3.457 \(\int \frac{A+B \tan (c+d x)}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx\)

Optimal. Leaf size=256 \[ \frac{2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}+\frac{2 (4 A b-5 a B) \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}}-\frac{2 A \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)} \]

[Out]

((I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) - ((I*A + B)
*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d) - (2*A*Sqrt[a + b*Tan
[c + d*x]])/(5*a*d*Tan[c + d*x]^(5/2)) + (2*(4*A*b - 5*a*B)*Sqrt[a + b*Tan[c + d*x]])/(15*a^2*d*Tan[c + d*x]^(
3/2)) + (2*(15*a^2*A - 8*A*b^2 + 10*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/(15*a^3*d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 1.06206, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3609, 3649, 3616, 3615, 93, 203, 206} \[ \frac{2 \left (15 a^2 A+10 a b B-8 A b^2\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}+\frac{2 (4 A b-5 a B) \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{(-B+i A) \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{(B+i A) \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}}-\frac{2 A \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

((I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) - ((I*A + B)
*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d) - (2*A*Sqrt[a + b*Tan
[c + d*x]])/(5*a*d*Tan[c + d*x]^(5/2)) + (2*(4*A*b - 5*a*B)*Sqrt[a + b*Tan[c + d*x]])/(15*a^2*d*Tan[c + d*x]^(
3/2)) + (2*(15*a^2*A - 8*A*b^2 + 10*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/(15*a^3*d*Sqrt[Tan[c + d*x]])

Rule 3609

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*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n
+ 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e +
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ
[a, 0])))

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3616

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 + I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 -
I*Tan[e + f*x]), x], x] + Dist[(A - I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 + I*Tan[e +
f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2
+ B^2, 0]

Rule 3615

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^2/f, Subst[Int[((a + b*x)^m*(c + d*x)^n)/(A - B*x), 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] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 0]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B \tan (c+d x)}{\tan ^{\frac{7}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 \int \frac{\frac{1}{2} (4 A b-5 a B)+\frac{5}{2} a A \tan (c+d x)+2 A b \tan ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{5 a}\\ &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A b-5 a B) \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\frac{1}{4} \left (-15 a^2 A+8 A b^2-10 a b B\right )-\frac{15}{4} a^2 B \tan (c+d x)+\frac{1}{2} b (4 A b-5 a B) \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A b-5 a B) \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}-\frac{8 \int \frac{\frac{15 a^3 B}{8}-\frac{15}{8} a^3 A \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{15 a^3}\\ &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A b-5 a B) \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}-\frac{1}{2} (-i A+B) \int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx-\frac{1}{2} (i A+B) \int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A b-5 a B) \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}+\frac{(i A-B) \operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac{2 A \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A b-5 a B) \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}+\frac{(i A-B) \operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{(i A+B) \operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}\\ &=\frac{(i A-B) \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{i a-b} d}-\frac{(i A+B) \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{i a+b} d}-\frac{2 A \sqrt{a+b \tan (c+d x)}}{5 a d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 (4 A b-5 a B) \sqrt{a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (15 a^2 A-8 A b^2+10 a b B\right ) \sqrt{a+b \tan (c+d x)}}{15 a^3 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 5.44007, size = 227, normalized size = 0.89 \[ \frac{\frac{2 \sqrt{a+b \tan (c+d x)} \left (\left (15 a^2 A+10 a b B-8 A b^2\right ) \tan ^2(c+d x)-3 a^2 A-a (5 a B-4 A b) \tan (c+d x)\right )}{a^3 \tan ^{\frac{5}{2}}(c+d x)}+\frac{15 \sqrt [4]{-1} (A+i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}-\frac{15 \sqrt [4]{-1} (A-i B) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

((15*(-1)^(1/4)*(A + I*B)*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sq
rt[-a - I*b] - (15*(-1)^(1/4)*(A - I*B)*ArcTanh[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c
+ d*x]]])/Sqrt[a - I*b] + (2*Sqrt[a + b*Tan[c + d*x]]*(-3*a^2*A - a*(-4*A*b + 5*a*B)*Tan[c + d*x] + (15*a^2*A
- 8*A*b^2 + 10*a*b*B)*Tan[c + d*x]^2))/(a^3*Tan[c + d*x]^(5/2)))/(15*d)

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Maple [B]  time = 0.853, size = 1890924, normalized size = 7386.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x)

[Out]

result too large to display

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))**(1/2)/tan(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Exception raised: TypeError