3.14.14 \(\int \frac {(a+b \tan (e+f x))^m}{\sqrt {c+d \tan (e+f x)}} \, dx\) [1314]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.19, antiderivative size = 261, normalized size of antiderivative = 1.00, 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 = {3656, 926, 142, 141} \begin {gather*} \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)}} \end {gather*}

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 141

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

Rule 142

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 926

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 3656

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]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (e+f x))^m}{\sqrt {c+d \tan (e+f x)}} \, dx &=\frac {\text {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 {\text {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 \text {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 \text {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 ) \text {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 ) \text {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 = 5.45, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b \tan (e+f x))^m}{\sqrt {c+d \tan (e+f x)}} \, dx \end {gather*}

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]

________________________________________________________________________________________

Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{m}}{\sqrt {c +d \tan \left (f x +e \right )}}\, dx\]

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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

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((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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________