3.408 \(\int \frac{(d \tan (e+f x))^m}{a+b \sqrt{c \tan (e+f x)}} \, dx\)

Optimal. Leaf size=460 \[ -\frac{b \left (a^2-b^2 \sqrt{-c^2}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,\frac{1}{2} (2 m+3),\frac{1}{2} (2 m+5),-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{c f (2 m+3) \left (a^4+b^4 c^2\right )}-\frac{b \left (a^2+b^2 \sqrt{-c^2}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,\frac{1}{2} (2 m+3),\frac{1}{2} (2 m+5),\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{c f (2 m+3) \left (a^4+b^4 c^2\right )}+\frac{a \left (a^2-b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1) \left (a^4+b^4 c^2\right )}+\frac{a \left (a^2+b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1) \left (a^4+b^4 c^2\right )}+\frac{b^4 c^2 \tan (e+f x) (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,2 (m+1),2 m+3,-\frac{b \sqrt{c \tan (e+f x)}}{a}\right )}{a f (m+1) \left (a^4+b^4 c^2\right )} \]

[Out]

(a*(a^2 - b^2*Sqrt[-c^2])*Hypergeometric2F1[1, 1 + m, 2 + m, -((c*Tan[e + f*x])/Sqrt[-c^2])]*Tan[e + f*x]*(d*T
an[e + f*x])^m)/(2*(a^4 + b^4*c^2)*f*(1 + m)) + (a*(a^2 + b^2*Sqrt[-c^2])*Hypergeometric2F1[1, 1 + m, 2 + m, (
c*Tan[e + f*x])/Sqrt[-c^2]]*Tan[e + f*x]*(d*Tan[e + f*x])^m)/(2*(a^4 + b^4*c^2)*f*(1 + m)) + (b^4*c^2*Hypergeo
metric2F1[1, 2*(1 + m), 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x]])/a)]*Tan[e + f*x]*(d*Tan[e + f*x])^m)/(a*(a^4 + b^4
*c^2)*f*(1 + m)) - (b*(a^2 - b^2*Sqrt[-c^2])*Hypergeometric2F1[1, (3 + 2*m)/2, (5 + 2*m)/2, -((c*Tan[e + f*x])
/Sqrt[-c^2])]*(c*Tan[e + f*x])^(3/2)*(d*Tan[e + f*x])^m)/(c*(a^4 + b^4*c^2)*f*(3 + 2*m)) - (b*(a^2 + b^2*Sqrt[
-c^2])*Hypergeometric2F1[1, (3 + 2*m)/2, (5 + 2*m)/2, (c*Tan[e + f*x])/Sqrt[-c^2]]*(c*Tan[e + f*x])^(3/2)*(d*T
an[e + f*x])^m)/(c*(a^4 + b^4*c^2)*f*(3 + 2*m))

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Rubi [A]  time = 1.28252, antiderivative size = 460, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {3670, 15, 6725, 64, 1831, 1286, 364} \[ -\frac{b \left (a^2-b^2 \sqrt{-c^2}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \, _2F_1\left (1,\frac{1}{2} (2 m+3);\frac{1}{2} (2 m+5);-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{c f (2 m+3) \left (a^4+b^4 c^2\right )}-\frac{b \left (a^2+b^2 \sqrt{-c^2}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \, _2F_1\left (1,\frac{1}{2} (2 m+3);\frac{1}{2} (2 m+5);\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{c f (2 m+3) \left (a^4+b^4 c^2\right )}+\frac{a \left (a^2-b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,m+1;m+2;-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1) \left (a^4+b^4 c^2\right )}+\frac{a \left (a^2+b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,m+1;m+2;\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1) \left (a^4+b^4 c^2\right )}+\frac{b^4 c^2 \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,2 (m+1);2 m+3;-\frac{b \sqrt{c \tan (e+f x)}}{a}\right )}{a f (m+1) \left (a^4+b^4 c^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^2 - b^2*Sqrt[-c^2])*Hypergeometric2F1[1, 1 + m, 2 + m, -((c*Tan[e + f*x])/Sqrt[-c^2])]*Tan[e + f*x]*(d*T
an[e + f*x])^m)/(2*(a^4 + b^4*c^2)*f*(1 + m)) + (a*(a^2 + b^2*Sqrt[-c^2])*Hypergeometric2F1[1, 1 + m, 2 + m, (
c*Tan[e + f*x])/Sqrt[-c^2]]*Tan[e + f*x]*(d*Tan[e + f*x])^m)/(2*(a^4 + b^4*c^2)*f*(1 + m)) + (b^4*c^2*Hypergeo
metric2F1[1, 2*(1 + m), 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x]])/a)]*Tan[e + f*x]*(d*Tan[e + f*x])^m)/(a*(a^4 + b^4
*c^2)*f*(1 + m)) - (b*(a^2 - b^2*Sqrt[-c^2])*Hypergeometric2F1[1, (3 + 2*m)/2, (5 + 2*m)/2, -((c*Tan[e + f*x])
/Sqrt[-c^2])]*(c*Tan[e + f*x])^(3/2)*(d*Tan[e + f*x])^m)/(c*(a^4 + b^4*c^2)*f*(3 + 2*m)) - (b*(a^2 + b^2*Sqrt[
-c^2])*Hypergeometric2F1[1, (3 + 2*m)/2, (5 + 2*m)/2, (c*Tan[e + f*x])/Sqrt[-c^2]]*(c*Tan[e + f*x])^(3/2)*(d*T
an[e + f*x])^m)/(c*(a^4 + b^4*c^2)*f*(3 + 2*m))

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 15

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rule 1831

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[((c*x)^(m + ii)*(Coeff[Pq,
x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2)))/(c^ii*(a + b*x^n)), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; Fr
eeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n

Rule 1286

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2))/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, -
Dist[e/2 + (c*d)/(2*q), Int[(f*x)^m/(q - c*x^2), x], x] + Dist[e/2 - (c*d)/(2*q), Int[(f*x)^m/(q + c*x^2), x],
x]] /; FreeQ[{a, c, d, e, f, m}, x]

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 \frac{(d \tan (e+f x))^m}{a+b \sqrt{c \tan (e+f x)}} \, dx &=\frac{c \operatorname{Subst}\left (\int \frac{\left (\frac{d x}{c}\right )^m}{\left (a+b \sqrt{x}\right ) \left (c^2+x^2\right )} \, dx,x,c \tan (e+f x)\right )}{f}\\ &=\frac{(2 c) \operatorname{Subst}\left (\int \frac{x \left (\frac{d x^2}{c}\right )^m}{(a+b x) \left (c^2+x^4\right )} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}\\ &=\frac{\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m}}{(a+b x) \left (c^2+x^4\right )} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}\\ &=\frac{\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \left (\frac{b^4 x^{1+2 m}}{\left (a^4+b^4 c^2\right ) (a+b x)}+\frac{x^{1+2 m} \left (a^3-a^2 b x+a b^2 x^2-b^3 x^3\right )}{\left (a^4+b^4 c^2\right ) \left (c^2+x^4\right )}\right ) \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}\\ &=\frac{\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m} \left (a^3-a^2 b x+a b^2 x^2-b^3 x^3\right )}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}+\frac{\left (2 b^4 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m}}{a+b x} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}\\ &=\frac{b^4 c^2 \, _2F_1\left (1,2 (1+m);3+2 m;-\frac{b \sqrt{c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a \left (a^4+b^4 c^2\right ) f (1+m)}+\frac{\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \left (\frac{x^{1+2 m} \left (a^3+a b^2 x^2\right )}{c^2+x^4}+\frac{x^{2+2 m} \left (-a^2 b-b^3 x^2\right )}{c^2+x^4}\right ) \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}\\ &=\frac{b^4 c^2 \, _2F_1\left (1,2 (1+m);3+2 m;-\frac{b \sqrt{c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a \left (a^4+b^4 c^2\right ) f (1+m)}+\frac{\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m} \left (a^3+a b^2 x^2\right )}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}+\frac{\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{2+2 m} \left (-a^2 b-b^3 x^2\right )}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}\\ &=\frac{b^4 c^2 \, _2F_1\left (1,2 (1+m);3+2 m;-\frac{b \sqrt{c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a \left (a^4+b^4 c^2\right ) f (1+m)}+\frac{\left (a c \left (b^2-\frac{a^2}{\sqrt{-c^2}}\right ) (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m}}{\sqrt{-c^2}+x^2} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}-\frac{\left (b c \left (b^2-\frac{a^2}{\sqrt{-c^2}}\right ) (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{2+2 m}}{\sqrt{-c^2}+x^2} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}-\frac{\left (a c \left (b^2+\frac{a^2}{\sqrt{-c^2}}\right ) (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m}}{\sqrt{-c^2}-x^2} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}+\frac{\left (b c \left (b^2+\frac{a^2}{\sqrt{-c^2}}\right ) (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{2+2 m}}{\sqrt{-c^2}-x^2} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}\\ &=\frac{a \left (a^2-b^2 \sqrt{-c^2}\right ) \, _2F_1\left (1,1+m;2+m;-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 \left (a^4+b^4 c^2\right ) f (1+m)}+\frac{a \left (a^2+b^2 \sqrt{-c^2}\right ) \, _2F_1\left (1,1+m;2+m;\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 \left (a^4+b^4 c^2\right ) f (1+m)}+\frac{b^4 c^2 \, _2F_1\left (1,2 (1+m);3+2 m;-\frac{b \sqrt{c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a \left (a^4+b^4 c^2\right ) f (1+m)}-\frac{b \left (a^2-b^2 \sqrt{-c^2}\right ) \, _2F_1\left (1,\frac{1}{2} (3+2 m);\frac{1}{2} (5+2 m);-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right ) f (3+2 m)}-\frac{b \left (a^2+b^2 \sqrt{-c^2}\right ) \, _2F_1\left (1,\frac{1}{2} (3+2 m);\frac{1}{2} (5+2 m);\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right ) f (3+2 m)}\\ \end{align*}

Mathematica [A]  time = 6.27851, size = 385, normalized size = 0.84 \[ \frac{2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m \left (\frac{a^3 (c \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(e+f x)\right )}{2 c^2 (m+1) \left (a^4+b^4 c^2\right )}+\frac{a b^2 (c \tan (e+f x))^{m+2} \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(e+f x)\right )}{2 c^2 (m+2) \left (a^4+b^4 c^2\right )}-\frac{a^2 b (c \tan (e+f x))^{\frac{1}{2} (2 m+3)} \text{Hypergeometric2F1}\left (1,\frac{1}{4} (2 m+3),\frac{1}{4} (2 m+7),-\tan ^2(e+f x)\right )}{c^2 (2 m+3) \left (a^4+b^4 c^2\right )}-\frac{b^3 (c \tan (e+f x))^{\frac{1}{2} (2 m+5)} \text{Hypergeometric2F1}\left (1,\frac{1}{4} (2 m+5),\frac{1}{4} (2 m+9),-\tan ^2(e+f x)\right )}{c^2 (2 m+5) \left (a^4+b^4 c^2\right )}+\frac{b^4 (c \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,2 (m+1),2 m+3,-\frac{b \sqrt{c \tan (e+f x)}}{a}\right )}{2 a (m+1) \left (a^4+b^4 c^2\right )}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*(d*Tan[e + f*x])^m*((a^3*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[e + f*x]^2]*(c*Tan[e + f*x])^(1
+ m))/(2*c^2*(a^4 + b^4*c^2)*(1 + m)) + (b^4*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x]
])/a)]*(c*Tan[e + f*x])^(1 + m))/(2*a*(a^4 + b^4*c^2)*(1 + m)) + (a*b^2*Hypergeometric2F1[1, (2 + m)/2, (4 + m
)/2, -Tan[e + f*x]^2]*(c*Tan[e + f*x])^(2 + m))/(2*c^2*(a^4 + b^4*c^2)*(2 + m)) - (a^2*b*Hypergeometric2F1[1,
(3 + 2*m)/4, (7 + 2*m)/4, -Tan[e + f*x]^2]*(c*Tan[e + f*x])^((3 + 2*m)/2))/(c^2*(a^4 + b^4*c^2)*(3 + 2*m)) - (
b^3*Hypergeometric2F1[1, (5 + 2*m)/4, (9 + 2*m)/4, -Tan[e + f*x]^2]*(c*Tan[e + f*x])^((5 + 2*m)/2))/(c^2*(a^4
+ b^4*c^2)*(5 + 2*m))))/(f*(c*Tan[e + f*x])^m)

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Maple [F]  time = 0.201, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d\tan \left ( fx+e \right ) \right ) ^{m} \left ( a+b\sqrt{c\tan \left ( fx+e \right ) } \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^m/(a+b*(c*tan(f*x+e))^(1/2)),x)

[Out]

int((d*tan(f*x+e))^m/(a+b*(c*tan(f*x+e))^(1/2)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((sqrt(c*tan(f*x + e))*(d*tan(f*x + e))^m*b - (d*tan(f*x + e))^m*a)/(b^2*c*tan(f*x + e) - a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \tan{\left (e + f x \right )}\right )^{m}}{a + b \sqrt{c \tan{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**m/(a+b*(c*tan(f*x+e))**(1/2)),x)

[Out]

Integral((d*tan(e + f*x))**m/(a + b*sqrt(c*tan(e + f*x))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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