3.1315 \(\int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=283 \[ \frac {b (a+b \tan (e+f x))^{m+1} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} F_1\left (m+1;\frac {3}{2},1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {b (a+b \tan (e+f x))^{m+1} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} F_1\left (m+1;\frac {3}{2},1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a) (b c-a d) \sqrt {c+d \tan (e+f x)}} \]

[Out]

1/2*b*AppellF1(1+m,1,3/2,2+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)/(-a*d+b*c)/f/(1+m)/(c+d*tan(f*x+e))^(1/2)-1/2*b*AppellF1(1+m,1,3/
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)/(-a*d+b*c)/f/(1+m)/(c+d*tan(f*x+e))^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 283, 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 = {3575, 912, 137, 136} \[ \frac {b (a+b \tan (e+f x))^{m+1} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} F_1\left (m+1;\frac {3}{2},1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (b c-a d) \sqrt {c+d \tan (e+f x)}}-\frac {b (a+b \tan (e+f x))^{m+1} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} F_1\left (m+1;\frac {3}{2},1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a) (b c-a d) \sqrt {c+d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

(b*AppellF1[1 + m, 3/2, 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)*(b*c - a*d)*f*(1 + m)*Sqrt[
c + d*Tan[e + f*x]]) - (b*AppellF1[1 + m, 3/2, 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)*(b*c
- a*d)*f*(1 + m)*Sqrt[c + d*Tan[e + f*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])

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 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 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]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^m}{(c+d x)^{3/2} \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) (c+d x)^{3/2}}+\frac {i (a+b x)^m}{2 (i+x) (c+d x)^{3/2}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i \operatorname {Subst}\left (\int \frac {(a+b x)^m}{(i-x) (c+d x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {i \operatorname {Subst}\left (\int \frac {(a+b x)^m}{(i+x) (c+d x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac {\left (i b \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) \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{2 (b c-a d) f \sqrt {c+d \tan (e+f x)}}+\frac {\left (i b \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) \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{2 (b c-a d) f \sqrt {c+d \tan (e+f x)}}\\ &=\frac {b F_1\left (1+m;\frac {3}{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) (b c-a d) f (1+m) \sqrt {c+d \tan (e+f x)}}-\frac {b F_1\left (1+m;\frac {3}{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) (b c-a d) f (1+m) \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

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Mathematica [F]  time = 8.84, size = 0, normalized size = 0.00 \[ \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

Integrate[(a + b*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^(3/2), x]

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fricas [F]  time = 1.30, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \tan \left (f x + e\right ) + c} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}}, x\right ) \]

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))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 1.15, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \tan \left (f x +e \right )\right )^{m}}{\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]

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))^(3/2),x)

[Out]

int((a+b*tan(f*x+e))^m/(c+d*tan(f*x+e))^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^m/(d*tan(f*x + e) + c)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]

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))^(3/2),x)

[Out]

int((a + b*tan(e + f*x))^m/(c + d*tan(e + f*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{m}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

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))**(3/2),x)

[Out]

Integral((a + b*tan(e + f*x))**m/(c + d*tan(e + f*x))**(3/2), x)

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