3.1314 \(\int \frac{(a+b \tan (e+f x))^m}{\sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=261 \[ \frac{(a+b \tan (e+f x))^{m+1} \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}} F_1\left (m+1;\frac{1}{2},1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d},\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) \sqrt{c+d \tan (e+f x)}}-\frac{(a+b \tan (e+f x))^{m+1} \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}} F_1\left (m+1;\frac{1}{2},1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d},\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a) \sqrt{c+d \tan (e+f x)}} \]

[Out]

(AppellF1[1 + m, 1/2, 1, 2 + m, -((d*(a + b*Tan[e + f*x]))/(b*c - a*d)), (a + b*Tan[e + f*x])/(a - I*b)]*(a +
b*Tan[e + f*x])^(1 + m)*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(2*(I*a + b)*f*(1 + m)*Sqrt[c + d*Tan[e +
f*x]]) - (AppellF1[1 + m, 1/2, 1, 2 + m, -((d*(a + b*Tan[e + f*x]))/(b*c - a*d)), (a + b*Tan[e + f*x])/(a + I*
b)]*(a + b*Tan[e + f*x])^(1 + m)*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(2*(I*a - b)*f*(1 + m)*Sqrt[c + d
*Tan[e + f*x]])

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Rubi [A]  time = 0.277998, antiderivative size = 261, 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 = {3575, 912, 137, 136} \[ \frac{(a+b \tan (e+f x))^{m+1} \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}} F_1\left (m+1;\frac{1}{2},1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d},\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) \sqrt{c+d \tan (e+f x)}}-\frac{(a+b \tan (e+f x))^{m+1} \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}} F_1\left (m+1;\frac{1}{2},1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d},\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a) \sqrt{c+d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(AppellF1[1 + m, 1/2, 1, 2 + m, -((d*(a + b*Tan[e + f*x]))/(b*c - a*d)), (a + b*Tan[e + f*x])/(a - I*b)]*(a +
b*Tan[e + f*x])^(1 + m)*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(2*(I*a + b)*f*(1 + m)*Sqrt[c + d*Tan[e +
f*x]]) - (AppellF1[1 + m, 1/2, 1, 2 + m, -((d*(a + b*Tan[e + f*x]))/(b*c - a*d)), (a + b*Tan[e + f*x])/(a + I*
b)]*(a + b*Tan[e + f*x])^(1 + m)*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(2*(I*a - b)*f*(1 + m)*Sqrt[c + d
*Tan[e + f*x]])

Rule 3575

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

Rule 912

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[c*d^2 + a*e^2,
0] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin{align*} \int \frac{(a+b \tan (e+f x))^m}{\sqrt{c+d \tan (e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^m}{\sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{i (a+b x)^m}{2 (i-x) \sqrt{c+d x}}+\frac{i (a+b x)^m}{2 (i+x) \sqrt{c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{(a+b x)^m}{(i-x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{i \operatorname{Subst}\left (\int \frac{(a+b x)^m}{(i+x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{\left (i \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^m}{(i-x) \sqrt{\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}}} \, dx,x,\tan (e+f x)\right )}{2 f \sqrt{c+d \tan (e+f x)}}+\frac{\left (i \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^m}{(i+x) \sqrt{\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}}} \, dx,x,\tan (e+f x)\right )}{2 f \sqrt{c+d \tan (e+f x)}}\\ &=\frac{F_1\left (1+m;\frac{1}{2},1;2+m;-\frac{d (a+b \tan (e+f x))}{b c-a d},\frac{a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m} \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}}{2 (i a+b) f (1+m) \sqrt{c+d \tan (e+f x)}}-\frac{F_1\left (1+m;\frac{1}{2},1;2+m;-\frac{d (a+b \tan (e+f x))}{b c-a d},\frac{a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m} \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}}}{2 (i a-b) f (1+m) \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [F]  time = 4.3566, size = 0, normalized size = 0. \[ \int \frac{(a+b \tan (e+f x))^m}{\sqrt{c+d \tan (e+f x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.37, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m}{\frac{1}{\sqrt{c+d\tan \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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