3.1187 \(\int (a+i a \tan (e+f x))^m \sqrt {c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=116 \[ -\frac {i (a+i a \tan (e+f x))^m \sqrt {c+d \tan (e+f x)} F_1\left (m;-\frac {1}{2},1;m+1;-\frac {d (i \tan (e+f x)+1)}{i c-d},\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3564, 137, 136} \[ -\frac {i (a+i a \tan (e+f x))^m \sqrt {c+d \tan (e+f x)} F_1\left (m;-\frac {1}{2},1;m+1;-\frac {d (i \tan (e+f x)+1)}{i c-d},\frac {1}{2} (i \tan (e+f x)+1)\right )}{2 f m \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*AppellF1[m, -1/2, 1, 1 + m, -((d*(1 + I*Tan[e + f*x]))/(I*c - d)), (1 + I*Tan[e + f*x])/2]*(a + I*a*Ta
n[e + f*x])^m*Sqrt[c + d*Tan[e + f*x]])/(f*m*Sqrt[(c + d*Tan[e + f*x])/(c + I*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 3564

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

Rubi steps

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

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Mathematica [F]  time = 2.48, size = 0, normalized size = 0.00 \[ \int (a+i a \tan (e+f x))^m \sqrt {c+d \tan (e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\frac {2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/
(e^(2*I*f*x + 2*I*e) + 1)), 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+I*a*tan(f*x+e))^m*(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 1.34, size = 0, normalized size = 0.00 \[ \int \left (a +i a \tan \left (f x +e \right )\right )^{m} \sqrt {c +d \tan \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*a*(tan(e + f*x) - I))**m*sqrt(c + d*tan(e + f*x)), x)

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