3.1685 \(\int \frac{(c+d x)^{5/4}}{(a+b x)^{25/4}} \, dx\)

Optimal. Leaf size=136 \[ \frac{512 d^3 (c+d x)^{9/4}}{13923 (a+b x)^{9/4} (b c-a d)^4}-\frac{128 d^2 (c+d x)^{9/4}}{1547 (a+b x)^{13/4} (b c-a d)^3}+\frac{16 d (c+d x)^{9/4}}{119 (a+b x)^{17/4} (b c-a d)^2}-\frac{4 (c+d x)^{9/4}}{21 (a+b x)^{21/4} (b c-a d)} \]

[Out]

(-4*(c + d*x)^(9/4))/(21*(b*c - a*d)*(a + b*x)^(21/4)) + (16*d*(c + d*x)^(9/4))/(119*(b*c - a*d)^2*(a + b*x)^(
17/4)) - (128*d^2*(c + d*x)^(9/4))/(1547*(b*c - a*d)^3*(a + b*x)^(13/4)) + (512*d^3*(c + d*x)^(9/4))/(13923*(b
*c - a*d)^4*(a + b*x)^(9/4))

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Rubi [A]  time = 0.0289592, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {45, 37} \[ \frac{512 d^3 (c+d x)^{9/4}}{13923 (a+b x)^{9/4} (b c-a d)^4}-\frac{128 d^2 (c+d x)^{9/4}}{1547 (a+b x)^{13/4} (b c-a d)^3}+\frac{16 d (c+d x)^{9/4}}{119 (a+b x)^{17/4} (b c-a d)^2}-\frac{4 (c+d x)^{9/4}}{21 (a+b x)^{21/4} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/4)/(a + b*x)^(25/4),x]

[Out]

(-4*(c + d*x)^(9/4))/(21*(b*c - a*d)*(a + b*x)^(21/4)) + (16*d*(c + d*x)^(9/4))/(119*(b*c - a*d)^2*(a + b*x)^(
17/4)) - (128*d^2*(c + d*x)^(9/4))/(1547*(b*c - a*d)^3*(a + b*x)^(13/4)) + (512*d^3*(c + d*x)^(9/4))/(13923*(b
*c - a*d)^4*(a + b*x)^(9/4))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

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

Rubi steps

\begin{align*} \int \frac{(c+d x)^{5/4}}{(a+b x)^{25/4}} \, dx &=-\frac{4 (c+d x)^{9/4}}{21 (b c-a d) (a+b x)^{21/4}}-\frac{(4 d) \int \frac{(c+d x)^{5/4}}{(a+b x)^{21/4}} \, dx}{7 (b c-a d)}\\ &=-\frac{4 (c+d x)^{9/4}}{21 (b c-a d) (a+b x)^{21/4}}+\frac{16 d (c+d x)^{9/4}}{119 (b c-a d)^2 (a+b x)^{17/4}}+\frac{\left (32 d^2\right ) \int \frac{(c+d x)^{5/4}}{(a+b x)^{17/4}} \, dx}{119 (b c-a d)^2}\\ &=-\frac{4 (c+d x)^{9/4}}{21 (b c-a d) (a+b x)^{21/4}}+\frac{16 d (c+d x)^{9/4}}{119 (b c-a d)^2 (a+b x)^{17/4}}-\frac{128 d^2 (c+d x)^{9/4}}{1547 (b c-a d)^3 (a+b x)^{13/4}}-\frac{\left (128 d^3\right ) \int \frac{(c+d x)^{5/4}}{(a+b x)^{13/4}} \, dx}{1547 (b c-a d)^3}\\ &=-\frac{4 (c+d x)^{9/4}}{21 (b c-a d) (a+b x)^{21/4}}+\frac{16 d (c+d x)^{9/4}}{119 (b c-a d)^2 (a+b x)^{17/4}}-\frac{128 d^2 (c+d x)^{9/4}}{1547 (b c-a d)^3 (a+b x)^{13/4}}+\frac{512 d^3 (c+d x)^{9/4}}{13923 (b c-a d)^4 (a+b x)^{9/4}}\\ \end{align*}

Mathematica [A]  time = 0.0638237, size = 118, normalized size = 0.87 \[ \frac{4 (c+d x)^{9/4} \left (357 a^2 b d^2 (4 d x-9 c)+1547 a^3 d^3+21 a b^2 d \left (117 c^2-72 c d x+32 d^2 x^2\right )+b^3 \left (468 c^2 d x-663 c^3-288 c d^2 x^2+128 d^3 x^3\right )\right )}{13923 (a+b x)^{21/4} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/4)/(a + b*x)^(25/4),x]

[Out]

(4*(c + d*x)^(9/4)*(1547*a^3*d^3 + 357*a^2*b*d^2*(-9*c + 4*d*x) + 21*a*b^2*d*(117*c^2 - 72*c*d*x + 32*d^2*x^2)
+ b^3*(-663*c^3 + 468*c^2*d*x - 288*c*d^2*x^2 + 128*d^3*x^3)))/(13923*(b*c - a*d)^4*(a + b*x)^(21/4))

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Maple [A]  time = 0.008, size = 171, normalized size = 1.3 \begin{align*}{\frac{512\,{b}^{3}{d}^{3}{x}^{3}+2688\,a{b}^{2}{d}^{3}{x}^{2}-1152\,{b}^{3}c{d}^{2}{x}^{2}+5712\,{a}^{2}b{d}^{3}x-6048\,a{b}^{2}c{d}^{2}x+1872\,{b}^{3}{c}^{2}dx+6188\,{a}^{3}{d}^{3}-12852\,{a}^{2}cb{d}^{2}+9828\,a{b}^{2}{c}^{2}d-2652\,{b}^{3}{c}^{3}}{13923\,{a}^{4}{d}^{4}-55692\,{a}^{3}bc{d}^{3}+83538\,{a}^{2}{c}^{2}{b}^{2}{d}^{2}-55692\,a{b}^{3}{c}^{3}d+13923\,{b}^{4}{c}^{4}} \left ( dx+c \right ) ^{{\frac{9}{4}}} \left ( bx+a \right ) ^{-{\frac{21}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/4)/(b*x+a)^(25/4),x)

[Out]

4/13923*(d*x+c)^(9/4)*(128*b^3*d^3*x^3+672*a*b^2*d^3*x^2-288*b^3*c*d^2*x^2+1428*a^2*b*d^3*x-1512*a*b^2*c*d^2*x
+468*b^3*c^2*d*x+1547*a^3*d^3-3213*a^2*b*c*d^2+2457*a*b^2*c^2*d-663*b^3*c^3)/(b*x+a)^(21/4)/(a^4*d^4-4*a^3*b*c
*d^3+6*a^2*b^2*c^2*d^2-4*a*b^3*c^3*d+b^4*c^4)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b x + a\right )}^{\frac{25}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(25/4),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(25/4), x)

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Fricas [B]  time = 4.62618, size = 1354, normalized size = 9.96 \begin{align*} \frac{4 \,{\left (128 \, b^{3} d^{5} x^{5} - 663 \, b^{3} c^{5} + 2457 \, a b^{2} c^{4} d - 3213 \, a^{2} b c^{3} d^{2} + 1547 \, a^{3} c^{2} d^{3} - 32 \,{\left (b^{3} c d^{4} - 21 \, a b^{2} d^{5}\right )} x^{4} + 4 \,{\left (5 \, b^{3} c^{2} d^{3} - 42 \, a b^{2} c d^{4} + 357 \, a^{2} b d^{5}\right )} x^{3} -{\left (15 \, b^{3} c^{3} d^{2} - 105 \, a b^{2} c^{2} d^{3} + 357 \, a^{2} b c d^{4} - 1547 \, a^{3} d^{5}\right )} x^{2} - 2 \,{\left (429 \, b^{3} c^{4} d - 1701 \, a b^{2} c^{3} d^{2} + 2499 \, a^{2} b c^{2} d^{3} - 1547 \, a^{3} c d^{4}\right )} x\right )}{\left (b x + a\right )}^{\frac{3}{4}}{\left (d x + c\right )}^{\frac{1}{4}}}{13923 \,{\left (a^{6} b^{4} c^{4} - 4 \, a^{7} b^{3} c^{3} d + 6 \, a^{8} b^{2} c^{2} d^{2} - 4 \, a^{9} b c d^{3} + a^{10} d^{4} +{\left (b^{10} c^{4} - 4 \, a b^{9} c^{3} d + 6 \, a^{2} b^{8} c^{2} d^{2} - 4 \, a^{3} b^{7} c d^{3} + a^{4} b^{6} d^{4}\right )} x^{6} + 6 \,{\left (a b^{9} c^{4} - 4 \, a^{2} b^{8} c^{3} d + 6 \, a^{3} b^{7} c^{2} d^{2} - 4 \, a^{4} b^{6} c d^{3} + a^{5} b^{5} d^{4}\right )} x^{5} + 15 \,{\left (a^{2} b^{8} c^{4} - 4 \, a^{3} b^{7} c^{3} d + 6 \, a^{4} b^{6} c^{2} d^{2} - 4 \, a^{5} b^{5} c d^{3} + a^{6} b^{4} d^{4}\right )} x^{4} + 20 \,{\left (a^{3} b^{7} c^{4} - 4 \, a^{4} b^{6} c^{3} d + 6 \, a^{5} b^{5} c^{2} d^{2} - 4 \, a^{6} b^{4} c d^{3} + a^{7} b^{3} d^{4}\right )} x^{3} + 15 \,{\left (a^{4} b^{6} c^{4} - 4 \, a^{5} b^{5} c^{3} d + 6 \, a^{6} b^{4} c^{2} d^{2} - 4 \, a^{7} b^{3} c d^{3} + a^{8} b^{2} d^{4}\right )} x^{2} + 6 \,{\left (a^{5} b^{5} c^{4} - 4 \, a^{6} b^{4} c^{3} d + 6 \, a^{7} b^{3} c^{2} d^{2} - 4 \, a^{8} b^{2} c d^{3} + a^{9} b d^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(25/4),x, algorithm="fricas")

[Out]

4/13923*(128*b^3*d^5*x^5 - 663*b^3*c^5 + 2457*a*b^2*c^4*d - 3213*a^2*b*c^3*d^2 + 1547*a^3*c^2*d^3 - 32*(b^3*c*
d^4 - 21*a*b^2*d^5)*x^4 + 4*(5*b^3*c^2*d^3 - 42*a*b^2*c*d^4 + 357*a^2*b*d^5)*x^3 - (15*b^3*c^3*d^2 - 105*a*b^2
*c^2*d^3 + 357*a^2*b*c*d^4 - 1547*a^3*d^5)*x^2 - 2*(429*b^3*c^4*d - 1701*a*b^2*c^3*d^2 + 2499*a^2*b*c^2*d^3 -
1547*a^3*c*d^4)*x)*(b*x + a)^(3/4)*(d*x + c)^(1/4)/(a^6*b^4*c^4 - 4*a^7*b^3*c^3*d + 6*a^8*b^2*c^2*d^2 - 4*a^9*
b*c*d^3 + a^10*d^4 + (b^10*c^4 - 4*a*b^9*c^3*d + 6*a^2*b^8*c^2*d^2 - 4*a^3*b^7*c*d^3 + a^4*b^6*d^4)*x^6 + 6*(a
*b^9*c^4 - 4*a^2*b^8*c^3*d + 6*a^3*b^7*c^2*d^2 - 4*a^4*b^6*c*d^3 + a^5*b^5*d^4)*x^5 + 15*(a^2*b^8*c^4 - 4*a^3*
b^7*c^3*d + 6*a^4*b^6*c^2*d^2 - 4*a^5*b^5*c*d^3 + a^6*b^4*d^4)*x^4 + 20*(a^3*b^7*c^4 - 4*a^4*b^6*c^3*d + 6*a^5
*b^5*c^2*d^2 - 4*a^6*b^4*c*d^3 + a^7*b^3*d^4)*x^3 + 15*(a^4*b^6*c^4 - 4*a^5*b^5*c^3*d + 6*a^6*b^4*c^2*d^2 - 4*
a^7*b^3*c*d^3 + a^8*b^2*d^4)*x^2 + 6*(a^5*b^5*c^4 - 4*a^6*b^4*c^3*d + 6*a^7*b^3*c^2*d^2 - 4*a^8*b^2*c*d^3 + a^
9*b*d^4)*x)

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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((d*x+c)**(5/4)/(b*x+a)**(25/4),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b x + a\right )}^{\frac{25}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(25/4),x, algorithm="giac")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(25/4), x)