∫ sin3(a + bx)(d tan(a + bx))3∕2 dx

3.69 \(\int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=110 \[ \frac {7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}-\frac {7 d^2 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b} \]

[Out]

7/2*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))*sin(b*x+a)/b/sin(2*

b*x+2*a)^(1/2)/(d*tan(b*x+a))^(1/2)+2*d*sin(b*x+a)^3*(d*tan(b*x+a))^(1/2)/b+7/3*d^3*sin(b*x+a)^3/b/(d*tan(b*x+

a))^(3/2)

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Rubi [A] time = 0.14, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2594, 2598, 2601, 2572, 2639} \[ \frac {7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}-\frac {7 d^2 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Sin[a + b*x]^3*(d*Tan[a + b*x])^(3/2),x]

[Out]

(7*d^3*Sin[a + b*x]^3)/(3*b*(d*Tan[a + b*x])^(3/2)) - (7*d^2*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(2*b*S

qrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]]) + (2*d*Sin[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/b

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +

f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},

Rule 2594

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sin[e

+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(n - 1)), x] - Dist[(b^2*(m + n - 1))/(n - 1), Int[(a*Sin[e + f*x])^m*(

b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n] && !(GtQ[m,

1] && !IntegerQ[(m - 1)/2])

Rule 2598

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[(b*(a*Sin[

e + f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] + Dist[(a^2*(m + n - 1))/m, Int[(a*Sin[e + f*x])^(m - 2)*(b*Ta

n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && IntegersQ[2

*m, 2*n]

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x

]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b

, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -

1/2])

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{

c, d}, x]

Rubi steps

\begin {align*} \int \sin ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-\left (7 d^2\right ) \int \frac {\sin ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=\frac {7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}+\frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-\frac {1}{2} \left (7 d^2\right ) \int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=\frac {7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}+\frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-\frac {\left (7 d^2 \sqrt {\sin (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)} \, dx}{2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=\frac {7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}+\frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}-\frac {\left (7 d^2 \sin (a+b x)\right ) \int \sqrt {\sin (2 a+2 b x)} \, dx}{2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\\ &=\frac {7 d^3 \sin ^3(a+b x)}{3 b (d \tan (a+b x))^{3/2}}-\frac {7 d^2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{2 b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}+\frac {2 d \sin ^3(a+b x) \sqrt {d \tan (a+b x)}}{b}\\ \end {align*}

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Mathematica [C] time = 0.60, size = 90, normalized size = 0.82 \[ \frac {(d \tan (a+b x))^{3/2} \left (2 \cos (a+b x) (\cos (2 (a+b x))+13) \sqrt {\sec ^2(a+b x)}-28 \sec (a+b x) \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(a+b x)\right )\right )}{12 b \sqrt {\sec ^2(a+b x)}} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sin[a + b*x]^3*(d*Tan[a + b*x])^(3/2),x]

[Out]

((-28*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b*x] + 2*Cos[a + b*x]*(13 + Cos[2*(a + b*x)])*

Sqrt[Sec[a + b*x]^2])*(d*Tan[a + b*x])^(3/2))/(12*b*Sqrt[Sec[a + b*x]^2])

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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (d \cos \left (b x + a\right )^{2} - d\right )} \sqrt {d \tan \left (b x + a\right )} \sin \left (b x + a\right ) \tan \left (b x + a\right ), x\right ) \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

integral(-(d*cos(b*x + a)^2 - d)*sqrt(d*tan(b*x + a))*sin(b*x + a)*tan(b*x + a), x)

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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(sin(b*x+a)^3*(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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