1\left (\frac{1}{2};1,-m;\frac{3}{2};\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}},\frac{a+b \tan (c+d x)}{a}\right ) \tan ^m(c+d x) \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} \sqrt{a+b \tan (c+d x)}}{b \left (a+\sqrt{-b^2}\right ) d}+\frac{2 C \, _2F_1\left (\frac{1}{2},-m;\frac{3}{2};\frac{a+b \tan (c+d x)}{a}\right ) \tan ^m(c+d x) \left (-\frac{b \tan (c+d x)}{a}\right )^{-m} \sqrt{a+b \tan (c+d x)}}{b d}\\ \end{align*}

Mathematica [F]  time = 27.1726, size = 0, normalized size = 0. \[ \int \frac{\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt{a+b \tan (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.6, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( A+B\tan \left ( dx+c \right ) +C \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{a+b\tan \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^m*(A+B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^(1/2),x)

[Out]

int(tan(d*x+c)^m*(A+B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^(1/2),x)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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 )}}{\sqrt{a + b \tan{\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*(A+B*tan(d*x+c)+C*tan(d*x+c)**2)/(a+b*tan(d*x+c))**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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