Optimal. Leaf size=401 \[ \frac {4 \sqrt {2} d^{15/4} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{231 b^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (b c-a d)^{3/2} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{3/4} (b c-a d)^2}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{7/4} (b c-a d)}-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}} \]
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Rubi [A] time = 0.43, antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {47, 51, 62, 623, 220} \[ \frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{3/4} (b c-a d)^2}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (a+b x)^{7/4} (b c-a d)}+\frac {4 \sqrt {2} d^{15/4} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{231 b^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (b c-a d)^{3/2} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}} \]
Antiderivative was successfully verified.
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Rule 47
Rule 51
Rule 62
Rule 220
Rule 623
Rubi steps
\begin {align*} \int \frac {(c+d x)^{5/4}}{(a+b x)^{19/4}} \, dx &=-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {d \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{15/4}} \, dx}{3 b}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {d^2 \int \frac {1}{(a+b x)^{11/4} (c+d x)^{3/4}} \, dx}{33 b^2}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}-\frac {\left (2 d^3\right ) \int \frac {1}{(a+b x)^{7/4} (c+d x)^{3/4}} \, dx}{77 b^2 (b c-a d)}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (b c-a d)^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {\left (4 d^4\right ) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx}{231 b^2 (b c-a d)^2}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (b c-a d)^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {\left (4 d^4 ((a+b x) (c+d x))^{3/4}\right ) \int \frac {1}{\left (a c+(b c+a d) x+b d x^2\right )^{3/4}} \, dx}{231 b^2 (b c-a d)^2 (a+b x)^{3/4} (c+d x)^{3/4}}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (b c-a d)^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {\left (16 d^4 ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{231 b^2 (b c-a d)^2 (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x)}\\ &=-\frac {4 d \sqrt [4]{c+d x}}{33 b^2 (a+b x)^{11/4}}-\frac {4 d^2 \sqrt [4]{c+d x}}{231 b^2 (b c-a d) (a+b x)^{7/4}}+\frac {8 d^3 \sqrt [4]{c+d x}}{231 b^2 (b c-a d)^2 (a+b x)^{3/4}}-\frac {4 (c+d x)^{5/4}}{15 b (a+b x)^{15/4}}+\frac {4 \sqrt {2} d^{15/4} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{231 b^{9/4} (b c-a d)^{3/2} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}\\ \end {align*}
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Mathematica [C] time = 0.05, size = 73, normalized size = 0.18 \[ -\frac {4 (c+d x)^{5/4} \, _2F_1\left (-\frac {15}{4},-\frac {5}{4};-\frac {11}{4};\frac {d (a+b x)}{a d-b c}\right )}{15 b (a+b x)^{15/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {5}{4}}}{b^{5} x^{5} + 5 \, a b^{4} x^{4} + 10 \, a^{2} b^{3} x^{3} + 10 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x + a^{5}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {19}{4}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {19}{4}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {19}{4}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{19/4}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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