3.1312 \(\int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.36, 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} \[ \frac {(b c-a d) \sqrt {c+d \tan (e+f x)} (a+b \tan (e+f x))^{m+1} F_1\left (m+1;-\frac {3}{2},1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 b f (m+1) (b+i a) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}-\frac {(b c-a d) \sqrt {c+d \tan (e+f x)} (a+b \tan (e+f x))^{m+1} F_1\left (m+1;-\frac {3}{2},1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 b f (m+1) (-b+i a) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [F]  time = 7.59, size = 0, normalized size = 0.00 \[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx \]

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 = 1.08, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

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((d*tan(f*x + e) + c)^(3/2)*(b*tan(f*x + e) + a)^m, 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 = 1.33, size = 0, normalized size = 0.00 \[ \int \left (a +b \tan \left (f x +e \right )\right )^{m} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

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 \[ \int {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

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

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

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 \[ \int \left (a + b \tan {\left (e + f x \right )}\right )^{m} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \]

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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