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

Optimal. Leaf size=121 \[ \frac{a \tan (e+f x) (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(e+f x)\right )}{f (m+1)}+\frac{2 b (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,\frac{1}{4} (2 m+3),\frac{1}{4} (2 m+7),-\tan ^2(e+f x)\right )}{c f (2 m+3)} \]

[Out]

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

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Rubi [A]  time = 0.330969, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3670, 15, 1831, 364} \[ \frac{a \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(e+f x)\right )}{f (m+1)}+\frac{2 b (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \, _2F_1\left (1,\frac{1}{4} (2 m+3);\frac{1}{4} (2 m+7);-\tan ^2(e+f x)\right )}{c f (2 m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[e + f*x]^2]*Tan[e + f*x]*(d*Tan[e + f*x])^m)/(f*(1 + m)) +
(2*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 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 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 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 (d \tan (e+f x))^m \left (a+b \sqrt{c \tan (e+f x)}\right ) \, dx &=\frac{c \operatorname{Subst}\left (\int \frac{\left (a+b \sqrt{x}\right ) \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)}{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)}{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{a x^{1+2 m}}{c^2+x^4}+\frac{b x^{2+2 m}}{c^2+x^4}\right ) \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}\\ &=\frac{\left (2 a c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m}}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}+\frac{\left (2 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{a \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\tan ^2(e+f x)\right ) \tan (e+f x) (d \tan (e+f x))^m}{f (1+m)}+\frac{2 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*}

Mathematica [C]  time = 0.603302, size = 304, normalized size = 2.51 \[ \frac{\tan (e+f x) (d \tan (e+f x))^m \left (\left (a-b \sqrt [4]{-c^2}\right ) \text{Hypergeometric2F1}\left (1,2 (m+1),2 m+3,-\frac{\sqrt{c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )+\left (a+i b \sqrt [4]{-c^2}\right ) \text{Hypergeometric2F1}\left (1,2 (m+1),2 m+3,-\frac{i \sqrt{c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )+a \text{Hypergeometric2F1}\left (1,2 (m+1),2 m+3,\frac{i \sqrt{c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )+a \text{Hypergeometric2F1}\left (1,2 (m+1),2 m+3,\frac{\sqrt{c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )-i b \sqrt [4]{-c^2} \text{Hypergeometric2F1}\left (1,2 (m+1),2 m+3,\frac{i \sqrt{c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )+b \sqrt [4]{-c^2} \text{Hypergeometric2F1}\left (1,2 (m+1),2 m+3,\frac{\sqrt{c \tan (e+f x)}}{\sqrt [4]{-c^2}}\right )\right )}{4 f (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((a - b*(-c^2)^(1/4))*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, -(Sqrt[c*Tan[e + f*x]]/(-c^2)^(1/4))] + (a + I
*b*(-c^2)^(1/4))*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, ((-I)*Sqrt[c*Tan[e + f*x]])/(-c^2)^(1/4)] + a*Hyperg
eometric2F1[1, 2*(1 + m), 3 + 2*m, (I*Sqrt[c*Tan[e + f*x]])/(-c^2)^(1/4)] - I*b*(-c^2)^(1/4)*Hypergeometric2F1
[1, 2*(1 + m), 3 + 2*m, (I*Sqrt[c*Tan[e + f*x]])/(-c^2)^(1/4)] + a*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, Sq
rt[c*Tan[e + f*x]]/(-c^2)^(1/4)] + b*(-c^2)^(1/4)*Hypergeometric2F1[1, 2*(1 + m), 3 + 2*m, Sqrt[c*Tan[e + f*x]
]/(-c^2)^(1/4)])*Tan[e + f*x]*(d*Tan[e + f*x])^m)/(4*f*(1 + m))

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Maple [F]  time = 0.198, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sqrt{c\tan \left ( fx+e \right ) } \right ) \left ( d\tan \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\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, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*tan(f*x+e))**(1/2))*(d*tan(f*x+e))**m,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{\left (\sqrt{c \tan \left (f x + e\right )} b + a\right )} \left (d \tan \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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