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

Optimal. Leaf size=105 \[ -\frac {i a 2^{\frac {1}{2}-\frac {m}{2}} (1+i \tan (c+d x))^{\frac {m+1}{2}} (e \cos (c+d x))^m \, _2F_1\left (-\frac {m}{2},\frac {m+1}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-I*2^(1/2-1/2*m)*a*(e*cos(d*x+c))^m*hypergeom([-1/2*m, 1/2+1/2*m],[1-1/2*m],1/2-1/2*I*tan(d*x+c))*(1+I*tan(d*x
+c))^(1/2+1/2*m)/d/m/(a+I*a*tan(d*x+c))^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3515, 3505, 3523, 70, 69} \[ -\frac {i a 2^{\frac {1}{2}-\frac {m}{2}} (1+i \tan (c+d x))^{\frac {m+1}{2}} (e \cos (c+d x))^m \, _2F_1\left (-\frac {m}{2},\frac {m+1}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*2^(1/2 - m/2)*a*(e*Cos[c + d*x])^m*Hypergeometric2F1[-m/2, (1 + m)/2, 1 - m/2, (1 - I*Tan[c + d*x])/2]*(
1 + I*Tan[c + d*x])^((1 + m)/2))/(d*m*Sqrt[a + I*a*Tan[c + d*x]])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 3505

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*S
ec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/2)*(a - b*Tan[e + f*x])^(m/2)), Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a
- b*Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]

Rule 3515

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rule 3523

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

Rubi steps

\begin {align*} \int (e \cos (c+d x))^m \sqrt {a+i a \tan (c+d x)} \, dx &=\left ((e \cos (c+d x))^m (e \sec (c+d x))^m\right ) \int (e \sec (c+d x))^{-m} \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\left ((e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2}\right ) \int (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{\frac {1}{2}-\frac {m}{2}} \, dx\\ &=\frac {\left (a^2 (e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{m/2}\right ) \operatorname {Subst}\left (\int (a-i a x)^{-1-\frac {m}{2}} (a+i a x)^{-\frac {1}{2}-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (2^{-\frac {1}{2}-\frac {m}{2}} a^2 (e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{\frac {1}{2}+\frac {m}{2}}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-\frac {1}{2}-\frac {m}{2}} (a-i a x)^{-1-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {a+i a \tan (c+d x)}}\\ &=-\frac {i 2^{\frac {1}{2}-\frac {m}{2}} a (e \cos (c+d x))^m \, _2F_1\left (-\frac {m}{2},\frac {1+m}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1+m}{2}}}{d m \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.88, size = 106, normalized size = 1.01 \[ \frac {i 2^{-m} \left (1+e^{2 i (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)} \left (e e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \, _2F_1\left (1,\frac {m+2}{2};\frac {3-m}{2};-e^{2 i (c+d x)}\right )}{d (m-1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*(1 + E^((2*I)*(c + d*x)))*((e*(1 + E^((2*I)*(c + d*x))))/E^(I*(c + d*x)))^m*Hypergeometric2F1[1, (2 + m)/2,
(3 - m)/2, -E^((2*I)*(c + d*x))]*Sqrt[a + I*a*Tan[c + d*x]])/(2^m*d*(-1 + m))

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {2} \left (\frac {1}{2} \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} e^{\left (-i \, d x - i \, c\right )}\right )^{m} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(2)*(1/2*(e*e^(2*I*d*x + 2*I*c) + e)*e^(-I*d*x - I*c))^m*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d
*x + I*c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.47, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{m} \sqrt {a +i a \tan \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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