3.5.9 \(\int \frac {(d \tan (e+f x))^m}{(a+b \sqrt {c \tan (e+f x)})^2} \, dx\) [409]

Optimal. Leaf size=617 \[ \frac {\left (a^6-3 a^2 b^4 c^2-3 a^4 b^2 \sqrt {-c^2}-b^6 \left (-c^2\right )^{3/2}\right ) \, _2F_1\left (1,1+m;2+m;-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 \left (a^4+b^4 c^2\right )^2 f (1+m)}+\frac {\left (a^6-3 a^2 b^4 c^2+3 a^4 b^2 \sqrt {-c^2}+b^6 \left (-c^2\right )^{3/2}\right ) \, _2F_1\left (1,1+m;2+m;\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 \left (a^4+b^4 c^2\right )^2 f (1+m)}+\frac {4 a^2 b^4 c^2 \, _2F_1\left (1,2 (1+m);3+2 m;-\frac {b \sqrt {c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{\left (a^4+b^4 c^2\right )^2 f (1+m)}+\frac {b^4 c^2 \, _2F_1\left (2,2 (1+m);3+2 m;-\frac {b \sqrt {c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a^2 \left (a^4+b^4 c^2\right ) f (1+m)}-\frac {2 a b \left (a^4-b^4 c^2-2 a^2 b^2 \sqrt {-c^2}\right ) \, _2F_1\left (1,\frac {1}{2} (3+2 m);\frac {1}{2} (5+2 m);-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right )^2 f (3+2 m)}-\frac {2 a b \left (a^4-b^4 c^2+2 a^2 b^2 \sqrt {-c^2}\right ) \, _2F_1\left (1,\frac {1}{2} (3+2 m);\frac {1}{2} (5+2 m);\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right )^2 f (3+2 m)} \]

[Out]

4*a^2*b^4*c^2*hypergeom([1, 2+2*m],[3+2*m],-b*(c*tan(f*x+e))^(1/2)/a)*tan(f*x+e)*(d*tan(f*x+e))^m/(b^4*c^2+a^4
)^2/f/(1+m)+b^4*c^2*hypergeom([2, 2+2*m],[3+2*m],-b*(c*tan(f*x+e))^(1/2)/a)*tan(f*x+e)*(d*tan(f*x+e))^m/a^2/(b
^4*c^2+a^4)/f/(1+m)+1/2*hypergeom([1, 1+m],[2+m],-c*tan(f*x+e)/(-c^2)^(1/2))*(a^6-3*a^2*b^4*c^2-b^6*(-c^2)^(3/
2)-3*a^4*b^2*(-c^2)^(1/2))*tan(f*x+e)*(d*tan(f*x+e))^m/(b^4*c^2+a^4)^2/f/(1+m)+1/2*hypergeom([1, 1+m],[2+m],c*
tan(f*x+e)/(-c^2)^(1/2))*(a^6-3*a^2*b^4*c^2+b^6*(-c^2)^(3/2)+3*a^4*b^2*(-c^2)^(1/2))*tan(f*x+e)*(d*tan(f*x+e))
^m/(b^4*c^2+a^4)^2/f/(1+m)-2*a*b*hypergeom([1, 3/2+m],[5/2+m],-c*tan(f*x+e)/(-c^2)^(1/2))*(a^4-b^4*c^2-2*a^2*b
^2*(-c^2)^(1/2))*(c*tan(f*x+e))^(3/2)*(d*tan(f*x+e))^m/c/(b^4*c^2+a^4)^2/f/(3+2*m)-2*a*b*hypergeom([1, 3/2+m],
[5/2+m],c*tan(f*x+e)/(-c^2)^(1/2))*(a^4-b^4*c^2+2*a^2*b^2*(-c^2)^(1/2))*(c*tan(f*x+e))^(3/2)*(d*tan(f*x+e))^m/
c/(b^4*c^2+a^4)^2/f/(3+2*m)

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Rubi [A]
time = 1.07, antiderivative size = 617, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3751, 15, 6857, 66, 1845, 1300, 371} \begin {gather*} \frac {4 a^2 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 )}{f (m+1) \left (a^4+b^4 c^2\right )^2}+\frac {b^4 c^2 \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (2,2 (m+1);2 m+3;-\frac {b \sqrt {c \tan (e+f x)}}{a}\right )}{a^2 f (m+1) \left (a^4+b^4 c^2\right )}-\frac {2 a b \left (a^4-2 a^2 b^2 \sqrt {-c^2}-b^4 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 )^2}-\frac {2 a b \left (a^4+2 a^2 b^2 \sqrt {-c^2}-b^4 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 )^2}+\frac {\left (a^6-3 a^4 b^2 \sqrt {-c^2}-3 a^2 b^4 c^2-b^6 \left (-c^2\right )^{3/2}\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 )^2}+\frac {\left (a^6+3 a^4 b^2 \sqrt {-c^2}-3 a^2 b^4 c^2+b^6 \left (-c^2\right )^{3/2}\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 )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^m/(a + b*Sqrt[c*Tan[e + f*x]])^2,x]

[Out]

((a^6 - 3*a^2*b^4*c^2 - 3*a^4*b^2*Sqrt[-c^2] - b^6*(-c^2)^(3/2))*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)^2*f*(1 + m)) + ((a^6 - 3*a^2*b^4*c^2
+ 3*a^4*b^2*Sqrt[-c^2] + b^6*(-c^2)^(3/2))*Hypergeometric2F1[1, 1 + m, 2 + m, (c*Tan[e + f*x])/Sqrt[-c^2]]*Ta
n[e + f*x]*(d*Tan[e + f*x])^m)/(2*(a^4 + b^4*c^2)^2*f*(1 + m)) + (4*a^2*b^4*c^2*Hypergeometric2F1[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^4 + b^4*c^2)^2*f*(1 + m)) + (b^
4*c^2*Hypergeometric2F1[2, 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^2*(a^4 + b^4*c^2)*f*(1 + m)) - (2*a*b*(a^4 - b^4*c^2 - 2*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
)^2*f*(3 + 2*m)) - (2*a*b*(a^4 - b^4*c^2 + 2*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)^2*f*(3 + 2*m))

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 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1300

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2))/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, D
ist[-(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 1845

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 3751

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 6857

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]

Rubi steps

\begin {align*} \int \frac {(d \tan (e+f x))^m}{\left (a+b \sqrt {c \tan (e+f x)}\right )^2} \, dx &=\frac {c \text {Subst}\left (\int \frac {\left (\frac {d x}{c}\right )^m}{\left (a+b \sqrt {x}\right )^2 \left (c^2+x^2\right )} \, dx,x,c \tan (e+f x)\right )}{f}\\ &=\frac {(2 c) \text {Subst}\left (\int \frac {x \left (\frac {d x^2}{c}\right )^m}{(a+b x)^2 \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 ) \text {Subst}\left (\int \frac {x^{1+2 m}}{(a+b x)^2 \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 ) \text {Subst}\left (\int \left (\frac {b^4 x^{1+2 m}}{\left (a^4+b^4 c^2\right ) (a+b x)^2}+\frac {4 a^3 b^4 x^{1+2 m}}{\left (a^4+b^4 c^2\right )^2 (a+b x)}+\frac {x^{1+2 m} \left (a^2 \left (a^4-3 b^4 c^2\right )-2 a b \left (a^4-b^4 c^2\right ) x+b^2 \left (3 a^4-b^4 c^2\right ) x^2-4 a^3 b^3 x^3\right )}{\left (a^4+b^4 c^2\right )^2 \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 ) \text {Subst}\left (\int \frac {x^{1+2 m} \left (a^2 \left (a^4-3 b^4 c^2\right )-2 a b \left (a^4-b^4 c^2\right ) x+b^2 \left (3 a^4-b^4 c^2\right ) x^2-4 a^3 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 )^2 f}+\frac {\left (8 a^3 b^4 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \text {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 )^2 f}+\frac {\left (2 b^4 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \text {Subst}\left (\int \frac {x^{1+2 m}}{(a+b x)^2} \, dx,x,\sqrt {c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}\\ &=\frac {4 a^2 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}{\left (a^4+b^4 c^2\right )^2 f (1+m)}+\frac {b^4 c^2 \, _2F_1\left (2,2 (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^2 \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 ) \text {Subst}\left (\int \left (\frac {x^{2+2 m} \left (-2 a b \left (a^4-b^4 c^2\right )-4 a^3 b^3 x^2\right )}{c^2+x^4}+\frac {x^{1+2 m} \left (a^2 \left (a^4-3 b^4 c^2\right )+b^2 \left (3 a^4-b^4 c^2\right ) 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 )^2 f}\\ &=\frac {4 a^2 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}{\left (a^4+b^4 c^2\right )^2 f (1+m)}+\frac {b^4 c^2 \, _2F_1\left (2,2 (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^2 \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 ) \text {Subst}\left (\int \frac {x^{2+2 m} \left (-2 a b \left (a^4-b^4 c^2\right )-4 a^3 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 )^2 f}+\frac {\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \text {Subst}\left (\int \frac {x^{1+2 m} \left (a^2 \left (a^4-3 b^4 c^2\right )+b^2 \left (3 a^4-b^4 c^2\right ) x^2\right )}{c^2+x^4} \, dx,x,\sqrt {c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right )^2 f}\\ &=\frac {4 a^2 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}{\left (a^4+b^4 c^2\right )^2 f (1+m)}+\frac {b^4 c^2 \, _2F_1\left (2,2 (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^2 \left (a^4+b^4 c^2\right ) f (1+m)}+\frac {\left (c \left (3 a^4 b^2-b^6 c^2-\frac {a^2 \left (a^4-3 b^4 c^2\right )}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \text {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 )^2 f}-\frac {\left (c \left (3 a^4 b^2-b^6 c^2+\frac {a^2 \left (a^4-3 b^4 c^2\right )}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \text {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 )^2 f}-\frac {\left (2 a b c \left (2 a^2 b^2-\frac {a^4-b^4 c^2}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \text {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 )^2 f}+\frac {\left (2 a b c \left (2 a^2 b^2+\frac {a^4-b^4 c^2}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \text {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 )^2 f}\\ &=\frac {\left (a^6-3 a^2 b^4 c^2-3 a^4 b^2 \sqrt {-c^2}-b^6 \left (-c^2\right )^{3/2}\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 )^2 f (1+m)}+\frac {\left (a^6-3 a^2 b^4 c^2+3 a^4 b^2 \sqrt {-c^2}+b^6 \left (-c^2\right )^{3/2}\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 )^2 f (1+m)}+\frac {4 a^2 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}{\left (a^4+b^4 c^2\right )^2 f (1+m)}+\frac {b^4 c^2 \, _2F_1\left (2,2 (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^2 \left (a^4+b^4 c^2\right ) f (1+m)}-\frac {2 a b \left (a^4-b^4 c^2-2 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 )^2 f (3+2 m)}-\frac {2 a b \left (a^4-b^4 c^2+2 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 )^2 f (3+2 m)}\\ \end {align*}

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Mathematica [A]
time = 4.05, size = 381, normalized size = 0.62 \begin {gather*} \frac {c (d \tan (e+f x))^m \left (\frac {a^2 \left (a^4-3 b^4 c^2\right ) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)}{c (1+m)}+\frac {4 a^2 b^4 c \, _2F_1\left (1,2 (1+m);3+2 m;-\frac {b \sqrt {c \tan (e+f x)}}{a}\right ) \tan (e+f x)}{1+m}+\frac {b^4 c \left (a^4+b^4 c^2\right ) \, _2F_1\left (2,2 (1+m);3+2 m;-\frac {b \sqrt {c \tan (e+f x)}}{a}\right ) \tan (e+f x)}{a^2 (1+m)}+\frac {b^2 \left (3 a^4-b^4 c^2\right ) \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-\tan ^2(e+f x)\right ) \tan ^2(e+f x)}{2+m}+\frac {4 a b \left (-a^4+b^4 c^2\right ) \, _2F_1\left (1,\frac {1}{4} (3+2 m);\frac {1}{4} (7+2 m);-\tan ^2(e+f x)\right ) (c \tan (e+f x))^{3/2}}{c^2 (3+2 m)}-\frac {8 a^3 b^3 \, _2F_1\left (1,\frac {1}{4} (5+2 m);\frac {1}{4} (9+2 m);-\tan ^2(e+f x)\right ) (c \tan (e+f x))^{5/2}}{c^2 (5+2 m)}\right )}{\left (a^4+b^4 c^2\right )^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^m/(a + b*Sqrt[c*Tan[e + f*x]])^2,x]

[Out]

(c*(d*Tan[e + f*x])^m*((a^2*(a^4 - 3*b^4*c^2)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[e + f*x]^2]*Tan[
e + f*x])/(c*(1 + m)) + (4*a^2*b^4*c*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x]])/a)]*T
an[e + f*x])/(1 + m) + (b^4*c*(a^4 + b^4*c^2)*Hypergeometric2F1[2, 2*(1 + m), 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x
]])/a)]*Tan[e + f*x])/(a^2*(1 + m)) + (b^2*(3*a^4 - b^4*c^2)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[e
+ f*x]^2]*Tan[e + f*x]^2)/(2 + m) + (4*a*b*(-a^4 + b^4*c^2)*Hypergeometric2F1[1, (3 + 2*m)/4, (7 + 2*m)/4, -T
an[e + f*x]^2]*(c*Tan[e + f*x])^(3/2))/(c^2*(3 + 2*m)) - (8*a^3*b^3*Hypergeometric2F1[1, (5 + 2*m)/4, (9 + 2*m
)/4, -Tan[e + f*x]^2]*(c*Tan[e + f*x])^(5/2))/(c^2*(5 + 2*m))))/((a^4 + b^4*c^2)^2*f)

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Maple [F]
time = 0.37, size = 0, normalized size = 0.00 \[\int \frac {\left (d \tan \left (f x +e \right )\right )^{m}}{\left (a +b \sqrt {c \tan \left (f x +e \right )}\right )^{2}}\, dx\]

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))^2,x)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^2,x, algorithm="fricas")

[Out]

integral(-(2*sqrt(c*tan(f*x + e))*(d*tan(f*x + e))^m*a*b - (b^2*c*tan(f*x + e) + a^2)*(d*tan(f*x + e))^m)/(b^4
*c^2*tan(f*x + e)^2 - 2*a^2*b^2*c*tan(f*x + e) + a^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{m}}{\left (a + b \sqrt {c \tan {\left (e + f x \right )}}\right )^{2}}\, dx \end {gather*}

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))**2,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^m}{{\left (a+b\,\sqrt {c\,\mathrm {tan}\left (e+f\,x\right )}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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