3.406 \(\int (d \tan (e+f x))^m (a+b \sqrt {c \tan (e+f x)})^2 \, dx\)

Optimal. Leaf size=212 \[ \frac {\left (a^2-b^2 \sqrt {-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,m+1;m+2;-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right )}{2 f (m+1)}+\frac {\left (a^2+b^2 \sqrt {-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,m+1;m+2;\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right )}{2 f (m+1)}+\frac {4 a b (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \, _2F_1\left (1,\frac {1}{4} (2 m+3);\frac {1}{4} (2 m+7);-\tan ^2(e+f x)\right )}{c f (2 m+3)} \]

[Out]

1/2*hypergeom([1, 1+m],[2+m],-c*tan(f*x+e)/(-c^2)^(1/2))*(a^2-b^2*(-c^2)^(1/2))*tan(f*x+e)*(d*tan(f*x+e))^m/f/
(1+m)+1/2*hypergeom([1, 1+m],[2+m],c*tan(f*x+e)/(-c^2)^(1/2))*(a^2+b^2*(-c^2)^(1/2))*tan(f*x+e)*(d*tan(f*x+e))
^m/f/(1+m)+4*a*b*hypergeom([1, 3/4+1/2*m],[7/4+1/2*m],-tan(f*x+e)^2)*(c*tan(f*x+e))^(3/2)*(d*tan(f*x+e))^m/c/f
/(3+2*m)

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Rubi [A]  time = 0.71, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3670, 15, 1831, 364, 1286} \[ \frac {\left (a^2-b^2 \sqrt {-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,m+1;m+2;-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right )}{2 f (m+1)}+\frac {\left (a^2+b^2 \sqrt {-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,m+1;m+2;\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right )}{2 f (m+1)}+\frac {4 a b (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \, _2F_1\left (1,\frac {1}{4} (2 m+3);\frac {1}{4} (2 m+7);-\tan ^2(e+f x)\right )}{c f (2 m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]))

Rubi steps

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

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Mathematica [A]  time = 1.43, size = 151, normalized size = 0.71 \[ \frac {\tan (e+f x) (d \tan (e+f x))^m \left (\frac {a^2 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(e+f x)\right )}{m+1}+b \left (\frac {4 a \sqrt {c \tan (e+f x)} \, _2F_1\left (1,\frac {1}{4} (2 m+3);\frac {1}{4} (2 m+7);-\tan ^2(e+f x)\right )}{2 m+3}+\frac {b c \tan (e+f x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(e+f x)\right )}{m+2}\right )\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tan[e + f*x]*(d*Tan[e + f*x])^m*((a^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[e + f*x]^2])/(1 + m) +
b*((b*c*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(2 + m) + (4*a*Hypergeometri
c2F1[1, (3 + 2*m)/4, (7 + 2*m)/4, -Tan[e + f*x]^2]*Sqrt[c*Tan[e + f*x]])/(3 + 2*m))))/f

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (2 \, \sqrt {c \tan \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{m} a b + {\left (b^{2} c \tan \left (f x + e\right ) + a^{2}\right )} \left (d \tan \left (f x + e\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.93, size = 0, normalized size = 0.00 \[ \int \left (a +b \sqrt {c \tan \left (f x +e \right )}\right )^{2} \left (d \tan \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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