∫ (a+ia tan(c+dx))5∕2(A+B tan(c+dx))__________________________

3.172 \(\int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=196 \[ \frac {(-1)^{3/4} a^{5/2} (2 A-5 i B) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (-B+2 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}} \]

[Out]

(-1)^(3/4)*a^(5/2)*(2*A-5*I*B)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+(4+4*I)*

a^(5/2)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+a^2*(2*I*A-B)*tan(d*x+c)^(1

/2)*(a+I*a*tan(d*x+c))^(1/2)/d-2*a*A*(a+I*a*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(1/2)

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Rubi [A] time = 0.70, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {3593, 3594, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac {(-1)^{3/4} a^{5/2} (2 A-5 i B) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (-B+2 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}} \]

Antiderivative was successfully veriﬁed.

[In]

Int[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(3/2),x]

[Out]

((-1)^(3/4)*a^(5/2)*(2*A - (5*I)*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]

)/d + ((4 + 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d

+ (a^2*((2*I)*A - B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d - (2*a*A*(a + I*a*Tan[c + d*x])^(3/2))/

(d*Sqrt[Tan[c + d*x]])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub

st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &

& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,

b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;

FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]

&& PosQ[a/b]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,

b}, x] && !GtQ[a, 0]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[

(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]

/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3593

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_

.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(a^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^

(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +

d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -

1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ

[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_

.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f

*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)

+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;

FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] && !LtQ[n, -1]

Rule 3599

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_

.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x

]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,

Rule 3601

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_

.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]

, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), x], x] /; FreeQ[{a, b

, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx &=-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+2 \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {1}{2} a (4 i A+B)+\frac {1}{2} a (2 A+i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+2 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^2 (2 i A+B)-\frac {1}{4} a^2 (2 A-5 i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\left (4 a^2 (i A+B)\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx-\frac {1}{2} (a (2 i A+5 B)) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}+\frac {\left (8 a^4 (A-i B)\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {\left (a^3 (2 i A+5 B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}-\frac {\left (a^3 (2 i A+5 B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}-\frac {\left (a^3 (2 i A+5 B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {\sqrt [4]{-1} a^{5/2} (2 i A+5 B) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (2 i A-B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [B] time = 10.33, size = 493, normalized size = 2.52 \[ \frac {\cos ^3(c+d x) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) (\csc (c) (-\cos (2 c)+i \sin (2 c)) (2 A \cos (c)+B \sin (c))+A \csc (c) (2 \cos (2 c)-2 i \sin (2 c)) \sin (d x) \csc (c+d x))}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac {e^{-2 i c} \sqrt {e^{i d x}} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} (a+i a \tan (c+d x))^{5/2} \left (32 (B+i A) \log \left (\sqrt {-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )-i \sqrt {2} (2 A-5 i B) \left (\log \left (-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )\right ) (A+B \tan (c+d x))}{4 \sqrt {2} d \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac {7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(3/2),x]

[Out]

(Sqrt[E^(I*d*x)]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(32*(I*A + B)*Log[E^(I*(c +

d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] - I*Sqrt[2]*(2*A - (5*I)*B)*(Log[1 - 3*E^((2*I)*(c + d*x)) - 2*Sqrt[2

]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] - Log[1 - 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*E^(I*(c + d*x))*

Sqrt[-1 + E^((2*I)*(c + d*x))]]))*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(4*Sqrt[2]*d*E^((2*I)*c)*

Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sec[c + d*x]^(7/2)*(Cos[d*x] +

I*Sin[d*x])^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^3*(Csc[c]*(2*A*Cos[c] + B*Sin[c])*(-Cos[2

*c] + I*Sin[2*c]) + A*Csc[c]*Csc[c + d*x]*(2*Cos[2*c] - (2*I)*Sin[2*c])*Sin[d*x])*Sqrt[Tan[c + d*x]]*(a + I*a*

Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [B] time = 0.70, size = 840, normalized size = 4.29 \[ \text {result too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(2)*((-4*I*A - 2*B)*a^2*e^(3*I*d*x + 3*I*c) + (-4*I*A + 2*B)*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x

+ 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + sqrt((32*I*A^2 + 64*A*B - 32*I*B

^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a

^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + sqrt((32*

I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((4*I*A + 4*B)*a^2)) - sqrt((32*I*A^2

+ 64*A*B - 32*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c)

+ (4*I*A + 4*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c)

+ 1)) - sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((4*I*A + 4*B)*a^2))

- sqrt((4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*((2*I*A + 5*B)*a^2*e^(2

*I*d*x + 2*I*c) + (2*I*A + 5*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2

*I*d*x + 2*I*c) + 1)) + 2*sqrt((4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/((2*

I*A + 5*B)*a^2)) + sqrt((4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*((2*I*

A + 5*B)*a^2*e^(2*I*d*x + 2*I*c) + (2*I*A + 5*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x +

2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) - 2*sqrt((4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-

I*d*x - I*c)/((2*I*A + 5*B)*a^2)))/(d*e^(2*I*d*x + 2*I*c) - d)

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab